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2 years ago

What is a conclusion?

Physics

2 answers:

Vikki [24]2 years ago

8 0

The end or finish of an event or process.

sleet_krkn [62]2 years ago

4 0

A conclusion is when lets say you read a story and they say whats the conclusion its like whats it about and what did you learn if it was a actual question with answer just show me a picture

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A 2.964 kilogram truck strikes a fence at 7.00 meters/second and comes to

Makovka662 [10]

10.92N

Explanation:

Given parameters:

Mass of truck = 2.964kg

Velocity of truck = 7m/s

Time taken = 1.9s

Unknown:

Average force on the car = ?

Solution:

According to newton's third law of motion "action and reaction are equal and opposite".

The force with which the truck struck the fence is the same as the force the fence acted on the truck with but in another direction.

From newton's second law:

Force = mass x acceleration

We know that acceleration is the change in velocity with time;

acceleration = $\frac{change in velocity }{time }$

Force = mass x $\frac{change in velocity }{time }$

Force = $\frac{2.964 x 7}{1.9}$ = 10.92N

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

3 0

2 years ago

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

2 years ago

In any energy transformation, there is always some energy that gets wasted as non-useful heat.

Nady [450]

It is a completely false statement that in <span>any energy transformation, there is always some energy that gets wasted as non-useful heat. The correct option among the two options that are given in the question is the second option. I hope that this is the answer that has actually come to your desired help.</span>

8 0

3 years ago

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When the distance between two interacting objects doubles, the gravitational force is

Umnica [9.8K]

The gravitational force will be one quarter.

The gravitational force between two objects is given by the formula

F=GMm/r^2

here, r is the distance between the objects.

Thus the gravitational force is inversely proportional to the square of the distance between the objects, Therefore if the distance between two objects is doubled the force will be one quarter.

5 0

2 years ago

A girl is standing still in the middle of a baseball field. The girl catches the baseball and begins to move in the same directi

marusya05 [52]

A. inelastic, since the girl moves in the same direction as the thrown ball

5 0

2 years ago

Read 2 more answers