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3 years ago

6

How fast must a 1.88-9 ping-pong ball move (in m/s) in order to have the same kinetic energy as a 230.5 g baseball moving

1 answer:

Andrew [12]3 years ago

7 0

Explanation:

KE = KE

½ mv² = ½ mv²

½ (1.88 g) v² = ½ (230.5 g) (58.5 m/s)²

v = 648 m/s

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A comet is traveling through space with speed 3.01 ✕ 104 m/s when it encounters an asteroid that was at rest. The comet and the

Tcecarenko [31]

Answer: $8.493(10)^{-3} m/s$

Explanation:

According to the conservation of linear momentum principle, the initial momentum $p_{i}$ (before the collision) must be equal to the final momentum $p_{f}$ (after the collision):

$p_{i}=p_{f}$ (1)

In addition, the initial momentum is:

$p_{i}=m_{1}V_{1}+m_{2}V_{2}$ (2)

Where:

$m_{1}=1.71(10)^{14} kg$ is the mass of the comet

$m_{2}=6.06(10)^{20} kg$ is the mass of the asteroid

$V_{1}=3.01(10)^{4} m/s$ is the velocity of the comet, which is positive

$V_{2}=0 m/s$ is the velocity of the asteroid, since it is at rest

And the final momentum is:

$p_{f}=(m_{1}+m_{2})V_{f}$ (3)

Where:

$V_{f}$ is the final velocity

Then :

$m_{1}V_{1}+m_{2}V_{2}=(m_{1}+m_{2})V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1}V_{1}}{m_{1}+m_{2}}$ (5)

$V_{f}=\frac{(1.71(10)^{14} kg)(3.01(10)^{4} m/s)}{1.71(10)^{14} kg+6.06(10)^{20} kg}$

Finally:

$V_{f}=8.493(10)^{-3} m/s$ This is the final velocity, which is also in the positive direction.

8 0

4 years ago

Question:<br>Define Force

ArbitrLikvidat [17]

<em>Force</em><em> </em><em>is</em><em> </em><em>push</em><em> </em><em>or</em><em> </em><em>pull</em><em>.</em><em>.</em><em>.</em><em>.</em>

<em>thank</em><em> </em><em>you</em>

7 0

3 years ago

The Achilles tendon connects the muscles in your calf to the back of your foot. When you are sprinting, your Achilles ten-don al

lisov135 [29]

<em>A) We know that the gravitational Weight of a Body is given by,</em>

$F=8mg$

The expression of young's modulus is given by,

$Y=\frac{Fl}{A\Delta l}$

Replacing the value of the force in the equation of young's modulis we have,

$Y = \frac{8mg*l}{A\Delta l}$

Re-arrange the equation for the rate of l, we have,

$\Delta l = \frac{8mgl}{AY}$

Our values here are giving by,

$m=70kg$

$g=9.8m/s^2$

$l=15*10^{-2}m$

$A=110*10^6m^2$

$Y = 1.5*10^{10}N/m^2$

Replacing in the previous equation we have that,

$\Delta l = \frac{8(70)(9.8)(15*10^{-2})}{110*10^6*1.5*{10}}$

$\Delta l = 4.98*10^{-3}m$

We note that Achilles tendon stretch around 5mm

<em>B) The strain is given by the equation,</em>

$\epsilon = \frac{\Delta l}{l}$

The fraction of the tendon's lenght is the ratio of the change in lenght to the streched lenght, that is

$\alpha = \frac{\Delta l}{l}$

$\alpha = \frac{5*10^{-3}}{15*10^{-2}}$

$\alpha = 0.033$

The fraction of the tendon's lenght is 0.0333

3 0

4 years ago

A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T

slamgirl [31]

Answer:

$u=14.48m/s$

Explanation:

From the question we are told that:

Height of window $h=2m$

Height of window off the ground $h_g=7.5m$

Time to fall and drop $t=1.3s$

Generally the Newton's equation motion is mathematically given by

$s=ut+\frac{1}{2}at^2$

Where

$h=ut+\frac{1}{2}at^2$

$2=u1.3-\frac{1}{2}*9.8*1.3^2$

$2=u1.3-8.281$

$u=7.91m/s^2$

Generally the Newton's equation motion is mathematically given by

$2as=v^2-u^2$

Where

$-2gh_g=v^2-u^2$

$-2*9.8*7.5=(7.91)^2-u^2$

$-147=62.5681-u^2$

$u=\sqrt{209.5681}$

$u=14.48m/s$

Therefore the ball’s initial speed

$u=14.48m/s$

8 0

3 years ago

By which method does heat travel from the sun to Earth? A. Conduction B. Convection C. Evaporation D. Radiation

qaws [65]

Answer:

radiation

Explanation:

7 0

3 years ago

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