Answer:
3.54
Explanation:
some nerd thing I found it on Yahoo answers
Answer:
3.54º
Explanation:
Find the blue θ first
sin⁻¹(540x10⁻⁹/2.88x10⁻⁶)=8.99°
Then find the orange θ
sin⁻¹(625x10⁻⁹/2.88x10⁻⁶)=12.53°
Take the differences and subtract
12.53°－8.99°=3.54°
Although planets orbit the Sun in ellipses, all the planetary orbits are fairly close to circular and not very eccentric.
True
False
Answer:
False
Explanation:
The Sun rotates in this same, right-hand-rule direction. All planetary orbits lie in nearly the same plane. All planetary orbits are nearly circular (eccentricity near zero).
Rate as Brainliest please
Which one the answer to this question
In an experiment, a disk is set into motion such that it rotates with a constant angular speed. As the disk spins, a small sphere of clay is dropped onto the disk, and the sphere sticks to the disk. All frictional forces are negligible. What would happened to the angular momentum and the total kinetic energy of the disk-sphere system immediately before and after the collision?
Answer:
L₀ = L_f , K_f < K₀
Explanation:
For this exercise we start as the angular momentum, with the friction force they are negligible and if we define the system as formed by the disk and the clay sphere, the forces during the collision are internal and therefore the angular momentum is conserved.
This means that the angular momentum before and after the collision changes.
Initial instant. Before the crash
L₀ = I₀ w₀
Final moment. Right after the crash
L_f = (I₀ + mr²) w
we treat the clay sphere as a point particle
how the angular momentum is conserved
L₀ = L_f
I₀ w₀ = (I₀ + mr²) w
w = [tex]\frac{I_o}{I_o + m r^2}[/tex] w₀
having the angular velocities we can calculate the kinetic energy
starting point. Before the crash
K₀ = ½ I₀ w₀²
final point. After the crash
K_f = ½ (I₀ + mr²) w²
sustitute
K_f = ½ (I₀ + mr²) ( [tex]\frac{I_o}{I_o + m r^2}[/tex] w₀)²
Kf = ½ [tex]\frac{I_o^2}{ I_o + m r^2}[/tex] w₀²
we look for the relationship between the kinetic energy
[tex]\frac{K_f}{K_o}[/tex]= [tex]\frac{I_o}{I_o + m r^2}[/tex]
[tex]\frac{K_f}{K_o } < 1[/tex]
K_f < K₀
we see that the kinetic energy is not constant in the process, this implies that part of the energy is transformed into potential energy during the collision
The energy goes from _____ to the _____ above it.
Answer:
The energy goes from the ground state to the excited states above it.
Which of the following best describes our
atmosphere?
A. envelope of gases that surround Earth
B. a specific range of altitude where plant life flourishes
C. The air, water, and land that form our planet
D. the water vapor in the air surrounding our planet
Why do herbivores of the Serengeti migrate year after year
Answer:
The main reason is that very young calves are more noticeable to predators when mixed with older calves from the previous year
Explanation:
They are forced to migrate by carnivores that hunt them.
An object is projected from a height of 100m above the ground at an angle of 300to the
horizontal with a velocity of 100m/s.
Calculate
(4)
(4)
The maximum height reached above the ground
Time of flight
The velocity and the direction of the object 1 sec before it hit the ground
(4)
Answer:
a) y = 127.6 m, b) 11.9s, c) v = 103.6 m / s, θ’= 326.7º
Explanation:
This is a missile throwing exercise
let's start by breaking down the initial velocity
sin 30 = [tex]\frac{v_{oy} }{v_o}[/tex]
cos 30 = v₀ₓ / v₀
v_{oy} = v₀ go sin 30
v₀ₓ = v₀ cos 30
v_{oy} = 100 sin 30 = 50 m / s
v₀ₓ = 100 cos 30 = 86.6 m / s
a) the maximum height is requested.
