Answer:
The actual height is [tex]A =3.158 \ m[/tex]
Explanation:
From the question we are told that
The depth of the person is [tex]d = 1.1 \ m[/tex]
The apparent height is [tex]D = 4.2 \ m[/tex]
Generally
The refractive index of water is [tex]n_w = 1.33[/tex]
The refractive index of the air is [tex]n_a = 1[/tex]
The apparent depth is mathematically represented as
[tex]D = A [\frac{n_w}{n_a} ][/tex]
substituting values
[tex]4.2 = A [\frac{1.33}{1} ][/tex]
=> [tex]A = \frac{4.2 }{1.33}[/tex]
[tex]A =3.158 \ m[/tex]
The ball was dropped at the height of "3.158 m". To understand the calculation, check below.
Refractive IndexAccording to the question,
Water's refractive index, [tex]n_w[/tex] = 1.33
Air's refractive index, [tex]n_a[/tex] = 1
Apparent height, D = 4.2 m
Person's depth, d = 1.1 m
We know the relation,
→ D = A[[tex]\frac{n_w}{n_a}[/tex]]
By substituting the values, we get
4.2 = A[[tex]\frac{1.33}{1}[/tex]]
By applying cross-multiplication,
A = [tex]\frac{4.2}{1.33}[/tex]
= 3.158 m
Thus the approach above is correct.
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A roller coaster car is going over the top of a 15-m-radius circular rise. The passenger in the roller coaster has a true weight of 600 N (therefore a mass of 61.2 kg). At the top of the hill, the passengers "feel light," with an apparent weight of only 360 N. How fast is the coaster moving
Answer:
v = 7.67 m/s
Explanation:
The equation for apparent weight in the situation of weightlessness is given as:
Apparent Weight = m(g - a)
where,
Apparent Weight = 360 N
m = mass passenger = 61.2 kg
a = acceleration of roller coaster
g = acceleration due to gravity = 9.8 m/s²
Therefore,
360 N = (61.2 kg)(9.8 m/s² - a)
9.8 m/s² - a = 360 N/61.2 kg
a = 9.8 m/s² - 5.88 m/s²
a = 3.92 m/s²
Since, this acceleration is due to the change in direction of velocity on a circular path. Therefore, it can b represented by centripetal acceleration and its formula is given as:
a = v²/r
where,
a = centripetal acceleration = 3.92 m/s²
v = speed of roller coaster = ?
r = radius of circular rise = 15 m
Therefore,
3.92 m/s² = v²/15 m
v² = (3.92 m.s²)(15 m)
v = √(58.8 m²/s²)
v = 7.67 m/s
A woman with mass 50 kg is standing on the rim of a large disk that is rotating at 0.80 rev/s about an axis through its center. The disk has mass 110 kg and radius 4.0 m. Calculate the magnitude of the total angular momentum of the woman–disk system. (Assume that you can treat the woman as a point.)
Answer:
The angular momentum is [tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]
Explanation:
From the question we are told that
The mass of the woman is [tex]m = 50 \ kg[/tex]
The angular speed of the rim is [tex]w = 0.80 \ rev/s = 0.8 * [\frac{2 \pi}{1} ] = 5.024 \ rad \cdot s^{-1}[/tex]
The mass of the disk is [tex]m_d = 110 \ kg[/tex]
The radius of the disk is [tex]r_d = 4.0 \ m[/tex]
The moment of inertia of the disk is mathematically represented as
[tex]I_D = \frac{1}{2} m_d r^2_d[/tex]
substituting values
[tex]I_D = \frac{1}{2} * 110 * 4^2[/tex]
[tex]I_D = 880 \ kg \cdot m^2[/tex]
The moment of inertia of the woman is
[tex]I_w = m * r_d^2[/tex]
substituting values
[tex]I_w = 50 * 4^2[/tex]
[tex]I_w =800\ kg[/tex]
The moment of inertia of the system (the woman + the large disk ) is
[tex]I_t = I_w + I_D[/tex]
substituting values
[tex]I_t = 880 +800[/tex]
[tex]I_t =1680 \ kg \cdot m^2[/tex]
The angular momentum of the system is
[tex]L = I_t w[/tex]
substituting values
[tex]L = 1680 * 5.024[/tex]
[tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]
A type of friction that occurs when air pushes against a moving object causing it to negatively accelerate
a) surface area
b) air resistance
c) descent velocity
d) gravity
Answer:
Air resistance
Answer B is correct
Explanation:
The friction that occurs when air pushes against a moving object causing it to negatively accelerate is called as air resistance.
