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2 years ago

9

You hit a hockey puck and it slides across the ice at nearly a constant speed.Is a force keeping it in motion?Explain.

1 answer:

Klio2033 [76]2 years ago

7 0

Answer:

Explanation:

When the puck is sliding on the ice, there is no force being exerted on the puck to keep it moving forward. Instead, inertia keeps the puck moving forward. Friction between the puck and the ice gradually slows the puck down. You hit a hockey puck and it slides across the ice at nearly a constant speed

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The speed of a 180-g toy car at the bottom of a vertical circular portion of track is 7.75 m/s. If the radius of curvature of th

Bezzdna [24]

Answer:

11.28 N toward the center of the track

Explanation:

Centripetal force: This is the force that tend to draw a body close to the center of a circle, during circular motion.

The formula for centripetal force is given as,

F = mv²/r................................ Equation 1

Where F = force, m = mass of the toy car, v = velocity, r = radius

Given: m = 108 g = 0.108 kg, v = 7.75 m/s, r = 57.5 cm = 0.575 m

Substitute into equation 1

F = 0.108(7.75²)/0.575

F = 11.28 N

Hence the magnitude and direction of the force = 11.28 N toward the center of the track

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2 years ago

What happens if you add additional,solid NaCl after the maximum has been reached?

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<span>it would bond to the phosphate

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3 years ago

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3 years ago

How to find new pressure of gas if bottle explodes

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3 years ago

The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on

alekssr [168]

Answer:

The change in temperature is $\Delta T = 1795 K$

Explanation:

From the question we are told that

The temperature coefficient is $\alpha = 4 * 10^{-3 }\ k^{-1 }$

The resistance of the filament is mathematically represented as

$R = R_o [1 + \alpha \Delta T]$

Where $R_o$ is the initial resistance

Making the change in temperature the subject of the formula

$\Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]$

Now from ohm law

$I = \frac{V}{R}$

This implies that current varies inversely with current so

$\frac{R}{R_o} = \frac{I_o}{I}$

Substituting this we have

$\Delta T = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]$

From the question we are told that

$I = \frac{I_o}{8}$

Substituting this we have

$\Delta T = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]$

=> $\Delta T = \frac{1}{3.9 * 10^{-3}} (8 -1 )$

$\Delta T = 1795 K$

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3 years ago