Which of the following terms best describes air

On the moon, which objects would fall with the same acceleration?

I am Lyosha [343]

On a planet/celestial body with gravity, the acceleration of gravity is the same for everything. Therefore, all three options are correct.

Hope this helps!

6 0

4 years ago

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What advantage does a modern camera offer a pinhole camera

AleksAgata [21]

Answer:

Done

Explanation:

The differences are significant. The main disadvantages of a pinhole camera compared to a modern camera are:

Proper exposure requires a trial and error approach.

Requires manual control of shutter speed (difficult to control)

Poor image quality (due to diffraction effects).

Long exposures (lead to blurry images if the subject moves).

Fixed aperture or f-stop.

Advantages include:

Large depth of field.

Inexpensive

Robust and nearly 100% reliable.

If used properly can create artistic effects.

4 0

4 years ago

A ring is attached at the center of the underside of a trampoline. A sneaky teenager crawls under the trampoline and uses the ri

IRINA_888 [86]

Answer:

E = 301.5 J

Explanation:

We have,

Mass of mother, m = 67 kg

Here, a sneaky teenager crawls under the trampoline and uses the ring to pull the trampoline slowly down. As she passes through the position at which she was before her son stretched the trampoline, her speed is 3 m/s.

It is required to find the elastic potential energy the son add to the trampoline by pulling it down. It is based on the conservation of energy.

The elastic potential energy of the mother = the elastic potential energy the son add to the trampoline.

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

So, the elastic potential energy is :

$E=\dfrac{1}{2}mv^2\\\\E=\dfrac{1}{2}\times 67\times 3^2\\\\E=301.5\ J$

So, the elastic potential energy of 301.5 J the son add to the trampoline by pulling it down.

6 0

4 years ago

Two 2.3 kg bodies, A and B, collide. The velocities before the collision are and . After the collision, . What are (a) the x-com

myrzilka [38]

Missing details in the question:

The velocities before collision are

$u_A=(40i+49j) m/s$

$u_B=(35i+11j) m/s$

After the collision:

$v_A=(14i+21j) m/s$

(a) 61 m/s

We can solve the problem by simply treating separately the x- and the y-components of the motion.

Here we want to analzye the motion along x. We have:

$u_A = 40 m/s$ is the initial velocity of A along the x-direction

$u_B = 35 m/s$ is the initial velocity of B along the x-direction

$v_A = 14 m/s$ is the final velocity of A along the x-direction

$v_B = ?$ is the final velocity of B along the x-direction

Since the total momentum along the x-direction must be conserved, we can write

$mu_A + mu_B = mv_A + mv_B$

where

m = 2.3 kg is the mass of the two bodies. Since the mass is the same, we can eliminate it from the equation,

$u_A + u_B = v_A + v_B$

And so, we find the final velocity of B along the x-direction:

$v_B = u_A + u_B - v_A=40+35-14=61 m/s$

(b) 39 m/s

Similarly to what we did in part a), here we analyze the conservation of momentum along the y-direction.

We have:

$u_A = 49 m/s$ is the initial velocity of A along the y-direction

$u_B = 11 m/s$ is the initial velocity of B along the y-direction

$v_A = 21 m/s$ is the final velocity of A along the y-direction

$v_B = ?$ is the final velocity of B along the y-direction

Since the total momentum along the y-direction must be conserved, we can write

$mu_A + mu_B = mv_A + mv_B$

Since the mass is the same, we can eliminate it from the equation,

$u_A + u_B = v_A + v_B$

And so, we find the final velocity of B along the y-direction:

$v_B = u_A + u_B - v_A=49+11-21=39 m/s$

c) +615 J

Here we have to find the total kinetic energy before and after the collision first.

First, we have to find the speed of each object before and after the collision. We have:

$u_A = \sqrt{40^2+49^2}=63.2 m/s\\u_B = \sqrt{35^2+11^2}=36.7 m/s\\v_A = \sqrt{14^2+21^2}=25.2 m/s\\v_B = \sqrt{61^2+39^2}=72.4 m/s$

So, the total kinetic energy before the collision was

$K_i = \frac{1}{2}mu_A^2+\frac{1}{2}mu_B^2 = \frac{1}{2}(2.3)(63.2)^2+\frac{1}{2}(2.3)(36.7)^2=6143 J$

While after the collision

$K_f = \frac{1}{2}mv_A^2+\frac{1}{2}mv_B^2 = \frac{1}{2}(2.3)(25.2)^2+\frac{1}{2}(2.3)(72.4)^2=6758 J$

So, the change in kinetic energy is

$\Delta K = K_f - K_i = 6758-6143 = +615 J$

(note that the system cannot gain kinetic energy in the collision, unless there is an external force acting on it)

3 0

3 years ago

Write 3.5 seconds as milliseconds

Phantasy [73]

3.5 seconds is 3500 milliseconds.

5 0

3 years ago

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