At this point the vertical velocity is zero (v_y = 0)
v_y² = [tex]v_{oy}^2[/tex] - 2 g y
0 = v_{oy}^2 - 2g y
y = [tex]\frac{v_{oy}^2 }{2g}[/tex]
y = 50² / (2 9.8)
y = 127.6 m
b) Flight time
this is the time it takes to reach the ground, the reference system for this movement is taken on the ground this is a height of y = 0 m and the body is at an initial height of i = 100m
y = y₀ + v₀ t - ½ g t²
0 = 100 + 50 t - ½ 9.8 t²
we solve the quadratic equation
4.9 t² - 50 t - 100 = 0
t = [tex]\frac{50 \pm \sqrt{50^2 + 4 \ 4.9 \ 100} }{2 \ 4.9}[/tex]
t = [tex]\frac{50 \ \pm \ 66.8}{9.8}[/tex]
t₁ = 11.9 s
t₂ = -8.4 s
flight time is 11.9s
c) The time 1 s before hitting the ground is
t1 = 11.9 -1
t1 = 10.9 s
let's find the vertical speed
v_y =[tex]v_{oy}[/tex] - g t
v_y = 50 - 9.8 10.9
v_y = -56.8 m / s
the negative sign indicates that the direction of the velocity is downward.
On the x-axis there is no acceleration therefore the speed is constant.
Let's use the Pythagorean theorem
v = [tex]\sqrt{v_x^2+v_y^2}[/tex]
v = [tex]\sqrt { 86.6^2 + 56.8^2}[/tex]
v = 103.6 m / s
let's use trigonometry
tan θ = [tex]\frac{v_y}{v_x}[/tex]
θ = tan⁻¹ \frac{v_y}{v_x}
θ = tan⁻¹ (-56.8 / 86.60)
θ = -33.3º
the negative sign indicates that it is measured clockwise from the x-axis
for a counterclockwise measurement
θ’= 360 - θ
θ' = 360 - 33.3
θ’= 326.7º
A go-cart is traveling at 15 mi/hr. How long does it take the go-cart to travel 3 miles?
Answer:
12 min
Explanation:every 4 minutes is 1 mile
What causes the Coriolis effect?
A
The sun's position relative to Earth
B.
Earth's orbit around the sun
с
Moon phases
D
Earth's rotation
A go-cart is traveling at a rate of 25 m/sec for 20 seconds. How far will the go cart travel?
Answer:
Distance travel by go-cart = 500 meter
Explanation:
Given:
Speed of go cart = 25 m/s
Time travel = 20 seconds
Find:
Distance travel by go-cart
Computation:
Distance = Speed x time
Distance travel by go-cart = Speed of go cart x Time travel
Distance travel by go-cart = 25 x 20
Distance travel by go-cart = 500 meter
A tabletop two-slit interference experiment consists of a laser pointer shining through two narrow, closely spaced slits in an opaque barrier. The light from the slits produces a characteristic pattern of alternating bright and dark regions on a nearby wall. Listed are five ways to modify the parameters of the experiment. Match each modification according to the effect it would have on the spacing between the bright spots on the wall. In each case, assume that everything else is held constant.
a. Causes the spacing between the bright spots on the wall to increase.
b. Causes the spacing between the bright spots on the wall to decrease.
c. Has effed on the spacing between the bright spots on the wall.
1. increasing the distance d between the two slits in the barrier
2. increasing the brightness of the laser light
3. increasing the distance L from the slits to the wall
4. increasing the distance x from the laser to the slits
5. increasing the wavelength of the laser light
Answer:
a - 3, 5
b - 1
c - 2, 4
Explanation:
a). The distance between fringes on the screen
[tex]$y = \frac{m \lambda L}{d}$[/tex]
where, λ is wavelength, m is order of pattern, L is distance between slit and screen, d is separation between slits.
