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3 years ago

Someone answer these questions please???

Physics

2 answers:

chubhunter [2.5K]3 years ago

8 0

Yeah sure someone else in answering rn

Fofino [41]3 years ago

4 0

Answer:

a. same

b. less

c. less

Explanation:

you can use the position, time, acceleration equation

xf=xi + vi*t+(at^2)/2

you can say the final distance will greater for X than Y

using acceleration for gravity a=9.81m/s^2

a. both X and Y will have the same acceleration due to gravity

b. to solve for final speed vf=vi+gt

since the initial velocity is 0 and X will take longer to fall, Y will hit the ground with less speed

c. as mentioned before, X will take longer to fall; therefore Y will have less time of descent.

You might be interested in

The formula to calculate velocity is

SashulF [63]

speed = distance/time

7 0

3 years ago

A police officer uses a radar gun to determine the speed of a car. A specialized radar gun uses ultraviolet light to determine s

lakkis [162]

Answer:

Radio Waves

Explanation:

Radio wave is a type of electromagnetic radiation notorious for their use in technologies used for communication like air traffic control, mobile phones, televisions and remote-controlled toys. Radio waves have the longest wavelength in the electromagnetic spectrum and are easily transmitted through air.

A radar gun uses radio waves to tell how far away an object is. To do that, the radar tool emits a focused radio wave and listens for any echo. If there is an item, say a car within the route of the radio wave, it's going to replicate some of the electromagnetic energy, and the radio wave will return to the radar gun. Radio waves flow via the air at a steady pace, so the radar gun can calculate how far the object is based totally on how long it takes the radio signal to return.

5 0

3 years ago

camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Usin

son4ous [18]

The question is incomplete. Here is the complete question.

The image below was taken with a camera that can shoot anywhere between one and two frames per second. A continuous series of photos was combined for this image, so the cars you see are in fact the same car, but photographed at differene times.

Let's assume that the camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Using this information and the photo itself, approximately how fast did the car drive?

Answer: v = 6.5 m/s

Explanation: The question asks for velocity of the car. Velocity is given by:

$v=\frac{\Delta x}{\Delta t}$

The camera took 7 pictures of the car and knowing its length is 5.3, the car's displacement was:

Δx = 7(5.3)

Δx = 37.1 m

The camera delivers 1.3 frames per second and it was taken 7 photos, so time the car drove was:

1.3 frames = 1 s

7 frames = Δt

Δt = 5.4 s

Then, the car was driving:

$v=\frac{37.1}{5.4}$

v = 6.87 m/s

The car drove at, approximately, a velocity of 6.87 m/s

7 0

3 years ago

A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mas

konstantin123 [22]

Answer:

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

Explanation:

The expression for conservation of the angular momentum (L) is

$L_{i} = L_{f} I_{i}\times\omega_{i} = I_{f}\times\omega_{f}$

Where

$I_{i}\ and \ \omega_{i}$ initial moment of inertia and angular velocity

$I_{f}\ and \ \omega_{f}$ is the final moment of inertia and angular velocity

The expression of moment of inertia of the satellite (a solid sphere) is

$I_{i} = \frac{2}{5}m_{s}r^{2}$

Where $m_{s}$ is the satellite mass

r is the radus of the sphere

Substititute 1900kg for m and 4.6m for r

$I_{i} = \frac{2}{5}m_{s}r^{2}\\\\ = \frac{2}{5}\times1900 kg\times (4.6 m)^{2} \\\\= 1.61 \cdot 10^{4} kgm^{2}$

The final moment of inertia of the satellite about the centre of mass

$I_{f} = I_{i} + 2\timesI_{x} \\\\= 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}m_{x}l^{2}$

Where $m_{x}$ is the antenna's mass and

I is the length of the antenna

$I_{f} = 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}150.0 kg\times(6.6 m)^{2} \\\\= 2.05 \cdot 10^{4} kgm^{2}$

So, the Final rotation rate of the satellite is:

$I_{i}\times\omega_{i} = I_{f}\times\omega_{f} \\\\\omega_{f} = \frac{I_{i}\times\omega_{i}}{I_{f}} \\\\= \frac{1.61 \cdot 10^{4} kgm^{2}\times8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kgm^{2}} \\\\= 6.3 rev/s$

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

5 0

3 years ago

Coil the wire around the nail. Connect one end of the wire to the positive battery terminal. Connect the other end of the wire t

Andrej [43]

Static i think
not sure
hope it helps

5 0

3 years ago