(a)The weight of two metallic spheres, each of 200 g and radius 4 cm and

How does earths rotation affect life on earth

dalvyx [7]

When watching the stars at night, they do appear to move very slowly. This is because the Earth is constantly moving. The Earth completes

6 0

3 years ago

What is the magnitude of the gravitational force acting on the sun due to the earth?The earth does not exert any gravitational f

Sholpan [36]

I haven't worked on Part-A, and I don't happen to know the magnitude of the gravitational force that the Sun exerts on the Earth.

But whatever it is, it's exactly, precisely, identical, the same, and equal to the magnitude of the gravitational force that the Earth exerts on the Sun.

I think that's the THIRD choice here, but I'm not sure of that either.

4 0

3 years ago

PLS ANSWER REAL ANSWERS ONLY

Julli [10]

Answer:

Who is moving the fastest: Green triangle

What is the slope of Blue Diamonds: 5/6 or 5/7, can’t tell completely

<h2><u>Hope This helped!</u></h2>

7 0

3 years ago

Calculate the initial (from rest) acceleration of a proton in a 5.00 x 10^6 N/C electric field (such as created by a research Va

julsineya [31]

Answer:

Acceleration of the proton will be equal to $4.79\times 10^{14}m/sec^2$

Explanation:

We have given electric field $E=5\times 10^6N/C$

Mass of proton is equal to $m=1.67\times 10^{-27}kg$

And charge on proton is equal to $e=1.6\times 10^{-19}C$

Electrostatic force will be responsible for the motion of proton

Electrostatic force will be equal to $F=qE=1.6\times 10^{-19}\times 5\times 10^6=8\times 10^{-13}N$

According to newton law force on the proton will be equal to F = ma, here m is mass of proton and a is acceleration

This newton force will be equal to electrostatic force

So $1.67\times 10^{-27}\times a=8\times 10^{-13}$

$a=4.79\times 10^{14}m/sec^2$

So acceleration of the proton will be equal to $4.79\times 10^{14}m/sec^2$

5 0

3 years ago

A shopper weighs 4.00 kg of apples on a supermarket scale whose spring obeys Hooke's law and notes that the spring stretches a d

Irina-Kira [14]

Answer:

a) 0.040625 m

b) 5.02272 J

Explanation:

k = Spring constant

x = Stretched length

F = Force

a)

$F=kx\\\Rightarrow k=\frac{F}{x}\\\Rightarrow k=\frac{4\times 9.81}{0.025}\\\Rightarrow k=1569.6\ N/m$

$F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{6.5\times 9.81}{1569.6}\\\Rightarrow x=0.040625\ m$

Extension of the spring would be 0.040625 m

b) Work done in a spring

$W=\frac{1}{2}kx^2\\\Rightarrow W=\frac{1}{2}\times 1569.6\times 0.08^2\\\Rightarrow W=5.02272\ J$

The work done by the shopper to stretch this spring a total distance of 8.00 cm is 5.02272 J

4 0

3 years ago