Answer:
A. Momentum of the grizzly bear is 3241.56 Kgm/s
B. The human will run at 28.56 m/s
Explanation:
A. Determination of the momentum of the grizzly bear.
Mass of grizzly bear = 420 lb
Velocity of grizzly bear = 17 m/s
momentum of the grizzly bear =?
Next, we shall convert 420 lb to Kg. This can be obtained as follow:
1 lb = 0.454 kg
Therefore,
420 lb = 420 lb × 0.454 Kg / 1 lb
420 lb = 190.68 Kg
Thus, 420 lb is equivalent to 190.68 Kg
Finally, we shall determine the momentum of the grizzly bear as follow:
Mass of grizzly bear = 190.68 Kg
Velocity of grizzly bear = 17 m/s
momentum of the grizzly bear =?
Momentum = mass × velocity
Momentum = 190.68 × 17
Momentum = 3241.56 Kgm/s
Thus, the momentum of the grizzly bear is 3241.56 Kgm/s
B. Determination of how fast the human will run.
Mass of human = 250 lb
Momentum of human = momentum of the grizzly bear
Momentum of the grizzly bear is 3241.56 Kgm/s
Momentum of human = 3241.56 Kgm/s
Velocity of human =?
Next, we shall convert 250 lb to Kg. This can be obtained as follow:
1 lb = 0.454 kg
Therefore,
250 lb = 250 lb × 0.454 Kg / 1 lb
250 lb = 113.5 Kg
Thus, 250 lb is equivalent to 113.5 Kg
Finally, we shall determine how fast the human will run as follow:
Mass of human = 113.5 Kg
Momentum of human = 3241.56 Kgm/s
Velocity of human =?
Momentum = mass × velocity
3241.56 = 113.5 × velocity
Divide both side by 113.5
Velocity = 3241.56 / 113.5
Velocity = 28.56 m/s
Therefore, the human will run at 28.56 m/s to match the momentum of the grizzly bear
A child and sled with a combined mass of 48.8 kg slide down a frictionless hill that is 7.05 m high. If the sled starts from rest, what is its speed at the bottom of the hill?
___m/s
Answer:
Speed at bottom of the hill (v) = 11.74 m/s
Explanation:
Given:
Combined mass = 48.8 kg
Height h = 7.05 m
Find:
Speed at bottom of the hill (v)
Computation:
v² = 2gh
v = √2 x 9.8 x 7.05
v = √138.18
v = 11.74 m/s
Speed at bottom of the hill (v) = 11.74 m/s
Examine the boxes below. Both boxes are identical to one another. The mass of each box is concentrated in the very center, so the center of mass is in the
middle of the box. (Remember that mass is similar to weight) The arrows represent gravity pulling down on the center of mass of each box. Which of the boxes
represents a stable system? if you were to give that box a small push, what would happen to its state of equilibrium? Describe how a disturbance to this box
would affect it. Use the terms center of mass and equilibrium in your response.
*A car is going through a dip in the road whose curvature approximates a circle of radius 150m. At what velocity will the occupants of the car appear to weigh 15% more than their normal weight
Answer:
v= 14.85 m/s
Explanation:
When at the bottom of the dip, the only force that keeps the car in the circular trajectory, is the centripetal force.This force is not a new force, is just the net force aiming to the center of the circle.In this case, is just the difference between the normal force (always perpendicular to the surface, pointing upward) and the force that gravity exerts on the car (which is known as the weight), pointing downward.So, we can write the following expression:[tex]F_{cent} = F_{n} - F_{g} (1)[/tex]
It can be showed that the centripetal force is related to the speed by the following expression:[tex]F_{cent} = m*\frac{v^{2}}{r} (2)[/tex]The normal force, it is called the apparent weight, because it would be the weight as measured by a scale.Replacing (2) in (1), and solving for Fn, we get:[tex]F_{n} = m*\frac{v^{2} }{r} + m*g (3)[/tex]
Now, we need to find the value of v that makes Fn, exactly 15% more than the weight m*g, so we can write the following equation:[tex]F_{n} = 1.15*F_{g} = m*\frac{v^{2}}{r} +F_{g} (4)[/tex]
Replacing Fg by its value, simplifying, and solving for v, we get:[tex]v = \sqrt{0.15*g*r} = \sqrt{9.8 m/s2*0.15*150m} = 14.85 m/s (5)[/tex]
A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose mag
Answer:
The number of revolutions is 10.68 rev/min.
