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3 years ago

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1.A student pushes a 20.0 kg mass 10.0 m across a floor with a horizontal force of 80.0 N. Calculate the amount of work that the

student docs.
Work = Force x Distance

W = F x d

1 answer:

Alina [70]3 years ago

7 0

W = (10.0 m)(80.0 N)
W = 800 J

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A projectile is launched vertically from the surface of the Moon with an initial speed of 1360 m/s. At what altitude is the proj

LenaWriter [7]

Answer:

485520 m

Explanation:

$v_{o}$ = initial velocity of the projectile = 1360 m/s

$v_{f}$ = final velocity of the projectile = $\left ( \frac{2}{5} \right )v_{_{o}}$ = $\left ( \frac{2}{5} \right )(1360)$ = 544 m/s

a = acceleraton due to gravity on moon = - 1.6 m/s²

h = Altitude of the projectile

Using the kinematics equation

$v_{f}^{2} = v_{o}^{2} + 2 a h$

Inserting the values

$544^{2} = 1360^{2} + 2 (-1.6) h$

h = 485520 m

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4 years ago

A 1000 kg car is rolling slowly across a level surface at 1m/s, heading toward a group of small innocent children.

Ne4ueva [31]

Answer:

The force is -1620.73 N.

Explanation:

Given that,

Mass of car = 1000 kg

Velocity = 1 m/s

Distance = 2 m

Angle = 30°

We need to calculate the force

Using formula of work done

$W_{net}=K_{f}-K_{i}$

$F\times d=K_{f}-K_{i}$

$Fd\cos\theta=-K_{i}$

$F=\dfrac{-K_{f}}{d\cos\theta}$

$F=-\dfrac{\dfrac{1}{2}mv^2}{d\cos\theta}$

Put the value into the formula

$F=-\dfrac{\dfrac{1}{2}\times1000\times(1)^2}{2\times\cos30}$

$F=-1620.73\ N$

Hence, The force is -1620.73 N.

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4 years ago

Read 2 more answers

Difference between diurnal and annual motion in two points

Valentin [98]

Answer:

Explanation below:

Explanation:

Annual motion describes the changes in motion of the earth around the sun. Diurnal motion can be better understood as the change in motion caused by Earths rotation at the poles.

This might not be the answer you were looking for, your question is very vague.

5 0

3 years ago

A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete

Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is $6.25\times 10^{- 3} ft/s^{2}$

Solution:

As per the question:

Constant velocity of the horse in the horizontal, $v_{x} = 1 ft/s$

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

$20^{2} + 10^{2} = l^{2}$

$l^{2}= 500$

Now,

$x^{2} + y^{2} = 500$ (1)

Differentiating equation (1) w.r.t 't':

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$x\frac{dx}{dt} = - y\frac{dy}{dt}$

where

$\frac{dx}{dt}$ = Rate of change of displacement along the horizontal

$\frac{dy}{dt}$ = Rate of change of displacement along the vertical

$v_{x}$ = velocity along the x-axis.

$v_{y}$ = velocity along the y-axis

$xv_{x} = -yv_{y}$

$v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s$

$|v_{y}| = 0.5\ ft/s$

Acceleration of the hay bale is given by the kinematic equation:

$v_{y}^{2} = u_{y} + 2ay$

$(-0.5)^{2} =0 + 2ay$

$0.25 = 2ay$

$\frac{0.25}{2y} = a$

$a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}$

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3 years ago

Which of the following is an example of the Doppler effect?

Oliga [24]

b is your answers in this thread

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2 years ago