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Dafna11 [192]
3 years ago
14

A baseball is traveling (+30m/s) and is hit by a bat. It leaves the bat traveling (−40m/s). What is the change in its velocity?

Physics
1 answer:
Crazy boy [7]3 years ago
5 0

Answer:

Explanation:

The change must be 30 - - 40 which means it came in a 30 meters / second and went out in the opposite direction at 40 meters / second

The change is 70 m/sec.

You could show it to be - 70 meters per second as well. That's done by making the outgoing direction minus.

Delta v = vf - vi.

Now it depends on which way you define vf and vi.

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What is the density of a block of marble that
EleoNora [17]

Explanation:

Solution,

Volume (v)=287 cm^3

Mass(m)=816 g

Density(d)=m/v

=816/287

=2.84

So, the density of the block of marbles is 2.84 g/cm^3.

I hope it helped U

stay safe stay happy

5 0
3 years ago
Air escaping out from an air hose at a gas station always feels cool. Why? ​
kati45 [8]

Answer:

The air is contained at a high pressure in the tube. When it escapes from a small orifice, it suddenly expands. A large amount of its heat is absorbed in the process of expansion resulting in considerable fall in its temperature. This is why the escaping air feels cold.

8 0
2 years ago
Two simple pendulums are in two different places. The length of the second pendulum is 0.4 times the length of the first pendulu
faltersainse [42]

Answer:

\sqrt{\frac{4}{9}}

Explanation:

The frequency of a simple pendulum is given by:

f=\frac{1}{2\pi}\sqrt{\frac{g}{L}}

where

g is the acceleration of gravity

L is the length of the pendulum

Calling L_1 the length of the first pendulum and g_1 the acceleration of gravity at the location of the first pendulum, the frequency of the first pendulum is

f_1=\frac{1}{2\pi}\sqrt{\frac{g_1}{L_1}}

The length of the second pendulum is 0.4 times the length of the first pendulum, so

L_2 = 0.4 L_1

while the acceleration of gravity experienced by the second pendulum is 0.9 times the acceleration of gravity experienced by the first pendulum, so

g_2 = 0.9 g_1

So the frequency of the second pendulum is

f_2=\frac{1}{2\pi}\sqrt{\frac{g_2}{L_2}}=\frac{1}{2\pi} \sqrt{\frac{0.9 g_1}{0.4 L_1}}

Therefore the ratio between the two frequencies is

\frac{f_1}{f_2}=\frac{\frac{1}{2\pi}\sqrt{\frac{g_1}{L_1}}}{\frac{1}{2\pi} \sqrt{\frac{0.9 g_1}{0.4 L_1}}}=\sqrt{\frac{0.4}{0.9}}=\sqrt{\frac{4}{9}}

8 0
3 years ago
10 points
grandymaker [24]
The answer is B tell me if I am wrong.
6 0
3 years ago
A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly abov
Rudik [331]

Answer:

Apparent depth (Da) = 60.15 cm (Approx)

Explanation:

Given:

Distance from fish (D) = 80 cm

Find:

Apparent depth (Da)

Computation:

We know that,

Refractive index of water (n2) = 1.33

So,

Apparent depth (Da) = D(n1/n2)

Apparent depth (Da) = 80 (1/1.33)

Apparent depth (Da) = 60.15 cm (Approx)

5 0
3 years ago
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