Explanation:
Given that,
Radius = 8 m
Suppose, centripetal acceleration equal to the gravity
[tex]a_{c}=g=9.8[/tex]
We need to calculate the velocity
Using formula of centripetal acceleration
[tex]a_{c}=\dfrac{v^2}{r}[/tex]
[tex]v^2=a_{c}\times r[/tex]
Put the value into the formula
[tex]v=\sqrt{9.8\times8}[/tex]
[tex]v=8.85\ m/s[/tex]
We need to calculate the value of [tex]\omega[/tex]
Using formula of velocity
[tex]v=r\omega[/tex]
[tex]\omega=\dfrac{v}{r}[/tex]
Put the value into the formula
[tex]\omega=\dfrac{8.85}{8}[/tex]
[tex]\omega=1.12\rad/s[/tex]
We need to calculate the number of revolutions
Using formula of angular frequency
[tex]\omega=\dfrac{2\pi}{T}[/tex]
[tex]\omega=2\pi N[/tex]
[tex]N=\dfrac{\omega}{2\pi}[/tex]
Put the value into the formula
[tex]N=\dfrac{1.12}{2\pi}[/tex]
[tex]N=0.178\ rev/s[/tex]
Using conversion rev/s to rev/min
[tex]N=0.178\times 60[/tex]
[tex]N=10.68\ rev/min[/tex]
Hence, The number of revolutions is 10.68 rev/min.
c. Describe how to use right-hand rules to relate the directions of magnetic fields, currents, moving charges, and forces. (1 point)
Answer:
Physicists use a hand mnemonic known as the right-hand rule to help remember the direction of magnetic forces. To form the mnemonic, first make an L-shape with the thumb and first two fingers of your right hand. Then, point your middle finger perpendicular to your thumb and index finger.
Answer:
If you point your pointer finger in the direction the positive charge is moving, and then your middle finger in the direction of the magnetic field, your thumb points in the direction of the magnetic force pushing on the moving charge.
Explanation:
Got it right
If a car travels 60 mph for a distance of 180 miles, how much time
did it take?
Answer:
3 hours
Explanation:
180 divided by 60 (mph means miles per hours by the way)
An object travels in a circular path at speed vand experiences a centripetal acceleration ac. What will be the object's new centripetal
acceleration, if its speed is increased to 2.0V?
Answer:
Option B. 4a꜀
Explanation:
From the question given above, the following data were obtained:
Initial speed (v₁) = v
Initial acceleration (a₁) = a꜀
New speed (v₂) = 2v
New acceleration (a₂) =?
Centripetal acceleration is given by the following equation:
a = v²/ r
Next, we shall make r the subject.
a = v²/ r
Cross multiply
a × r = v²
Divide both side by a
r = v² / a
NOTE: We assume the radius (r) to be constant.
v₁² / a₁ = v₂² / a₂
Thus, we shall determine the new acceleration as follow:
Initial speed (v₁) = v
Initial acceleration (a₁) = a꜀
New speed (v₂) = 2v
New acceleration (a₂) =?
v₁² / a₁ = v₂² / a₂
v² / a꜀ = (2v)² / a₂
v² / a꜀ = 4v² / a₂
Cross multiply
v² × a₂ = a꜀ × 4v²
Divide both side by v²
a₂ = a꜀ × 4v² / v²
a₂ = a꜀ × 4
a₂ = 4a꜀
Therefore, the new centripetal acceleration of the object is 4a꜀
Answer:
4.0 · ac
Explanation:
Just did it
A car is stopped at a red light,
Which claim about the forces acting on the car must be true?
