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3 years ago

A baseball with a mass of 0.15 kg is moving at a speed of 40 m/s. What is

Physics

2 answers:

Afina-wow [57]3 years ago

6 0

Answer:

120 J

Explanation:

KE = mv²/2 = (0.15 kg * [40 m/s]²)/2 = 120 J

aev [14]3 years ago

3 0

Answer:

120 J

Explanation:

KE = 1/2 mv²

KE = 1/2 (0.15)(40)²

KE = 1/2 (0.15)(1600)

KE = 1/2 (240)

KE = 120 J

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A certain planet has an escape speed . If another planet has twice size and twice the mass of the first planet, its escape speed

solong [7]

I think the correct answer from the choices listed above is option A. A certain planet has an escape speed . If another planet has twice size and twice the mass of the first planet, its escape speed will be <span>Sqrt[2] V. Hope this answers the question.</span>

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3 years ago

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Two bullets have masses of 3.0 g and 6.0 g respectively. Both are fired with a speed of 40.0 m/s.(a) Which bullet has more kinet

Stolb23 [73]

Answer:

(a) The bullet with a mass of 6.0 g

(b) $\frac{K_2}{K_1}=2$

Explanation:

(a) Kinetic energy is defined as:

$K=\frac{mv^2}{2}$

So, we have:

$K_1=\frac{3*10^{-3}kg(40\frac{m}{s})^2}{2}\\K_1=2.4J$

$K_2=\frac{6*10^{-3}kg(40\frac{m}{s})^2}{2}\\K_2=4.8J$

The bullet with a mass of 6.0 g have more kinetic energy

(b) We have $m_2=2m_1(1)$ and $v_1=v_2(2)$ . So:

$K_1=\frac{m_1v_1^2}{2}\\K_2=\frac{m_2v_2^2}{2}\\\frac{K_2}{K_1}=\frac{\frac{m_2v_2^2}{2}}{\frac{m_1v_1^2}{2}}(3)$

Replacing (1) and (2) in (3):

$\frac{K_2}{K_1}=\frac{\frac{2m_1v_1^2}{2}}{\frac{m_1v_1^2}{2}}\\\frac{K_2}{K_1}=2$

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3 years ago

What is the force on a 1000 kg elevator that is free falling at 9.8 m/s/s?

Anton [14]

Answer:

Explanation:

Force = (mass) · (acceleration)

= (1,000 kg) · (9.8 m/s²)

= 9,800 newtons

4 0

3 years ago

A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used

zloy xaker [14]

Answer:

The compression is $\sqrt{2} \ d$ .

Explanation:

A Hooke's law spring compressed has a potential energy

$E_{potential} = \frac{1}{2} k (\Delta x)^2$

where k is the spring constant and $\Delta x$ the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

$E_{kinetic} = \frac{1}{2} m v^2$ .

So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity $v_1$ . Knowing that the energy is constant.

$\frac{1}{2} m v_1^2 = \frac{1}{2} k d^2$

If we want to double the kinetic energy, then, the knew kinetic energy for a obtained by compressing the spring a distance D, implies:

$2 * (\frac{1}{2} m v_1^2) = \frac{1}{2} k D^2$

But, in the left side we can use the previous equation to obtain:

$2 * (\frac{1}{2} k d^2) = \frac{1}{2} k D^2$

$D^2 = \frac{2 \ (\frac{1}{2} k d^2)}{\frac{1}{2} k}$

$D^2 = 2 \ d^2$

$D = \sqrt{2 \ d^2}$

$D = \sqrt{2} \ d$

And this is the compression we are looking for

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3 years ago

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Without inertia, how would an object that is experiencing a centripetal force behave?

nikitadnepr [17]

The correct option is this: IT WOULD MOVE IN A CURVED CIRCULAR PATH.
Objects that are travelling in circular paths change directions all the time as they move round the circle, but they are prevented from moving off in a straight line by centripetal force. The centripetal force keeps pulling the objects towards the center of the circle. <span />

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3 years ago

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