Now we analyze that spacing between bright spots increases as increasing distance L from slits to wall and increasing the wavelength of the laser light, λ.
b). The spacing between the bright spots decreases as the distance between the two slits increases.
c). Has no effect on the spacing between the bright spots on the wall by increasing the distance from the laser to the slits as well as increasing the brightness of the laser.
A dog runs 51 m west to fetch a ball and brings it back only 27 m before stopping.
The total displacement of the dog is:
The potential energy of a mass-spring system when the spring is fully compressed and the mass is at rest is 200 J. After releasing the mass, assuming there is no dissipative force, the system will oscillate. At a point during the oscillation the potential energy of the system is 50 J. What is the kinetic energy of the mass at that point
Answer:
150 J
Explanation:
If there is not any dissipative force in the system, the mass and the spring will oscillate eternally., but of course, we assume this is a theoretical situation. The conservation of energy in a system implies that the sum of the potential energy plus kinetic energy remains constant, therefore if in the initial point the mass has 200 J (potential energy) and is at rest ( kinetic energy = 0) the overall energy at the beginning is 200 J. At any point of the oscillation if the potential energy is 50 J the kinetic energy must be 150 J.
What is the term used to describe the
force pushing on an area or surface?
A. density
B. pressure
C. surface tension
D. volume
A 2.0 kg breadbox on a fric-
tionless incline of angle u 40 is
connected, by a cord that runs over a
pulley, to a light spring of spring con-
stant k 120 N/m, as shown in
Fig. 8-43. The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless. (a) What is the speed of the box when it has moved 10 cm down the in- cline? (b) How far down the incline from its point of release does the box slide before momentarily stopping, and what are the (c) magnitude and (d) direction (up or down the incline) of the box’s acceleration at the instant the box momentarily stops?
The water pressure to an apartment is increased by the water company. The water enters the apartment through an entrance valve at the front of the apartment. Where will the increase in the static water pressure be greatest when no water is flowing in the system
Answer:
Option C
Explanation:
Options for the question are as follows -
A. At a faucet close to entrance valve
B. At a faucet away from the entrance valve
C. It will be the same at all faucets
D. There will be no increase in the pressure at the faucets
Solution -
The static force will be the same at all faucets and also the area of the faucets be same.
Thus, the pressure created at all faucets will be the same.
Thus, option C is correct
a disk of radius 10 cm speeds up from rest. it turns 60 radians reaching an angular velocity of 15 rad/s. what was the angular acceleration?
b. how long did it take the disk to reach this velocity?
Answer:
a) The angular acceleration is 1.875 radians per square second.
b) The time taken by the disk to reach the final angular speed is 8 seconds.
Explanation:
a) Let suppose that the disk accelerates uniformly, given that initial and final angular speed ([tex]\omega_{o}[/tex], [tex]\omega_{f}[/tex]), in radians per second, and change in angular position ([tex]\Delta \theta[/tex]), in radians, are known. The angular acceleration ([tex]\alpha[/tex]), in radians per square second, are found by using this expression:
[tex]\alpha = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \Delta \theta}[/tex] (1)
If we know that [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega_{f} = 15\,\frac{rad}{s}[/tex] and [tex]\Delta \theta = 60\,rad[/tex], then the angular acceleration of the disk is:
[tex]\alpha = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \Delta \theta}[/tex]
[tex]\alpha = 1.875\,\frac{rad}{s^{2}}[/tex]
The angular acceleration is 1.875 radians per square second.
b) The time taken by the disk to reach the final angular velocity is determined by the following kinematic formula:
[tex]t = \frac{\omega_{f}-\omega_{o}}{\alpha}[/tex] (2)
Where [tex]t[/tex] is the time, in seconds.
If we know that [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega_{f} = 15\,\frac{rad}{s}[/tex] and [tex]\alpha = 1.875\,\frac{rad}{s^{2}}[/tex], then the time taken by the disk is:
[tex]t = \frac{\omega_{f}-\omega_{o}}{\alpha}[/tex]
[tex]t = 8\,s[/tex]
The time taken by the disk to reach the final angular speed is 8 seconds.
