2 years ago

This is a test oooppps

Physics

2 answers:

san4es73 [151]2 years ago

8 0

Explanation:

First of all friend attach a picture with the question then only I can Solve......

hope you might understood..

iris [78.8K]2 years ago

3 0

If you attach the question(s) i could probably help!

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2 Jupiter orbits the sun in a nearly circular path with radius 7.8x10^11m. The orbital period of Jupiter is 12 years.

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Mass of Jupiter=1.9×10

㎏=M

Mass of Sun=1.99×10

㎏=M

Mean distance of Jupiter from Sun=7.8×10

m=r

G=6.67×10

−11

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−2

Gravitational Force, F=

(7.8×10

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2 years ago

When the Sun is directly overhead, the intensity of sunlight reaching the ground is about 1000 W/m^{2} 2 . The Earth has a radiu

Hunter-Best [27]

Answer:

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3 years ago

A car changes its velocity from 40.0 m/s to -5.0 m/s in 5.0 seconds what was the car average acceleration

ValentinkaMS [17]

Answ40x-50 =2334

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3 years ago

How can people who live near rivers control flodding

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3 years ago

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A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh

alekssr [168]

(a) 392 N/m

Hook's law states that:

$F=k\Delta x$ (1)

where

F is the force exerted on the spring

k is the spring constant

$\Delta x$ is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

$F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N$

- The stretching of the spring is

$\Delta x=15 cm-10 cm=5 cm=0.05 m$

Solving eq.(1) for k, we find the spring constant:

$k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m$

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

$F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N$

And so, the stretch of the spring is

$\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm$

And since the initial lenght of the spring is

$x_0 = 10 cm$

The final length will be

$x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm$

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3 years ago

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