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The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic field used to align the spins, with proportionality constant 5 hz/T. If the strength of the applied field is known to be 20 T plus or minus 3 T, which of the following correctly describes the uncertainty in the INVERSE frequency (1/frequency)?
A. 3/2000s
B. 3/5s
C. 1/15s
D. 1/4
Complete Question
The complete question is shown on the first uploaded image
Answer:
The uncertainty in inverse frequency is [tex]\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s[/tex]
Explanation:
From the question we are told that
The value of the proportionality constant is [tex]k = 5 \frac{Hz }{T}[/tex]
The strength of the magnetic field is [tex]B = 20 \ T[/tex]
The change in this strength of magnetic field is [tex]\Delta B = 3 \ T[/tex]
The magnetic field is given as
[tex]B = \frac{k}{\frac{1}{w} }[/tex]
Where [tex]w[/tex] is frequency
The uncertainty or error of the field is given as
[tex]\Delta B = \frac{k }{[\frac{1}{w}^]^2 } \Delta [\frac{1}{w} ][/tex]
The uncertainty in inverse frequency is given as
[tex]\Delta [\frac{1}{w} ] = \frac{\Delta B}{k [\frac{1}{w^2} ]}[/tex]
[tex]\Delta [\frac{1}{w} ]= \frac{\Delta B}{k (B)^2 }[/tex]
substituting values
[tex]\Delta [\frac{1}{w} ]= \frac{3}{5 (20)^2 }[/tex]
[tex]\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s[/tex]
Richard is driving home to visit his parents. 150 mi of the trip are on the interstate highway where the speed limit is 65 mph . Normally Richard drives at the speed limit, but today he is running late and decides to take his chances by driving at 80 mph. How many minutes does he save?
Answer:
t = 25.5 min
Explanation:
To know how many minutes does Richard save, you first calculate the time that Richard takes with both velocities v1 = 65mph and v2 = 80mph.
[tex]t_1=\frac{x}{v_1}=\frac{150mi}{65mph}=2.30h\\\\t_2=\frac{x}{v_2}=\frac{150mi}{80mph}=1.875h[/tex]
Next, you calculate the difference between both times t1 and t2:
[tex]\Delta t=t_1-t_2=2.30h-1.875h=0.425h[/tex]
This is the time that Richard saves when he drives with a speed of 80mph. Finally, you convert the result to minutes:
[tex]0.425h*\frac{60min}{1h}=25.5min=25\ min\ \ 30 s[/tex]
hence, Richard saves 25.5 min (25 min and 30 s) when he drives with a speed of 80mph
An unstrained horizontal spring has a length of 0.36 m and a spring constant of 320 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.033 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.
Answer:
1.been both -ve charged or both +be charged particles
2. 3.52mC
Explanation:
For the charge particle to cause an extension or movement of the string from its unrestrained position they would have been both -ve charged or both +be charged particles that's because like charges repel.
Now the Force sustain by the extended string is
F = Ke;
Where K is the force constant of the string, 320 N/m
e is the extension,0.033 m
F = 320 × 0.033 =10.56N
2.But according to columns law of charge;
F = kQ1 Q2
But Q1=Q2{ since the charge are of the same magnitude}.
Hence F = KQ^2
Where K is columns constant =9×10^9F/m
Hence Q=√F/K
Q= √10.56/9×10^9
=3.52×10^-3C
= 3.52mC
An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?
