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1 year ago

A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?

Physics

1 answer:

ankoles [38]1 year ago

4 0

Given required solution

M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
=5m/s*2.4s
=12m
so W=12*100
W=1200J

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2 years ago

Why does carpet tend to produce differences in static electricity more that hardwood or tile floors

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Answer:

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Explanation:

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2 years ago

A parallel-plate capacitor is constructed of two square plates, size L×L, separated by distance d. The plates are given charge ±

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Answer: A) 2 B) 4 C) 1

Explanation:

The Electric field from a parallel-plate capacitor is given by:

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2 years ago

A girl and a boy are riding on a merry-go-round that is turning at a constant rate. The girl is near the outer edge, and the boy

skad [1K]

(a) Both the girl and the boy have the same nonzero angular displacement.

Explanation:

The angular displacement of an object moving in uniform circular motion, as the boy and the girl on the merry-go-round, is given by

$\theta= \omega t$

where

$\omega$ is the angular speed

t is the time interval

For a uniform object in uniform circular motion, all the points of the object have same angular speed. This means that the value of $\omega$ is the same for the boy and the girl.

Therefore, if we consider the same time interval t, the boy and the girl will also have same nonzero angular displacement.

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Explanation:

The linear (tangential) speed of a point along the merry-go-round is given by

$v=\omega r$

where

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1 year ago

If a 20.0 g object at a temperature of 35.0∘C has a specific heat of 2.89Jg∘C, and it releases 450. J into the atmosphere, what

nataly862011 [7]

Answer:

The final temperature of the object will be 42.785 °C

Explanation:

When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.

In other words, sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change), so that the temperature varies.

The equation for calculating the heat exchanges in this case is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature.

In this case:

Q= 450 J
c= 2.89 $\frac{J}{g*C}$
m= 20 g
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Replacing:

450 J= 2.89 $\frac{J}{g*C}$ *20 g* (Tfinal - 35°C)

Solving for Tfinal:

$\frac{450 J}{2.89\frac{J}{g*C}*20g} =Tfinal -35C$

7.785 °C=Tfinal - 35°C

7.785 °C + 35°C= Tfinal

42.785 °C=Tfinal

<u><em>The final temperature of the object will be 42.785 °C</em></u>

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2 years ago

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