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3 years ago

Here’s the picture of the problem

Physics

1 answer:

Dennis_Churaev [7]3 years ago

6 0

Answer:C is the answer

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True or False: <br><br> Solids always<br> have a higher density than<br> liquids and gases.

Hunter-Best [27]

Answer:

<h3>False</h3>

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4 years ago

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An Earth satellite is in a circular orbit at an altitude of 500 km.

Dafna1 [17]

Answer:

Speed Unchanged.

Explanation:

As work is defined as a product of force over a distance. If the distance in altitude is constant = 500km, there's 0 change in distance and force, no work would be done by the gravitational force.

Since potential energy of the satellite is unchanged, unless there's additional internal energy source, the kinetic energy would remain unchanged, so would its speed.

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3 years ago

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sertanlavr [38]

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3 years ago

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A cabbie is trying to stop when he notices a fare is whistling them over. The

liberstina [14]

K.E=18750J
Mass=m=2100kg
Velocity=v

$\boxed{\sf K.E=\dfrac{1}{2}mv^2}$

$\\ \sf\longmapsto 18750=\dfrac{1}{2}2100v^2$

$\\ \sf\longmapsto 18750=1050v^2$

$\\ \sf\longmapsto v^2=\dfrac{18750}{1050}$

$\\ \sf\longmapsto v^2=17.85m^2$

$\\ \sf\longmapsto v=\sqrt{17.85}$

$\\ \sf\longmapsto v=4.1m/s$

7 0

3 years ago

A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho

fiasKO [112]

Answer:

$v_f=8.17\frac{m}{s}$

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

$W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m$

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

$W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$

The box is initially at rest, so $v_i=0$ . Solving for $v_f$ :

$v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}$

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3 years ago