Sort the following quantities as known or unknown. Take the horizontal direction to be along the x axis.
mQ: the mass of the quarterback
mB: the mass of the football
(vQx)i: the horizontal velocity of quarterback before throwing the ball
(vBx)i: the horizontal velocity of football before being thrown
(vQx)f: the horizontal velocity of quarterback after throwing the ball
(vBx)f: the horizontal velocity of football after being thrown
Answer:
vBxf = 0.08625m/s
Explanation:
This is a problem about the momentum conservation law. The total momentum before equals the total momentum after.
[tex]p_f=p_i[/tex]
pf: final momentum
pi: initial momentum
The analysis of the momentum conservation is about a horizontal momentum (x axis). When the quarterback jumps straight up, his horizontal momentum is zero. Then, after the quarterback throw the ball the sum of the momentum of both quarterback and ball must be zero.
Then, you have:
[tex]m_Qv_{Qxi}+m_{Bxi}v_{Bxi}=m_Qv_{Qxf}+m_{Bxf}v_{Bxf}[/tex] (1)
mQ: the mass of the quarterback = 80kg
mB: the mass of the football = 0.43kg
(vQx)i: the horizontal velocity of quarterback before throwing the ball = 0m/s
(vBx)i: the horizontal velocity of football before being thrown = 0m/s
(vQx)f: the horizontal velocity of quarterback after throwing the ball = ?
(vBx)f: the horizontal velocity of football after being thrown = 15 m/s
You replace the values of the variables in the equation (1), and you solve for (vBx)f:
[tex]0\ kgm/s=-(80kg)(v_{Bxf})+(0.46kg)(15m/s)\\\\v_{Bxf}=\frac{(0.46kg)(15m/s)}{80kg}=0.08625\frac{m}{s}[/tex]
Where you have taken the speed of the quarterback as negative because is in the negative direction of the x axis.
Hence, the speed of the quarterback after he throws the ball is 0.08625m/s
A CD is spinning on a CD player. In 220 radians, the cd has reached an angular speed of 92 r a d s by accelerating with a constant acceleration of 14 r a d s 2 . What was the initial angular speed of the CD
Answer:
The initial angular speed is [tex]w_i = 48 \ rad/s[/tex]
Explanation:
From the question we are told that
The angular displacement is [tex]\theta = 220 \ rad[/tex]
The angular speed is [tex]w_f = 92 \ rad/s[/tex]
The acceleration is [tex]\alpha = 14 \ rad/s^2[/tex]
Generally the initial angular speed can be evaluated as
[tex]w_f ^2 = w_i ^2 + 2 * \alpha * \theta[/tex]
=> [tex]w_i ^2 = w_f ^2 - 2 * \alpha * \theta[/tex]
substituting values
=> [tex]w_i ^2 = 92 ^2 - 2 * 14 * 220[/tex]
=> [tex]w_i ^2 = 2304[/tex]
=> [tex]w_i = 48 \ rad/s[/tex]
You are watching an object that is moving in SHM. When the object is displaced 0.560 m to the right of its equilibrium position, it has a velocity of 2.45 m/s to the right and an acceleration of 8.60 m/s2 to the left. Part A How much farther from this point will the object move before it stops momentarily and then starts to move back to the left
Answer:
2.95m
Explanation:
The farthest distance the object can move is the radius of the circle of which the Simple harmonic motion is assumed to be a part
But V = w× r; where V is velocity,
w is angular velocity and r is radius.
Also,
a= w2r; where a is linear acceleration
but a = v× r ; by comparing both equations
Hence r = a/v =8.6/2.45 =3.51m
But the horizontal distance of the motion is given by:
X = rcosx ; where x is the angle
X is the distance covered.
We know that the maximum value of cos x is 1 which is 0°
When the object moves in a fashion directly parallel to an horizontal distance, maximum distance would be reached and hence:
X = r=3.51m
Meaning the object needs to travel 3.51-0.56=2.95m further.
Note: the acceleration of the motion is constant whether it is swinging towards the left or right.
When the object is displaced 0.560 m to the right of its equilibrium position, it has a velocity of 2.45 m/s to the right and an acceleration of 8.60 m/s2 to the left and the amplitude of motion A = 0.732 m.
What is Amplitude of motion?
The distance between the central and extreme points for a moving particle is known as the amplitude of motion.