A step-down transformer has 2500 turns on its primary and 5.0 x 10' tums on its secondary. If the potential difference across the primary is 4850 V, what is
the potential difference across the secondary?
Answer:
I dont know sorry
Explanation:
hehe
The weight of a column of air pushing
down over an area is called which of
Help Resources
these?
A. density
B. volume
C. mass
D. air
pressure

Answer:
air pressure
:::::::::::::::::
For the Earth, most of the information we've learned about its interior, including the mantle and core, comes from drilling down directly into it.
True
False
Which of the following is NOT something
the atmosphere does?
A. traps in warmth
B. keeps Earth in a spherical shape
C. provides oxygen to breathe
D. protects Earth from meteoroids
Answer:
I think the answer is B, keeps Earth in a spherical shape
A 10 kg box initially at rest is pulled with a 50 N horizontal force for 4 m across a level surface. The force of friction
acting on the box is a constant 20 N. How much work is done by the gravitational force?
A. 03
OB. 10 J
C. 100
D. 50 J
Answer:
B i think
Explanation:
...
Potential energy is best defined as which of the following?
A Mass energy
B Energy of Motion
C Stored Energy
D Energy of height
Answer: C. Stored energy
This photo shows a beam of light entering and exiting a piece of glass.
What happens when the light enters the glass?
A. The light is absorbed.
B. The light is reflected.
C. The light is scattered.
D. The light is refracted.
Answer:
D. The Light Is Refracted
Explanation:
<3
here is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted so that it can be rotated about its axis by turning a hand crank. The outer wall of the cylinder is perforated with small holes. You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is 10.7 cm. When the cylinder is rotating at 1.65 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall
Answer:
11.5 m/s²
Explanation:
The centripetal acceleration, a = rω² where r = radius of cylinder = 10.7 cm = 0.107 m and ω = angular speed = 2πN where N = number of revolutions per second = 1.65 rev/s
So, a = rω²
a = r(2πN)²
a = 4π²rN²
substituting the values of the variables into the equation, we have
a = 4π²rN²
a = 4π²(0.107 m)(1.65 rev/s)²
a = 4π²(0.107 m)(2.7225 rev²/s)²
a = 4π² × 0.2913075 mrev²/s)²
a = 11.5 m/s²
Help me please with both questions?
Answer:
question #1 is A
Question #2 is C
Explanation:
What are the five classes of objects that orbit the sun?
an elevator of mass 250kg is carrying two persons whose masses are 50kg and 100kg. if the forces exerted by the motor is 3000N. calculate the mass of the bodies in the elevator.... Taking g as 10m/s²
Explanation:
mass=force*acceleration
mass=3000*10
mass=30,000
The mass of the bodies in the elevator is 400 kg.
The acceleration of the elevator is 2.5 m/s².
What is acceleration?Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).
Given parameters:
Mass of the elevator: M = 250 kg.
Mass of two persons: m₁ = 50 kg and m₂ = 100 kg.
Force exerted by the motor: F = 3000N.
g = 10 m/s².
Let, the acceleration of the elevator = a.
the mass of the bodies in the elevator :m= 250 kg. + 50 kg +100 kg. = 400 kg.
Now, F = mg - ma
⇒ 3000 = 400×10 - 400a
⇒ a = 1000/400 = 2.5 m/s²
Hence, the acceleration of the elevator is 2.5 m/s².
Learn more about acceleration here:
brainly.com/question/12550364
#SPJ2
An airplane flies 1000 miles in 2 hours. What is its average speed in miles per hour?
Answer:
500km per hour
Explanation:
if in 2 hours the airplane flies 1000 km then 1000 divided by 2 is 500km per hour.
Easy question just don’t understand it please help.