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Nonamiya [84]
2 years ago
12

Two dogs fight over a bone. The larger of the two pulls on the bone to the right with a force of 42 N. The smaller one pulls to

the left with a force of 35 N
a. Draw the free-body diagram for thebone.
b. What is the net force acting on thebone?
c. If the bone has a mass of 2.5 kg, whatis its acceleration?
Physics
1 answer:
tamaranim1 [39]2 years ago
4 0

Answer:

Explanation:

A

35 N  Small Dog <=======BONE=========> Bigger Dog 42 N

B

Fnet = Large Dog - small dog The forces are subtracted because they are acting in opposite Directions.

Fnet = 42 - 35

Fnet = 7 N

C

m = 2.5 kg

F = 7 N

a = ?

F = m * a

7 = 2.5 a

a = 7 / 2.5

a = 2.8 m/s^2

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Mystical ride her scooter down the hill from the state of rest when she reaches the bottom of the hill 4.0 seconds later her spe
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The acceleration is 2.5 m/s^2

Explanation:

This is a uniformly accelerated motion (constant acceleration), therefore we can apply the following suvat equation:

v=u+at

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v is the final velocity

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t is the time

For the scooter in this problem, we have

u = 0 (it starts from rest)

v = 10.0 m/s is the final velocity

t = 4.0 s is the time interval

Solving for a, we find the acceleration:

a=\frac{v-u}{t}=\frac{10-0}{4}=2.5 m/s^2

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3 years ago
A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm ,
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Answer:

\vec{F}_{21}=-5.63\times 10^{-11}N\\\\\vec{F}_{21}=\\

Explanation:

Given that

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As both charges are negative so there exist force of repulsion in direction as shown in figure.

F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N

Angle at which force F12 is acting is

\theta=tan^{-1}\frac{3.2}{3.8}\\\\\theta=tan^{-1}\frac{y}{x}\\\\\theta= 40.1^o

F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N\\\\

\vec{F}_{12}=\vec{F}_{x}+\vec{F}_y\\\\\vec{F}_{12}=4.30\times 10^{-11}\,\hat{i} + 3.62\times 10^{-11}\,\hat{j}\\\\\vec{F}_{12}=

Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction

F_{21}=-5.63\times 10^{-11}N

\vec{F}_{21}=

7 0
3 years ago
Work is being done in which of these situations? All motions are at a constant velocity (including velocity = zero).
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Answer:

A

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Because Jessica pushes a 25 kilogram cart a distance of 10 meters.

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3 years ago
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A stone is dropped from a certain height, distance covered by it in one second is:
Alinara [238K]

Answer:

d = 4.9 m

Explanation:

It is mentioned that a stone is dropped from a certain height. It is required to find the distance covered by it in one second.

The initial speed of the stone is equal to 0 as it was at rest. Let d is the distance covered by the stone.

Using second equation of motion :

d=ut+\dfrac{1}{2}at^2

Put u = 0 and a = g

d=\dfrac{1}{2}gt^2\\\\d=\dfrac{1}{2}\times 9.8\times 1^2\\\\d=4.9\ m

So, the distance covered by it in one second is 4.9 m.

5 0
3 years ago
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Answer:

1.144 A

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given that;

the length of the wire = 2.0 mm

the diameter of the wire = 1.0 mm

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∴  the current if this wire if it is connected to the terminals of a 17.0V battery = 1.144 A

8 0
3 years ago
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