The given data to find the amplitude of motion,
Object displaced = 0.560 m
Velocity = 2.45 m/s
Acceleration = 8.60 m/s²
Starting with sine:
x(t) = Asin(ωt)
so that t = 0, x = 0
x(t) = 0.56 m = Asin(ωt)
v(t) = x(t)'= 2.45 m/s = Aωcos(ωt)
a(t) = v(t)'= -8.60 m/s² = -Aω²sin(ωt)
x(t) / a(t) = Asin(ωt) / -Aω²sin(ωt)
0.56m / -8.60 m/s² = -1 / ω²
ω² = 15.3571 rad^2/s^2
ω = 3.91881 rad/s
x(t) / v(t) = Asin(ωt) / Aωcos(ωt)
0.560m / 2.45m/s = tan(3.91t) / 3.91rad/s
0.8937= tan(3.91t)
t = 0.176 s
x(0.176) = Asin(3.59×0.176)
0.65 m= Asin(0.631)
A = 0.732 m is the amplitude of motion.
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A mechanic applies a force of 60N at a distance of 80 cm from the pivot on a wheel wrench. What is the size of the moment?
Answer:
48 Nm
Explanation:
Moment, or torque, is the cross product of radius and force vectors.
τ = r × F
τ = (0.80 m) (60 N)
τ = 48 Nm
A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance . For the next loading, the spring is compressed a distance . How much faster does the second dart leave the gun compared with the first
Complete question is;
A dart is inserted into a spring - loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is compressed a distance 2x. How much faster does the second dart leave the gun compared to the first?
Answer:
The second dart leaves the gun two times faster than the first one.
Explanation:
If we assume there was no energy loss during the spring - dart energy transfer, we can easily apply the principle of conservation of energy. So;
Potential energy = kinetic energy
Thus;
½kx² = ½mv²
Making velocity "v" the subject, we have;
v = √(kx²/m)
Since the initial distance is "x", thus initial launching velocity is;
v1 = √(kx²/m)
Since next distance is 2x, thus, second launch velocity is;
v2 = √(k(2x)²/m)
Expanding, we have;
v2 = √(4kx²/m)
v2 = 2√(kx²/m)
Comparing this to the one gotten for v1 earlier, we can see that it is double v1.
So, v2 = 2v1
Hence, The second dart leaves the gun two times faster than the first one.
Block 1, of mass m1 = 2.50 kg , moves along a frictionless air track with speed v1 = 27.0 m/s. It collides with block 2, of mass m2 = 33.0 kg , which was initially at rest. The blocks stick together after the collision.A. Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically.B. Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically.C. what is the change deltaK= Kfinal- K initial in the two block systems kinetic energy due to the collision ? Express your answer numerically in joules.
Answer:
a
The total initial momentum of the two-block system is [tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]
b
The magnitude of the final velocity of the two-block system [tex]v_f = 1.9014 \ m/s[/tex]
c
the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision is
[tex]\Delta KE =- 847.08 \ J[/tex]
Explanation:
From the question we are told that
The mass of first block is [tex]m_1 = 2.50 \ kg[/tex]
The initial velocity of first block is [tex]u_1 = 27.0 \ m/s[/tex]
The mass of second block is [tex]m_2 = 33.0\ kg[/tex]
initial velocity of second block is [tex]u_2 = 0 \ m/s[/tex]
The magnitude of the of the total initial momentum of the two-block system is mathematically repented as
[tex]p_i = (m_1 * u_1 ) + (m_2 * u_2)[/tex]
substituting values
[tex]p_i = (2.50* 27 ) + (33 * 0)[/tex]
[tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]
According to the law of linear momentum conservation
[tex]p_i = p_f[/tex]
Where [tex]p_f[/tex] is the total final momentum of the system which is mathematically represented as
[tex]p_f = (m_+m_2) * v_f[/tex]
Where [tex]v_f[/tex] is the final velocity of the system
[tex]p_i = (m_1 +m_2 ) v_f[/tex]
substituting values
[tex]67.5 = (2.50+33 ) v_f[/tex]
[tex]v_f = 1.9014 \ m/s[/tex]
The change in kinetic energy is mathematically represented as
[tex]\Delta KE = KE_f -KE_i[/tex]
Where [tex]KE_f[/tex] is the final kinetic energy of the two-body system which is mathematically represented as
