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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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6 0
3 years ago
Two charges are located in the xx–yy plane. If q1=−4.10 nCq1=−4.10 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.080 m)
Gala2k [10]

Answer:

Explanation:

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= - 31.6 j N/C

Due to second charge electric field at origin

= 8.99 x 10⁹ x 3.6  x 10⁻⁹ / 1.2²+ .6²

= 8.99 x 10⁹ x 3.6  x 10⁻⁹ / 1.8

= 18 N/C

It is making angle θ where

Tanθ = .6 / 1.2

= 26.55°

this field in vector form

= - 18 cos 26.55 i - 18 sin26.55 j

= - 16.10 i - 8.04 j

Total field

= - 16.10 i - 8.04 j + ( - 31.6 j )

= -16.1 i - 39.64 j .

Ex = - 16.1 i

Ey = - 39.64 j .

8 0
3 years ago
a 1.2x10^3 kilogram car is accelerated uniformly from 10. meters per second to 20 meters per second in 5.0 seconds. what is the
irinina [24]
Force , F = ma

F =  m(v - u)/t               

Where m = mass in kg, v= final velocity in m/s, u = initial velocity in m/s
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m= 1.2*10³ kg,  u = 10 m/s,  v = 20 m/s, t = 5s

F =  1.2*10³(20 - 10)/5

F = 2.4*10³ N = 2400 N


7 0
2 years ago
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