Answer:
there is no net force onthe car
Explanation:
khan academy
In order to track the movement of Earth’s tectonic plates over time, scientists use which of the following tools?
A) estimating
B) rulers
C) digitized signals from GPS
An 82 kg man drops from rest on a diving board 3.0 m above the surface of the water. What is his speed when he hits the water? On hitting the water, it takes him 0.55 s to slow down to a stop. Calculate the net force on him inside the water that brings him to a stop. What are the forces on him when he slows down inside the water?
Answer:1.1x10^3N upward
Explanation:
A dog running at a speed of 12 m/s has 1,080 J of kinetic energy. What is the mass of the dog
Answer:
The mass of the dog is 15 kg
Explanation:
We use the formula for the kinetic energy of the dog:
[tex]KE=\frac{1}{2} m\,v^2[/tex]
Then, in our case:
[tex]KE=\frac{1}{2} m\,v^2\\1080 \,=\,\frac{1}{2} m\,(12)^2\\m = \frac{2*1080}{144} \\m=15 \,\,kg[/tex]
How is a controlled variable different from a responding variable?
A. A controlled variable changes due to changes in the respondin
variable during the experiment.
B. A controlled variable stays the same due to changes in the
responding variable during the experiment.
C. A controlled variable changes throughout an experiment, but a
responding variable stays the same.
D. A controlled variable stays the same throughout an experiment,
but a responding variable changes.
An apple fell off a tree and is traveling to the ground.
Which claim about the forces acting on the apple must be true?
Answer:
GRAVITYYYYY
HOPE THIS HELPS
Answer: There is a net force acting on which is a force of gravity
Explanation:
Describe the motion of an object that has balanced forces acting on it.
Answer:
If it has balanced forces, then it is completely still.
Balanced forces do not cause a change in motion. So, the object that has balanced forces acting on it either moves with constant speed at a constant direction or remains rest.
What is balanced force?Two forces are said to be balanced when their strengths are equal but their directions of action are opposing. Once more, tug-of-war is a prime illustration. The forces are balanced if the pullers are exerting equal force but going in the opposite direction on either side of the rope. There is hence no motion.
Forces that are balanced can cancel one another out. The thing stays in place whenever there is a balanced force.
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_______ is the most common pollutant; we often simply refer to it
as "smog".
Ground-level ozone is the most common pollutant that is often referred to as "smog".
What is smog?Smog is a type of air pollution that is characterized by a mixture of smoke and fog.
It typically forms in urban areas with high levels of vehicle and industrial emissions, as well as natural factors such as temperature inversions.
Smog is composed of a complex mixture of pollutants, including ground-level ozone, nitrogen oxides, sulfur oxides, particulate matter, and volatile organic compounds (VOCs).
Exposure to smog can cause a range of health problems, including respiratory issues, heart disease, and cancer. It can also have negative impacts on the environment, including damage to crops, forests, and bodies of water. Efforts to reduce smog often involve reducing emissions from vehicles and industrial sources and promoting cleaner forms of transportation and energy production.
Therefore, The Ground-level ozone is the most common pollutant that is often referred to as "smog".
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What does a producer need to make its own food
Answer:
A producer needs the sun to make its own food, because producers use photosynthesis. In photosynthesis you use the sun to turn the air people breathe out and water into glucose and oxygen.
Explanation:
Hope this helps ^-^
a bus traveling at 60m/s brakes and accelerates to a stop at a rate of -6m/s^2. how far did it travel while it stopped?
Answer:
300 m
Explanation:
using constant acceleration equations:
v = vi + a * t, v = final velocity = 0m/s , vi = initial velocity = 60m/s,
a = acceleration = -6m/s², t = time
solve t
t = 10s
x = vi * t + .5 * a * t²
plug
x = 300 m
A tennis ball is thrown up to a height of 26 m. Neglecting air resistance, with what initial
velocity was it thrown up with? Use Energy equations only
Answer:
v = 22.58[m/s]
Explanation:
To solve this problem we will use the principle of energy conservation, which tells us that potential energy is transformed into kinetic energy or vice versa.
