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2 years ago

According to Newton’s Second Law of Motion, an object will accelerate if which kind of force is applied?

Physics

2 answers:

Kobotan [32]2 years ago

7 0

Answer:

unbalanced force

Explanation:

this is a guess so just look it up

Darina [25.2K]2 years ago

5 0

Answer is A unbalanced force

Explanation: I just looked it up

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What is NOT an example of Reflection

raketka [301]

Refraction as in a pencil going into water

8 0

3 years ago

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A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching i

Sunny_sXe [5.5K]

Answer:

114.44 J

Explanation:

From Hook's Law,

F = ke................. Equation 1

Where F = Force required to stretch the spring, k = spring constant, e = extension.

make k the subject of the equation

k = F/e.............. Equation 2

Given: F = 10 lb = (10×4.45) N = 44.5 N, e = 4 in = (4×0.254) = 1.016 m.

Substitute into equation 2

k = 44.5/1.016

k = 43.799 N/m

Work done in stretching the 9 in beyond its natural length

W = 1/2ke²................. Equation 3

Given: e = 9 in = (9×0.254) = 2.286 m, k = 43.799 N/m

Substitute into equation 3

W = 1/2×43.799×2.286²

W = 114.44 J

3 0

3 years ago

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A climber pulls herself 8 meters upwards with a force of 150 Newtons. If it takes her 16 seconds to cover the 8 meters, how much

mr Goodwill [35]

Answer:

P = 75 W

Explanation:

given,

Distance, L = 8 m

Force,F = 150 N

Time, t = 16 s

Work by the climber

Work done = Force x displacement

W = F. L

W = 150 x 8

W = 1200 J

We know,

$Power =\dfrac{Work\ done}{time}$

$P =\dfrac{1200}{16}$

P = 75 W

Hence, Power climber is using to climb is equal to 75 W.

3 0

3 years ago

What is the acceleration of a race car if it has a mass of 1200kg and is moving with an engine force of 400N

meriva

Answer:

0.34 m/s^2

Explanation:

force=mass × acceleration

400 =1200 × acceleration

acceleration=400/1200

=0.34 m/s^2

5 0

2 years ago

The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is $W=1.20 \cdot 10^{-3}J$

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: $W_e=q\Delta V$

where

q is the charge

$\Delta V$ is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

$K_f - K_i = W + W_e = W+q\Delta V$

and since the charge starts from rest, $K_i = 0$ , so the formula becomes

$K_f = W+q\Delta V$

In this problem, we have

$W=1.20 \cdot 10^{-3}J$ is the work done by the external force

$q=-6.70 \mu C=-6.7\cdot 10^{-6}C$ is the charge

$K_f = 4.72\cdot 10^{-4}J$ is the final kinetic energy

Solving the formula for $\Delta V$ , we find

$\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V$

4 0

3 years ago

Read 2 more answers