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Lelu [443]
2 years ago
8

How much force is needed to stretch a spring 1. 2 m if the spring constant is 8. 5 N/m? 7. 1 N 7. 3 N 9. 7 N 10. 2 N.

Physics
1 answer:
statuscvo [17]2 years ago
6 0

The amount of force required to stretch or compress the spring is known as the spring force. Its unit is Newton(N). Force is needed to stretch spring is 10.2 N.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

F is the spring force =?

K is the spring constant= 8.5 N/m

x is the length by which spring got stretched = 1.2m

\rm F_S=Kx \\\\ \rm F_S=8.5 \times 1.2 \\\\ \rm F_S=10.2 N

Hence the force is needed to stretch the spring is 10.2 N.

To learn more about the spring force refer to the link;

brainly.com/question/4291098

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Answer:

surface temperature of the chip located 120 mm Ts=42.5°C

surface temperature of the chip in Mexico Ts=46.9°C

Explanation:

from the energy balance equation we have to:

q=E=30W

from Newton´s law:

Ts=Tα+(q/(h*A)), where A=l^2

N=h/k=0.04*(Vl/V)^0.85*Pr^1/3

data given:

l=0.12 m

v=10 m/s

k=0.0269 W/(m*K)

Pr=0.703

Replacing:

h=0.04*(0.0269/0.12)*(10*0.12)/((16*69x10^-6))^0.85*(0.703^1/3) = 107 W/m^2*K

The surface temperature at sea level is equal to:

Ts=25+(30x10^-3/107*0.004^2)=42.5°C

h=0.04*(0.0269/0.12)*((10*0.12)/(21*81x10^-6))^0.85*(0.705^1/3)=85.32 W/(m*K)

the surface temperature at Mexico City is equal to:

Ts=25+(30x10^-3/85.32*0.004^2)=46.9°C

8 0
3 years ago
A toy car has a momentum of 3 kilogram meters per second south. The car has a 1-kilogram mass. Which is the velocity of the car?
antiseptic1488 [7]
Using p = mv 3 = 1× v v = 3m/s
6 0
3 years ago
A car is moving with an initial relocity of
MA_775_DIABLO [31]

Answer:

The final acceleration of the car, v = 70 m/s

Explanation:

Given,

The initial velocity of the car, u = 20 m/s

The acceleration of the car, a = 10 m/s²

The time period of travel, t = 5 s

Using the I equations of motion

                     v = u + at

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                        = 70 m/s

Hence, the final acceleration of the car, v = 70 m/s

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3 years ago
What is the student's kinetic energy at the bottom of the hill if he is moving
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Answer:

KE = 10530 J or 10.53 KJ

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The formula for kinetic energy is KE = 1/2 mv^2

Let's apply the formula:

KE = 1/2 mv^2

KE = 1/2 (65kg) (18m/s)^2

KE = 10530 J or 10.53 KJ

5 0
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Find the ratio of the new/old periods of a pendulum if the pendulum were transported from earth to the moon, where the accelerat
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Since the gravitational acceleration on Earth is g_e = 9.81 m/s^2 while on the Moon is g_m=1.63 m/s^2, the ratio between the period on the Moon and on Earth is
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3 0
3 years ago
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