An object moves 120 m in 15 seconds. Calculate the object's speed.

Dams hold collected water back from flowing at their normal rate. What type of energy is in the collected water above the dam? c

Andreyy89

It would be mechanical potential energy.

8 0

2 years ago

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A 1.70 m tall woman stands 5.00 m in front of a camera with a 50.00 cm focal

slamgirl [31]

Answer:

18.89cm

Explanation:

As we know that the person is standing 5m in front of the camera

$d_0=5m=500cm$

The focal length of the lens =50cm

f=50 cm

By Lens formula we have:

$\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm$

By the formula of magnification

$\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm$

The height of the image formed is 18.89cm.

5 0

2 years ago

A particle moves according to the equation x = 11t^2, where x is in meters and t is in seconds.

Savatey [412]

We are given the equation:

x = 11t^2

We use that equation to calculate for the distance traveled.
For (a)

At t=2.20 sec,
x =53.24 meters

At t=2.95 sec,
x =95.73 meters

Velocity = (95.73 meters - 53.24 meters) / (2.95 s - 2.20 s ) = 56.65 m/s

For (b)

At t=2.20 sec,
x =53.24 meters

At t=2.40 sec,
x =63.36 meters

Velocity = (63.36 meters - 53.24 meters) / (2.40 s - 2.20 s ) = 50.6 m/s

4 0

2 years ago

A person fires a 38 gram bullet straight up into the air. It rises, then falls straight back down, striking the ground with a sp

Y_Kistochka [10]

Answer:

Force exerted = 25.41 kN

Explanation:

We have equation of motion

v² = u²+2as

u = 345 m/s, s = 8.9 cm = 0.089 m, v = 0 m/s

0² = 345²+2 x a x 0.089

a = -668679.78 m/s²

Force exerted = Mass x Acceleration

Mass of bullet = 38 g = 0.038 kg

Acceleration = 668679.78 m/s²

Force exerted = 25409.83 N = 25.41 kN

4 0

3 years ago

An eagle is flying horizontally at a speed of 3.10 m/s when the fish in her talons wiggles loose and falls into the lake 6.10 m

Alinara [238K]

Answer:

10.93m/s with the assumption that the water in the lake is still (the water has a speed of zero)

Explanation:

The velocity of the fish relative to the water when it hits the water surface is equal to the resultant velocity between the fish and the water when it hits it.

The fish drops on the water surface vertically with a vertical velocity v. Nothing was said about the velocity of the water, hence we can safely assume that the velocity if the water in the lake is zero, meaning that it is still. Therefore the relative velocity becomes equal to the velocity v with which the fish strikes the water surface.

We use the first equation of motion for a free-falling body to obtain v as follows;

v = u + gt....................(1)

where g is acceleration due to gravity taken as 9.8m/s/s

It should also be noted that the horizontal and vertical components of the motion are independent of each other, hence we take u = 0 as the fish falls vertically.

To obtain t, we use the second equation of motion as stated;

$h=ut+gt^2/2.................(2)$