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3 years ago

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An athlete does one push-up period in the process, she moves half of her body weight, 250 N, a distance of 20 cm, this distance

is the distance her gravity moves when she fully extend her arms how much work did you do after one push-up

1 answer:

ioda3 years ago

3 0

In the push-up, the athlete's weight (or force) of 250 N moves through a distance of 20 cm.

By definition,
Work = force * distance.

Therefore, work done in one push-up is
$W=(250 \, N)*(20 \, cm)*(10^{-2} \, \frac{cm}{m}) = 50 \, J$

Note that 1 J = 1 N-m.

Answer: 1 J

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A 77.0−kg short-track ice skater is racing at a speed of 12.6 m/s when he falls down and slides across the ice into a padded wal

sukhopar [10]

Answer:

-6112.26 J

Explanation:

The initial kinetic energy, $KE_i$ is given by

$KE_i=0.5mv_1^{2$ } where m is the mass of a body and $v_i$ is the initial velocity

The final kinetic energy, $KE_f$ is given by

$KE_f=0.5mv_f^{2}$ where $v_f$ is the final velocity

Change in kinetic energy, $\triangle KE$ is given by

$\triangle KE=KE_f-KE_i=0.5mv_f^{2}-0.5mv_1^{2}=0.5m(v_f^{2}-v_i^{2})$

Since the skater finally comes to rest, the final velocity is zero. Substituting 0 for $v_f$ and 12.6 m/s for $v_i$ and 77 Kg for m we obtain

$\triangle KE=0.5*77*0^{2}-0.5*77*(0^{2}-12.6^{2})=-6112.26 J$

From work energy theorem, work done by a force is equal to the change in kinetic energy hence for this case work done equals <u>-6112.26 J</u>

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2 years ago

Is N2 a triple bond

krek1111 [17]

Yes, N2 is a triple bond.
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3 years ago

A box has a weight of 150 N and is being pulled across a horizontal floor by a force that has a magnitude of 110 N. The pulling

ivann1987 [24]

Answer:

42.99°

Explanation:

$F_h$ = Kinetic friction force

$F_{\theta}$ = Pulling force at angle $\theta$

$N_h$ = Weight of the box = 150 N

Kinetic friction force

$F_h=\muN_h$

Pulling force at angle $\theta$

$F_{\theta}=\muN_{\theta}$

N = Pulling force

According to question

$\frac{F_h}{F_{\theta}}=\frac{2}{1}\\\Rightarrow \frac{\muN_h}{\muN_{\theta}}=2\\\Rightarrow \frac{N_h}{N_{\theta}}=2\\\Rightarrow N_{\theta}=\frac{N_h}{2}\\\Rightarrow N_{\theta}=\frac{150}{2}\\\Rightarrow N_{\theta}=75\ N$

Applying Newton's second law in the vertical direction we get

$N_h-Nsin\theta=N_{\theta}\\\Rightarrow 150-110sin\theta=75\\\Rightarrow \theta=sin^{-1}\frac{75}{110}\\\Rightarrow \theta=42.99\ ^{\circ}$

The angle is 42.99°

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3 years ago

A basketball player passes a ball to a teammate at a velocity of 6 m/s. The ball has a mass of 0.51 kg. If the original player h

Vilka [71]

We can solve the problem by using conservation of momentum.

The player + ball system is an isolated system (there is no net force on it), therefore the total momentum must be conserved. Assuming the player is initially at rest with the ball, the total initial momentum is zero:

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The total final momentum is:

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where $p_p$ is the momentum of the player and $p_b$ is the momentum of the ball.

The momentum of the ball is: $p_b = mv=(0.51 kg)(6 m/s)=3.06 kg m/s$

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so we can write

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and we find

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and the negative sign means that it is in the opposite direction of the ball.

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