[tex]E_{kin}=E_{pot}[/tex]
We must remember that potential energy is defined as the product of mass by gravity by height.
[tex]E_{pot}=m*g*h[/tex]
And the kinetic energy can be calculated by means of the following expression.
[tex]E_{kin}=0.5*m*v^{2}\\[/tex]
Now equalizing the two equations we can determine the initial velocity.
[tex]m*9.81*26=0.5*m*v^{2}\\255.06=0.5*v^{2}\\v=\sqrt{255.06/0.5}\\v=22.58[m/s][/tex]
A ball attached to a string is whirled at a constant speed of 2.0 meters per second in a
horizontal circle of radius 0.50 meter. What is the magnitude of the ball's centripetal
acceleration?
Answer:8.0 m/s^2
Explanation: The ball is traveling in uniform circular motion at the speed of 2 m/s in a path of the radius that is 0.50 meters.
v^2/r
2.0 m/s^2/ 0.50 m= 8.0 m/s^2
The magnitude of the ball's centripetal acceleration is [tex]\rm 8 \;m/sec^2[/tex] and this can be determined by using the formula of the centripetal acceleration.
Given :
A ball attached to a string is whirled at a constant speed of 2.0 meters per second in a horizontal circle of a radius of 0.50 meters.
The following steps can be used in order to determine the magnitude of the ball's centripetal acceleration:
Step 1 - The formula of the centripetal acceleration can be used in order to determine the magnitude of the ball's centripetal acceleration.
Step 2 - The formula of the centripetal acceleration is given below:
[tex]\rm a_c=\dfrac{v^2}{r}[/tex]
Step 3 - Substitute the values of the known terms in the above formula.
[tex]\rm a_c=\dfrac{2^2}{0.5}[/tex]
Step 4 - Simplify the above expression.
[tex]\rm a_c = 8\;m/sec^2[/tex]
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i need help on number 6 pls due today
Explanation:
Graph A matches description 4 because the car is coming back.
Graph B matches description 3 because the speed of the car is decreasing.
Graph C matches the description 2 because the car is traveling at a constant rate.
Graph D matches the description 1 because the car is stopped.
what is letaral inversion?
Answer:
Images that are not aligned are laterally inverted
Explanation:
When a plane mirror reverses an image so that it becomes upside down and downside up or the other way round, we say that the image is laterally inverted.
Hence the effect produced when a plane mirror reverses an image is called lateral inversion
A good example of lateral inversion ins the lettering of an ambulance
0-kilogram child moving at 4.0 meters per second jumps onto a 50-kilogram sled that is initially at rest on a long, frictionless, horizontal sheet of ice. [INF] a. Determine the speed of the child-sled system after the child jumps onto the sled. b. After coasting at constant speed for a short time, the child jumps off the sled in such a way that she is at rest with respect to the ice. Determine the speed of the sled after the child jumps off
This question is incomplete, the complete question is;
A 30-kilogram child moving at 4.0 meters per second jumps onto a 50-kilogram sled that is initially at rest on a long, frictionless, horizontal sheet of ice.
a. Determine the speed of the child-sled system after the child jumps onto the sled.
b. After coasting at constant speed for a short time, the child jumps off the sled in such a way that she is at rest with respect to the ice. Determine the speed of the sled after the child jumps off
Answer:
a) the speed of the child-sled system after the child jumps onto the sled is 1.5 m/s
b) the speed of the sled after the child jumps off is 2.4 m/s
Explanation:
Given that;
Weight of the child m_c = 30 kg
u = 4 m/s
M = 50 kg
a) the speed of the child-sled system after the child jumps onto the sled.