[tex]KE_f = \frac{1}{2} (m_1 +m_2) * v_f^2[/tex]
substituting values
[tex]KE_f = \frac{1}{2} (2.50 +33) * (1.9014)^2[/tex]
[tex]KE_f =64.17 J[/tex]
While [tex]KE_i[/tex] is the initial kinetic energy of the two-body system
[tex]KE_i = \frac{1}{2} * m_1 * u_1^2[/tex]
substituting values
[tex]KE_i = \frac{1}{2} * 2.5 * 27^2[/tex]
[tex]KE_i = 911.25 \ J[/tex]
So
[tex]\Delta KE = 64.17 -911.25[/tex]
[tex]\Delta KE =- 847.08 \ J[/tex]
The lowest-pitch tone to resonate in a pipe of length L that is closed at one end and open at the other end is 200 Hz. Which one of the following frequencies will NOT resonate in the same pipe
a. 1800 Hz
b. 1000 Hz
c. 1400 Hz
d. 600 Hz
e. 400 Hz
Answer:
e. 400 Hz
Explanation:
In closed organ pipe, only odd harmonics of fundamental note is possible .
The fundamental frequency is 200 Hz . Then other overtones will be having following frequencies .
200 x 3 , 200 x 5 , 200 x 7 , 200 x 9 etc
600 Hz , 1000 Hz , 1400 Hz , 1800 Hz .
Frequency not possible is 400 Hz .
Compare the energy consumption of two commonly used items in the household. Calculate the energy used by a 1.40 kW toaster oven, Wtoaster , which is used for 5.40 minutes , and then calculate the amount of energy that an 11.0 W compact fluorescent light (CFL) bulb, Wlight , uses when left on for 10.50 hours .
Energy = (power) x (time)
-- For the toaster:
Power = 1.4 kW = 1,400 watts
Time = 5.4 minutes = 324 seconds
Energy = (1,400 W) x (324 s) = 453,600 Joules
-- For the CFL bulb:
Power = 11 watts
Time = 10.5 hours = 37,800 seconds
Energy = (11 W) x (37,800 s) = 415,800 Joules
-- The toaster uses energy at 127 times the rate of the CFL bulb.
-- The CFL bulb uses energy at 0.0079 times the rate of the toaster.
-- The toaster is used for 0.0086 times as long as the CFL bulb.
-- The CFL bulb is used for 116.7 times as long as the toaster.
-- The toaster uses 9.1% more energy than the CFL bulb.
-- The CFL bulb uses 8.3% less energy than the toaster.
How much work must be done on a 10 kg snowboard to increase its speed from 4 m/s to 6 m/s?
Answer:
100 J
Explanation:
Work = change in energy
W = ΔKE
W = ½ mv² − ½ mv₀²
W = ½ m (v² − v₀²)
W = ½ (10 kg) ((6 m/s)² − (4 m/s)²)
W = 100 J
A freight car moves along a frictionless level railroad track at constant speed. The freight car is open on top. A large load of coal is suddenly dumped into the car. What happens to the speed of the freight car
Answer:
The speed of the freight car decreases.
Explanation:
According to the law of conservation of momentum indicates that for colliding in an isolated system, the total momentum pre and post collision is same for the two objects this is done because the momentum that one item has lost is same for the momentum that the other received
In the given situation, the freight car travels at constant speed along a frictionless railroad line. The top floor freight car is open. Then a huge load of coal is dumped inside the car.
Therefore the speed of the freight car decreased by applying the law of conservation of momentum i
Davina accelerates a box across a smooth frictionless horizontal surface over a displacement of 18.0 m with a constant 25.0 N force angled at 23.0° below the horizontal. How much work does she do on the box? A. 176 J B. 414 J C. 450 J D. 511 J Group of answer choices
Answer:
W = 414 J, correct is B
Explanation:
Work is defined by
W = ∫ F .dx
where F is the force, x is the displacement and the point represents the dot product
this expression can also be written with the explicit scalar product
W = ∫ F dx cos θ
where is the angle between force and displacement
for this case as the force is constant
W = F x cos θ
calculate
W = 25.0 18.0 cos (-23)
W = 414 J
the correct answer is B
When solving vector addition problems you can use either the graphical
method or the
Answer :the resultant of two vectors can be found using either the parallelogram method or the triangle method. don't know if this was helpful ?
