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nexus9112 [7]
4 years ago
11

The builders of the pyramids used a long ramp to lift 20000-kg (20.0-ton) blocks. If a block rose 0.840 m in height while travel

ing 20.0 m along the ramp surface, how much uphill force was needed to push it up the ramp at constant velocity

Physics
1 answer:
olga_2 [115]4 years ago
8 0

Given Information:  

Mass = m =  20000 kg

Height of ramp = h = 0.840 m

Length of ramp = L = 20 m

Required Information:  

Uphill Force = F = ?

Answer:

F = 8232 N

Explanation:

The force can be found using the equation

F = mgsinθ

Where m is the mass of block, g is the acceleration due to gravity

Since the ramp can be modeled as an inclined plane so it can formed into a triangle, recall that in right angle triangle sinθ is equal to opposite over hypotenuse. The opposite is the height and hypotenuse is the ramp surface.

sinθ = h/L = 0.840/20 = 0.042

F = 20000*9.8*0.042

F = 8232 N

Therefore, an uphill force of 8232 N would be needed.

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3 years ago
The energy of atoms and molecules in an object due to their motion is_______ energy
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I think the answer is potential 
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3 years ago
I have my exam monday I need some help!
Lana71 [14]

Answer:

Explanation:

If the passage of the waves is one crest every 2.5 seconds, then that is the frequency of the wave, f.

The distance between the 2 crests (or troughs) is the wavelength, λ.

We want the velocity of this wave. The equation that relates these 3 things is

f=\frac{v}{\lambda} and filling in:

2.5=\frac{v}{2.0} so

v = 2.5(2.0) and

v = 5.0 m/s

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3 years ago
An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large o
Alekssandra [29.7K]

Answer:

0.53m/s^2

Explanation:

We are given that

Mass of wire=m=3.6 g=3.6\times 10^{-3} kg

1 kg=1000 g

Length of wire=l=1.6 m

Mass of object=m'=3 kg

Time,t=60.1 ms=60.1\times 10^{-3} s

1 ms=10^{-3} s

Speed,v=\frac{distance}{time}=\frac{1.6}{60.1\times 10^{-3}}=26.62 m/s

g=\frac{v^2m}{m'l}

Using the formula

g=\frac{(26.62)^2\times 3.6\times 10^{-3}}{3\times 1.6}=0.53m/s^2

8 0
4 years ago
You drag a crate across a floor with a rope. The force applied is 750 N and the angle of the rope is 25.0° above the horizontalH
frez [133]

a.

The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:

W=F\cos\theta\cdot d

where F is the magnitude of the force, theta is the angle between them and d is the distance.

The problen gives the following data:

The magnitude of the force 750 N.

The angle between the force and the displacement which is 25°

The distance, 26 m.

Plugging this in the formula we have:

\begin{gathered} W=\left(750\right)\left(\cos25\right)\left(26\right) \\ W=17673 \end{gathered}

Therefore the work done is 17673 J.

b)

The power is given by:

P=\frac{W}{t}

the problem states that the time it takes is 6 s. Then:

\begin{gathered} P=\frac{17673}{6} \\ P=2945.5 \end{gathered}

Therefore the power is 2945.5 W

5 0
1 year ago
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