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Dafna1 [17]
3 years ago
5

What are the effect of sound energy in air

Physics
1 answer:
ser-zykov [4K]3 years ago
5 0

This will take me a little bit let me research :)

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Two identical carts are free to move along a straight frictionless track. At time t1, cart X is moving at 2.0 m/s when it collid
jarptica [38.1K]

Answer:

Explanation:

Using the law of conservation of momentum;

m_1 u_1+m_2u_2 = m_1v_1+m_2v_2

here;

There is a need for conservation of the total momentum that occurred before and after the collision.

So;

m_1 = mass of cart X

m_2 = mas 9f cart Y

u_1 = velocity of cart X (before collision)

u_2 = velocity of cart Y (before collision)

v_1 = velocity of cart X (after collision)

v_2 = velocity of cart Y (after collision)

So;

m(u_1+0) =(m_1v+m_2)v

because the mass is identical and v represents the velocity of both carts.

Now;

u_1 = 2 m/s

u_2 = 0 ( at rest)

∴

m(2) = (2m)v

v = 1 m/s

Thus, we can see from the graphical image attached below that the velocity of X reduces to 1 m/s after collision with cart Y.

8 0
3 years ago
Suppose a person sits on a skateboard with her feet up and throws a ball. Explain why she will move as a result of throwing the
Snowcat [4.5K]
I’m guessing is because she uses force to throw the ball, allowing the energy to move the person.


sorry if it’s not 100% correct
8 0
3 years ago
Gravitational pull between 2 objects is determined by which of the following?
vampirchik [111]
Both distance between and mass of the 2 objects.
3 0
3 years ago
Read 2 more answers
In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a
SOVA2 [1]

Answer:0.084 m/s^2

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance d_1=8.13 km

time taken=25 min =1500 s

initial speed v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s

after 8.13 km mark steve started to accelerate

speed after 60 s

v_2=v_1+at

v_2=5.6+a\times 60

distance traveled in 60 sec

d_2=v_1\times 60+\frac{a60^2}{2}

d_2=336+1800 a

time taken in last part of journey

t_3=1663.6-1560=103.6 s

distance traveled in this time

d_3=v_2\times t_3

d_3=\left ( 5.6+a\times 60\right )103.6

and total distance=d_1+d_2+d_3

10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6

1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6

a=0.084 m/s^2

5 0
3 years ago
10 turns of wire are closely wound around a pencil as shown in the figure. when measured using a scale as shown, the length of t
Mila [183]

Answer:

a. The thickness of the wire is 2.5 mm.

b. The wire is 0.25 cm thick.

Explanation:

Number of turns of the wire = 10

The length of total turns = 25 mm

a. The thickness of the wire can be determined by;

thickness of the wire = \frac{length of total turns}{number of turns}

                           = \frac{25}{10}

                           = 2.5 mm

Therefore, the wire is 2.5 mm thick.

b. To determine the thickness of the wire in centimetre;

10 mm = 1 cm

So that,

2.5 mm = x

x  = \frac{2.5}{10}

   = 0.25 cm

The wire is 0.25 cm thick.

8 0
3 years ago
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