Explanation:
Answer:
Analytical method.
A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring
Answer:
a) The spring constant of the spring is [tex]28.381\,\frac{N}{m}[/tex], b) The angular frequency of the block is [tex]7.038\,\frac{rad}{s}[/tex].
Explanation:
This question is incomplete and complete version will be presented herein:
A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring (b) Find the angular frequency of the block 's vibrations.
a) Since spring is hung from the ceiling and is stretched by action of gravity on 0.573 kilogram block. According to the Hooke's Law, force experimented by the spring is directly proportional to elongation. An expression describing the phenomenon is presented and described below: (System at equilibrium - Newton's Second Law)
[tex]m\cdot g = k\cdot \Delta x[/tex]
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]\Delta x[/tex] - Spring linear deformation, measured in meters.
Now, the spring constant is cleared in this equation and outcome is computed: ([tex]m = 0.573\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta x = 0.198\,m[/tex])
[tex]k = \frac{m\cdot g}{\Delta x}[/tex]
[tex]k = \frac{(0.573\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.198\,m}[/tex]
[tex]k = 28.381\,\frac{N}{m}[/tex]
The spring constant of the spring is [tex]28.381\,\frac{N}{m}[/tex].
b) Let suppose that mass-spring system is experimenting a simple harmonic motion, so that angular frequency is equal to:
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
Given that [tex]k = 28.381\,\frac{N}{m}[/tex] and [tex]m = 0.573\,kg[/tex], the angular frequency, measured in radians per second, of the block is:
[tex]\omega = \sqrt{\frac{28.381\,\frac{N}{m} }{0.573\,kg} }[/tex]
[tex]\omega = 7.038\,\frac{rad}{s}[/tex]
The angular frequency of the block is [tex]7.038\,\frac{rad}{s}[/tex].
An object will sink in a liquid if the density of the object is greater than that of the liquid. The mass of a sphere is 0.723 g. If the volume of this sphere is less than ________ cm3, then the sphere will sink in liquid mercury (density
Answer:
= 0.0532 cm^3
Explanation:
The computation of volume of the sphere is shown below:-
[tex]Density = \frac{Mass}{Volume}[/tex]
Where,
Density = 13.6 g/cm^3
Mass of sphere = 0.723 g
now we will put the values into the above formula to reach volume of the sphere which is here below:-
[tex]Volume = \frac{0.723}{13.6}[/tex]
= 0.0532 cm^3
Therefore for computing the volume of the sphere we simply applied the above formula.
commune time to work ( physics) i need help pls :(
HOW CAN I SOLVE THIS QUESTION? PLEASE HELP The movement of a locomotive piston in the cylinder is limited to 0.76 m. Assume that the piston makes a simple harmonic movement that makes 180 revolutions per minute, and find its maximum speed.
Answer:
7.2 m/s
Explanation:
The maximum speed is the amplitude times the frequency.
v = Aω
v = (0.76 m / 2) (180 rev × 2π rad/rev / 60 s)
v = 7.2 m/s
An object with a mass of 1500 g (grams) accelerates 10.0 m/s2 when an
unknown force is applied to it. What is the amount of the force
Answer:
15N
Explanation:
F=ma
m=1500g = 1.5kg
a=10m/s2
1.5×10=15 N
Answer:15000gms^-2
Explanation:
F=m×a
m=1500g, a=10ms^-2
F=(1500×10)gms^-2
F=15000gms^-2
EASY! WILL GIVE BRAINLIEST!
Find the conductivity of a conduit with a cross-sectional area of 0.50 cm2 and a length of 15 cm if its conductance G is 0.050 ohm-1.
