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storchak [24]
3 years ago
13

A student drops a ball from the top of a 10-meter tall building. The ball leaves the thrower's hand with a zero speed. What is t

he speed of the ball at the moment just before it hits the ground?
Physics
1 answer:
Sergio [31]3 years ago
7 0

Answer:

14 m/s

Explanation:

u = 0, h = 10 m, g = 9.8 m/s^2

Use third equation of motion

v^2 = u^2 + 2 g h

Here, v be the velocity of ball as it just strikes with the ground

v^2 = 0 + 2 x 9.8 x 10

v^2 = 196

v = 14 m/s

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A car is traveling at 30 m/s in a straight line. The drive applies the brakes for 3 seconds and the car slows down to 12 m/s. Wh
jeyben [28]

Answer:

The answer to your question is: -6m/s²

Explanation:

Data

vo = 30 m/s

vf = 12 m/s

t = 3 s

a = ?

Formula

vf = vo + at

a = (vt - vo) / t

Process

                a = (12 - 30) / 3                substitution

                a = -18 / 3                         simplify

                a = -6 m/s²                        result, is negative because the car

                                                          is slowing down.

3 0
3 years ago
Is diet Pepsi heterogeneous or is it homogeneous
Lana71 [14]
Heterogeneous because it has less amount of sugar than the regular Pepsi has
6 0
3 years ago
Read 2 more answers
A line of charge starts at x = +x0 and extends to positive infinity. The linear charge density is λ = λ0x0/x. Determine the elec
kari74 [83]

Explanation:

it is given that, the linear charge density of a charge, \lambda=\dfrac{\lambda_ox_o}{x}

Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :

dE=\dfrac{k\ dq}{x^2}..........(1)

The linear charge density is given by :

\lambda=\dfrac{dq}{dx}

dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx

Integrating equation (1) from x = x₀ to x = infinity

E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx

E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}

E=\dfrac{k\lambda_o}{2x_o}

Hence, this is the required solution.

5 0
3 years ago
Read 2 more answers
Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?
Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

f = focal length

Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0
3 years ago
a feather is dropped on the moon from a height of 1.40meters. the acceleration of gravity on the moon is 1.67m/s^2. determine th
kykrilka [37]

Answer:

1min since there is no gravity on the moon so it will take time to drop on the moon.

Explanation:

6 0
3 years ago
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