Using conservation of momentum;
P_i = P_f
30 × 4 = 50(v) + 30(v)
120 = 80(v)
v = 120 / 80
v = 1.5 m/s
therefore, the speed of the child-sled system after the child jumps onto the sled is 1.5 m/s
b) the speed of the sled after the child jumps off .
Also using conservation of momentum;
mv1 = mv2
(30+50) × 1.5 = (30 × 0) + (50 × v2 )
80 × 1.5 = 50v2
v2 = 120 / 50
v2 = 2.4 m/s
Therefore the speed of the sled after the child jumps off is 2.4 m/s
A football is kicked with an initial speed of 11.0 m/s, at an angle of 35degrees above the horizontal across horizontal ground.
(Assume air resistance is negligible)
What is the range of the ball?
Answer:
The range of the ball is 11.6 meters
Explanation:
Projectile Motion
It's the type of motion that experiences an object launched with an initial angle and moves along a curved path exclusively under the action of gravity.
Being vo the initial speed of the object, θ the initial launch angle, and g the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:
[tex]\displaystyle d={\frac {v_o^{2}\sin(2\theta )}{g}}[/tex]
The football is kicked with an initial speed of vo=11 m/s at an angle of θ=35°.
Calculating the range:
[tex]\displaystyle d={\frac {11^{2}\sin(2\cdot 35^\circ)}{9.8}}[/tex]
[tex]\displaystyle d={\frac {121\sin(70^\circ)}{9.8}}[/tex]
d = 11.6 m
The range of the ball is 11.6 meters
A 2.0-kg ball with an initial velocity of (4i + 3j) m/s collides with a wall and rebounds with a velocity of (–4i + 3j) m/s. What is the impulse exerted on the ball by the wall?
Answer:
-16i kgm/s
Explanation:
Impulse I = m(v - u) where m = mass of ball = 2.0 kg, u = initial velocity of ball = (4i + 3j) m/s and v = final velocity of ball = (-4i + 3j) m/s.
So, the impulse is thus
I = m(v - u)
= 2.0 kg[(-4i + 3j) m/s - (4i + 3j) m/s]
= 2.0 kg[(-4i - 4i) + (3j - 3j) m/s]
= 2.0kg[-8i + 0j] m/s
= -16i kgm/s
The impulse on a 2.0-kg ball with an initial velocity of (4i + 3j) m/s collides with a wall and rebounds with a velocity of (–4i + 3j) m/s = 16i
Impulse: This can be defined as change in momentum of a body. The s.i unit of impulse is kgm/s²
The formula of impulse is
I = m(v-u)................ Equation 1
Where, I = impulse exerted on the ball by the wall, m = mass of the ball, v = final velocity of the ball, u = initial velocity of the ball
From the question,
Given: m = 2.0 kg, v = (-4i+3j) m/s, u = (4i+3j) m/s
Substitute these values into equation 1
I = 2.0[(-4i+3j)-(4i+3j)
I = 2.0(-4i-4i+3j-3j)
I = 2.0(-8i+0j)
I = -16i.
Hence the impulse exerted by the wall on the ball is -16i
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Suppose you are traveling down the highway at 100 km/hr, and want to pass a truck. You accelerate from 100 km/hr to 120 km/hr in 3 seconds in order to do so. What is your average acceleration, in meters per second squared (m/s 2 )
Answer:
1.87 ms-2
Explanation:
We need to convert the velocity from km/hr to m/s as follows;
100 × 1000/3600 = 27.7 m/s
Also
120 × 1000/3600 = 33.3 m/s
From;
v= u + at
v= 33.3 m/s
u= 27.7 m/s
t= 3 secs
v= u + at
a = v- u/t
a = 33.3 - 27.7/3
a= 1.87 ms-2
the mass of the earth is 6×10²⁴ kg and the it's radius is 6400 km . what is the mass of a man weighing 977 N in spring balance ? { G= 6.67×10-¹¹ }
Answer:
Mass of the man = 100 kg
Explanation:
Given that,
Mass of the Earth, M = 6×10²⁴ kg
The radius of the Earth, r = 6400 km
Force on man, F = 977 N
We need to find the mass of the man. Let the mass of the man be m. The gravitational force acting between two objects is given by :
[tex]F=G\dfrac{mM}{r^2}\\\\m=\dfrac{Fr^2}{GM}\\\\m=\dfrac{977\times (6400\times 10^3)^2}{6.67\times 10^{-11}\times 6\times 10^{24}}\\\\m=99.99\ kg[/tex]
or
m = 100 kg
So, the mass of the man is 100 kg.