σ = _____ ohm-1cm-l
3
75
1.5
0.0017
the answer is 1.5 hope this helps
Answer:
1.5
Explanation:
0.5=σ/(15/0.5)
σ=3/2 or 1.5
Unit conversion
The choices are in units
A,GA,MA,uA,kA,mA,nA,pA. Pick one the units
Answer:
1.234567 kA
Explanation:
The prefix k stands for kilo-, or 10³. The prefix m stands for milli-, or 10⁻³. The sum shown is ...
1.234 kA + 0.000567 kA = 1.234567 kA
On April 13, 2029 (Friday the 13th!), the asteroid 99942 mi Apophis will pass within 18600 mi of the earth-about 1/13 the distance to the moon! It has a density of 2600 kg/m^3, can be modeled as a sphere 320 m in diameter, and will be traveling at 12.6 km/s.
1)If, due to a small disturbance in its orbit, the asteroid were to hit the earth, how much kinetic energy would it deliver?
2)The largest nuclear bomb ever tested by the United States was the "Castle/Bravo" bomb, having a yield of 15 megatons of TNT. (A megaton of TNT releases 4.184x10^15 J of energy.) How many Castle/Bravo bombs would be equivalent to the energy of Apophis?
Answer:
Explanation:
Volume of asteroid = 4/3 x π x 160³
= 17.15 x 10⁶
mass = volume x density
= 17.15 x 10⁶ x 2600
= 445.9 x 10⁸ kg
kinetic energy
= 1/2 x 445.9 x 10⁸ x( 12.6 )² x 10⁶
= 35.4 x 10¹⁷ J .
2 )
energy of 15 megaton
= 4.184 x 10¹⁵ x 15 J
= 62.76 x 10¹⁵ J
No of bombs required
= 35.4 x 10¹⁷ / 62.76 x 10¹⁵
= 56.4 Bombs .
An object, with mass 70 kg and speed 21 m/s relative to an observer, explodes into two pieces, one 4 times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame
Answer:
K = 3.9 kJ
Explanation:
The kinetic energy ([tex]K_{T}[/tex]) added is given by the difference between the final kinetic energy and the initial kinetic energy:
[tex] K_{T} = K_{f} - K_{i} [/tex]
The initial kinetic energy is:
[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} [/tex]
Where m₁ is the mass of the object before the explosion and v₁ is its velocity
[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} = \frac{1}{2}70 kg*(21 m/s)^{2} = 1.54 \cdot 10^{4} J [/tex]
Now, the final kinetic energy is:
[tex] K_{f} = \frac{1}{2}m_{2}v_{2}^{2} + \frac{1}{2}m_{3}v_{3}^{2} [/tex]
Where m₂ and m₃ are the masses of the 2 pieces produced by the explosion and v₁ and v₂ are the speeds of these pieces
Since m₂ is 4 times as massive as m₃ and v₃ = 0, we have:
[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}v_{2}^{2} + \frac{1}{2}*\frac{1}{5}m_{1}*0 [/tex] (1)
By conservation of momentum we have:
[tex] p_{i} = p_{f} [/tex]
[tex] m_{1}v_{1} = m_{2}v_{2} + m_{3}v_{3} [/tex]
[tex] m_{1}v_{1} = \frac{4}{5}m_{1}v_{2} + \frac{1}{5}m_{1}*0 [/tex]
[tex] v_{2} = \frac{5}{4}v_{1} [/tex] (2)
By entering (2) into (1) we have:
[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}(\frac{5}{4}v_{1})^{2} = \frac{1}{2}*\frac{4}{5}70 kg(\frac{5}{4}*21 m/s)^{2} = 1.93 \cdot 10^{4} J [/tex]
Hence, the kinetic energy added is:
[tex] K_{T} = K_{f} - K_{i} = 1.93 \cdot 10^{4} J - 1.54 \cdot 10^{4} J = 3.9 \cdot 10^{3} J [/tex]
Therefore, the kinetic energy added to the system during the explosion is 3.9 kJ.
