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3 years ago

9

PLEASE HELP ME QUICK

1 answer:

OverLord2011 [107]3 years ago

8 0

I really don’t get this give me a second

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Friction is a force in which two objects __________.

balu736 [363]

The answer should be C) Slide against each other.

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3 years ago

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When observing an object, you determine it is moving at a constant velocity and in a straight line. Which of these BEST describe

Marina86 [1]

The answer would be C. Hope this helps XD

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3 years ago

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QUICK ITS DUE IN 6 MINUTES. create a scenario that applies the 3 laws of motion. Explain in complete sentences how each law if d

yKpoI14uk [10]

Answer:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

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2 years ago

A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in

dexar [7]

Answer:

K = 0.076 J

Explanation:

The height of the target, h = 0.860 m

The mass of the steel ball, m = 0.0120 kg

Distance moved, d = 1.50 m

We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.

$h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}}$

Put all the values,

$t=\sqrt{\dfrac{2\times 0.860 }{9.8}} \\\\=0.418\ s$

The velocity of the ball is :

$v=\dfrac{1.5}{0.418}\\\\= $$3.58\ m/s$

The kinetic energy of the ball is :

$K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 0.0120\times 3.58^2\\\\=0.076\ J$

So, the required kinetic energy is 0.076 J.

6 0

3 years ago

One cycle of the power dissipated by a resistor ( R = 800 Ω R=800 Ω) is given by P ( t ) = 60 W , 0 ≤ t < 5.0 s P(t)=60 W, 0≤

OLga [1]

Answer:

42.5W

Explanation:

To solve this problem we must go back to the calculations of a weighted average based on the time elapsed thus,

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

We need to calculate the average power dissipated by the 800\Omega resistor.

Our values are given by:

$P(t)=60 W, 0\leq t$

$P(t)=25 W, 5.0\leq t$

Aplying the values to the equation we have:

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

$Power_{avg} = \frac{60(5-0)+25(10-5)}{(5-0)+(10-5)}$

$Power_{avg} = 42.5W$

5 0

3 years ago