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bearhunter [10]

3 years ago

7

Clyde and Marilyn are riding a roller coaster. During which section of the track is their potential energy converted to kinetic

energy?

1 answer:

Alex17521 [72]3 years ago

5 0

Potential energy is when the roller coaster rises. Kinetic energy is when the roller coaster declines.

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Can a body have zero velocity and finite acceleration?Explain

sergejj [24]

Answer:

Kinda? Depends what the question is fully asking

Explanation:

Acceleration is a change in velocity. So I guess if the velocity of something is -2 m/s and its positively accelerating at a value of +1 m/s, then that means every second its velocity changes by +1m/s.

So that -2 m/s thing after one second will be going -1 m/s.

After another second it'll be going 0 m/s.

After another itll be going +1 m/s and so on.

So at one point for a brief moment, it can have an acceleration but be at 0 m/s velocity.

5 0

3 years ago

A saddle weighs 250 newtons. The mass of the saddle is __________ kilograms. Use g = 9.8 N/kg for gravity.

Irina18 [472]

Answer:

saddle weighs 250 newtons. The mass of the saddle is ____250/9.8 kg______ kilograms. Use g = 9.8 N/kg for gravity.

7 0

2 years ago

Convert i) 1 microgram into kilogram<br>ii) 100 decameter into cm

Kobotan [32]

Answer:

1×10^-9kilogram

100 decameter =100000cm

3 0

2 years ago

Read 2 more answers

Can’t someone help me

Aleonysh [2.5K]

Answer:

A

Explanation:

Objects with the same charge (Ex. North and north) repel each other.

3 0

3 years ago

If a car is taveling with a speed 6 and comes to a curve in a flat road with radius ???? 13.5 m, what is the minumum value the c

Lemur [1.5K]

Answer:

$\mu_s \geq 0.27$

Explanation:

The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:

$f_s=\frac{mv^{2}}{R}$

But we know that:

$f_s\leq \mu_s N$

And the normal force is given by the sum of the forces in the vertical direction:

$N-mg=0 \implies N=mg$

Finally, we have:

$f_s=\frac{mv^{2}}{R} \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR} \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27$

So, the minimum value for the coefficient of friction is 0.27.

4 0

3 years ago