I hope it helps you!
a ballistic pendulum is used to measure the speed of high-speed projectiles. A 6 g bullet A is fired into a 1 kg wood block B suspended by a cord of length l =2.2m. The block then swings through a maximum angle of theta = 60. Determine (a) the initial speed of the bullet vo, (b) the impulse imparted by the bullet on the block, (c) the force on the cord immediately after the impact
Answer:
(a) v-bullet = 399.04 m/s
(b) I = 2.38 kg m/s
(c) T = 2.59 N
Explanation:
(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.
The potential energy is given by:
[tex]U=(M+m)gh[/tex] (1)
U: potential energy
M: mass of the wood block = 1 kg
m: mass of the bullet = 6g = 6.0*10^-3 kg
g: gravitational constant = 9.8m/s^2
h: distance to the ground
The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:
[tex]h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m[/tex]
The potential energy is:
[tex]U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J[/tex]
Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:
[tex]U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}[/tex]
Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:
[tex]Mv_1+mv_2=(M+m)v[/tex] (2)
v1: initial velocity of the wood block = 0m/s
v2: initial speed of the bullet
v: speed of bullet and block = 2.38m/s
You solve the equation (2) for v2:
[tex]M(0)+mv_2=(M+m)v[/tex]
[tex]v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}[/tex]
The speed of the bullet before the impact with the wood block is 399.04 m/s
(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:
[tex]I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}[/tex]
The impulse is 2.38 kgm/s
(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:
[tex]T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N[/tex]
The force on the cord after the impact is 2.59N
Answer:
The initial speed of the bullet [tex]V_o = 777.97m/s[/tex]The force on the cord immediately after the impact = [tex]19.71N[/tex]Explanation:
Apply the law of conversion of energy
[tex]V_f = \sqrt{2gh}[/tex]
where,
h = height of which the bullet and block rise after impact
[tex]h = L - Lcos\theta\\\\h = 2.2 - (2.2*cos60)\\\\h = 1.1m[/tex]
Therefore,
[tex]V_f = \sqrt{2gh}\\\\V_f = \sqrt{2*9.8*1.1}\\\\V_f = 4.64m/s[/tex]
From conservation of momentum principle, [tex]m_Bv_B = 0[/tex]
[tex]m_ov_o + m_Bv_B = (m_b+m_B)V_f\\\\0.006V_o = (0.006+1)*4.64\\\\V_o = 777.97m/s[/tex]
C) The force in the cable is due to the centrfugal force of the system, which is due to the motion of the system is a curved path and weight of the system
[tex]F = \frac{m_b+m_B}{L}V_f^2 + (m_b+m_B)g\\\\F = \frac{0.006+1}{2.2}*4.64^2 + (0.006+1)9.81\\\\F = 19.71N[/tex]
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How many ohms of resistance are in a 120–volt hair dryer that draws 7.6 amps of current?
From Ohm's law . . . Resistance = (voltage) / (current)
Resistance = (120 volts) / (7.6 Amperes)
Resistance = 15.8 Ω
A worker with spikes on his shoes pulls on rope that is attached to a box that is resting on a flat, frictionless frozen lake. The box has mass m, and the worker pulls with a constant tension T at an angle θ = 40 ∘ above the horizontal. There is a strong headwind on the lake, which produces a horizontal force Fw that is pointed in the opposite direction than the box is being pulled. Draw a free-body diagram for this system. Assume that the worker pulls the box to the right. If the wind force has a magnitude of 30 N, with what tension must the worker pull in order to move the box at a constant velocity?
Answer:
a
The free body diagram is shown on the first uploaded image
b
The tension on the rope is [tex]T=39.16 \ N[/tex]
Explanation:
From the question we are told that
The mass of the box is m
The tension on the box is T
The angle at which it is pulled is [tex]\theta = 40^o[/tex]
The force produced by the strong head wind is [tex]Fw = 30 \ N[/tex]
At equilibrium the net force acting on the block along the horizontal axis is zero i.e
[tex]Tcos \theta -F_w = 0[/tex]
substituting values
[tex]Tcos (40) -30 = 0[/tex]
[tex]Tcos (40) = 30[/tex]
[tex]T(0.76604)) = 30[/tex]
[tex]T=39.16 \ N[/tex]