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3 years ago

A ball is dropped from 8.5 meters above the ground. If it begins at rest, how long does it take to hit the ground?

Physics

1 answer:

Anna35 [415]3 years ago

6 0

Answer:

Explanation:

Givens

d = 8.5 meters

vi = 0

a = 9.81

t = ?

Formula

d = vi * t + 1/2 a t^2

Solution

8.5 = 0 + 1/2 9.81 * t^2 multiply both sides by 2

8.5 = 4.095 t^2 Divide both sides by 4.095

8.5/4.095 = t^2

1.7329 = t^2 Take the square root of both sides

t = 1.316

It takes 1.316 seconds to hit the ground.

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3 years ago

An electron moves through a uniform electric field vector E = (2.80î + 5.20ĵ) V/m and a uniform magnetic field vector B = 0.400k

alina1380 [7]

Answer:

1.758820×10^11(-2.5i-0.8j) m/s^2

Explanation:

From the question, the parameters given are; E=(2.80i+ 5.20j) v/m, a uniform magnetic field,B= 0.400K T, acceleration, a= ??? and velocity vector, v= 11.0i metre per seconds (m/s)...

We can solve this problem using the formula below;

Ma= q[E+V × B] ---------------(1).

Note: q is negative, m= mass of electron.

Making acceleration,a the subject of the formula and substituting the parameters into equation (1);

a= -e/m × (2.5i + 5.2j +11.0i × 0.400K)

a= -e/m × (2.5i+5.2j-4.4j)

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Therefore, slotting in the value of charge to mass(e/m) ratio;

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7 0

3 years ago

A 38.2 kg wagon is towed up a hill inclined at 17.5 ◦ with respect to the horizontal. The tow rope is parallel to the incline an

Tema [17]

Answer:

v = 8.57 m/s

Explanation:

As we know that the wagon is pulled up by string system

So the net force on the wagon along the inclined is due to tension in the rope and component of weight along the inclined plane

So as per work energy theorem we know that

work done by tension force + work done by force of gravity = change in kinetic energy

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so we have

$F_t = 129 N$

$\theta = 17.5^o$

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so now we have

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$v = 8.57 m/s$

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tamaranim1 [39]

True ..........................

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