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AVprozaik [17]
3 years ago
11

A man does 1500 J of work to pull a box of 150 N. what is the displacement of the box?

Physics
1 answer:
Assoli18 [71]3 years ago
5 0
  • Work done=1500J
  • Force=150N
  • Displacement =?

We know

\boxed{\sf Work\:Done=Force\times Displacement}

\\ \sf\longmapsto 1500=150\times Displacement

\\ \sf\longmapsto Displacement=\dfrac{1500}{150}

\\ \sf\longmapsto Displacement=10m

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1. Which statement about friction is true?(1 point) A. Static friction and kinetic friction in a system always act in the same d
ArbitrLikvidat [17]

Answer:

true 1Ay 2D

Explanation:

1) In this exercise you are asked to investigate which statements are true

A) True. The friction force opposes the movement caused by the external force,

B) False. Mantuano in the opposite direction force

C) False. The static and scientific friction force act in the same direction, since the second appears when the movement does not start and the static friction decreases.

D) Fale the static and kinetic friction forces act in the same direction

2) How to overcome friction on a ramp

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B) False. Pressing the brick against the surface increases the normal and as the friction is proportional to the normal, it also increases

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3 0
3 years ago
A stone is thrown vertically upward with velocity 9.8ms 1)calculate the time taken to reach maximum height.2)what is the maximum
ira [324]

Answer:

1.t=-1.96sec

2.H=4.8m

3.T=1.96sec

4.R=19.2m

Explanation:

u=9.8,t=?,sin theta=1

using formula t=2usintheta/g

t=2x9.8x1=19.6/10

t=1.96seconds

using formula H=u(squared)sin(squared)theta/2g

H=9.8(squared)x1(squared)/2x10

H=96x1/20

H=96/20

H=4.8m

using formula T=2usintheta/g

T=2x9.8x1/10

T=19.6/10

T=1.96sec

using the formula R=u(squared)sin2theta/g

R=9.8(squared)x2/10

R=96x2/10

R=192.08/10

R=19.2

5 0
4 years ago
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A non-ideal 12.2 V battery is connected across a resistor R. The internal resistance of the battery is 1.9Ohm. Calculate the pot
Brums [2.3K]

Answer:

R=100 Ohm, V=11.97 volts and I=0.12 amperes

R=10 Ohm, V=10.25 volts and I=1.20 amperes

R=2 Ohm, V=6.26 volts

Explanation:

The potential difference (voltage) of a battery with internal resistance is:

V=\xi-Ir (1)

with \xi the electromotive force (the voltage the batteries say to has) , I the current and r the internal resistance. By Ohm's law the current that passes through the resistor is:

I=\frac{V}{R} (2)

using (2) on (1):

V=\xi-\frac{V*r}{R}

solving for V:

V+\frac{V*r}{R}=\xi

V=\frac{\xi}{1+\frac{r}{R}} (3)

R=100 Ohm

V=\frac{12.2}{1+\frac{1.9}{100}}=11.97 V

R=10 Ohm

V=\frac{12.2}{1+\frac{1.9}{10}}=10.25 V

R=2 Ohm

V=\frac{12.2}{1+\frac{1.9}{2}}=6.26 V

Because we have now the values of I on the circuit (is the same through all the components because is a series circuit)

We use back substitution on (1) to find the current:

R=100 Ohm

I=\frac{V}{R}=\frac{11.97}{100}=0.12 A

R=10 Ohm

I=\frac{V}{R}=\frac{11.97}{10}=1.20 A

7 0
4 years ago
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