3 years ago

Can some help me pliz

Physics

1 answer:

muminat3 years ago

3 0

1). a

2). g

3). b

4). h

5). e

6). c

7). d

8). f

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Starting from rest, a 2.3x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts

Harman [31]

Answer:

3.13 m/s

Explanation:

From the question,

Since the flea spring started from rest,

Ek = W................... Equation 1

Where Ek = Kinetic Energy of the flea spring, W = work done on the flea spring.

But,

Ek = 1/2mv²............ Equation 2

Where m = mass of the flea spring, v = flea's speed when it leaves the ground.

substitute equation 2 into equation 1

1/2mv² = W.................... Equation 3

make v the subject of the equation

v = √(2W/m)................. Equation 4

Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg

Substitute into equation 4

v = √[2×3.6×10⁻⁴ )/2.3×10⁻⁴]

v = 7.2/2.3

v = 3.13 m/s

Hence the flea's speed when it leaves the ground = 3.13 m/s

4 0

3 years ago

A 70 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 590 m/s . Part A Part complete Wha

Kisachek [45]

Answer:

The recoil velocity is 0.354 m/s.

Explanation:

Given that,

Mass of hunter = 70 kg

Mass of bullet = 42 g = 0.042 kg

Speed of bullet = 590 m/s

We need to calculate the recoil speed of hunter

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Where, $m_{1}$ = mass of hunter

$m_{2}$ = mass of bullet

u = initial velocity

v = recoil velocity

Put the value in the equation

$0+0=70\times v_{1}+0.042\times590$

$v_{1}=-\dfrac{0.042\times590}{70}$

$v=-0.354\ m/s$

Hence, The recoil velocity is 0.354 m/s.

8 0

3 years ago

If you hadto throw one of the fourtoys above, which one would take the MOST force to throw

ohaa [14]

Answer:

that one

Explanation:

cause its heavier

7 0

3 years ago

Which of the following statements describes an interaction between the geosphere and atmosphere?

german

B is the answer, I’m really good at this subject

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3 years ago

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At ground level g is 9.8m/s^2. Suppose the earth started to increase its angular velocity. How long would a day be when people o

velikii [3]

Let s = rate of rotation
Let r = radius of earth = 6,400km 
Then solving (s^2) r = g will give the desired rate, from which length of day is inferred. 
People would not be thrown off. They would simply move eastward in a straight line while the curved surface of earth fell away from beneath them.

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4 years ago

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Can some help me pliz ​

Can some help me pliz