A cannonball is launched off a 100 m cliff horizontally with an initial velocity of 50
m/s. The goal is to hit a target 200m away but the cannonball goes too far. Which of
these changes could make the cannonball travel a shorter distance?
a) Decrease the launch speed of the cannon.
UbNeither of these will decrease the range of the cannonball
d Both of these will decrease the range of the cannonball
Lower the launch height of the cannon.
Answer:
the correct one is b, Decrease the launch speed and Low the launch height
Explanation:
Let's find the solution to the problem in order to answer the questions.
The bullet is fired horizontally with a velocity vo, let's find the time or it takes to reach the ground
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
as it shoots horizontally v_{oy} = 0 and when I reach the floor y = 0
0 = y₀ - ½ g t2
t = [tex]\sqrt{\frac{2y_{o} }{g} }[/tex]
where y₀ is the height of the cliff
at this time the bullet recreates a distance of
x = v₀ t
we substitute
x = v₀ \sqrt{\frac{2y_{o} }{g} }
There, to reduce the range of the bullet, we must decrease its speed. This is the magnitude that most affects the distance and to a lesser degree we can decrease the height.
When checking the answers, the correct one is b
If an object doubles its speed, how much does its kinetic energy go up by? What if it triples its speed?
Answer:
Kinetic energy is the energy of motion. An object that has motion - whether it is vertical or horizontal motion - has kinetic energy. There are many forms of kinetic energy - vibrational (the energy due to vibrational motion), rotational (the energy due to rotational motion), and translational (the energy due to motion from one location to another). To keep matters simple, we will focus upon translational kinetic energy. The amount of translational kinetic energy (from here on, the phrase kinetic energy will refer to translational kinetic energy) that an object has depends upon two variables: the mass (m) of the object and the speed (v) of the object. The following equation is used to represent the kinetic energy (KE) of an object.
KE = 0.5 • m • v2
where m = mass of object
v = speed of object
This equation reveals that the kinetic energy of an object is directly proportional to the square of its speed. That means that for a twofold increase in speed, the kinetic energy will increase by a factor of four. For a threefold increase in speed, the kinetic energy will increase by a factor of nine. And for a fourfold increase in speed, the kinetic energy will increase by a factor of sixteen. The kinetic energy is dependent upon the square of the speed. As it is often said, an equation is not merely a recipe for algebraic problem solving, but also a guide to thinking about the relationship between quantities.
Kinetic energy is a scalar quantity; it does not have a direction. Unlike velocity, acceleration, force, and momentum, the kinetic energy of an object is completely described by magnitude alone. Like work and potential energy, the standard metric unit of measurement for kinetic energy is the Joule. As might be implied by the above equation, 1 Joule is equivalent to 1 kg*(m/s)^2.
1 Joule = 1 kg • m2/s2
Explanation:
what type of motion occurs when an object spins around an axis without altering its linear position?
An adjustable tennis ball launcher launches tennis balls into the air from level ground and that return to level ground. The tennis balls are first launched with an initial velocity (vi) of 8.0 meters per second at an angle of 50° above the horizontal. The ball has an initial horizontal velocity of 5.1 meters per second. [Neglect friction.]
Calculate the vertical component of the ball’s initial velocity.
Calculate the maximum height reached by the ball.
Calculate the time elapsed to reach its maximum height.
Calculate the total horizontal distance travelled during its flight.
Answer:
Explanation:
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