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zloy xaker [14]
3 years ago
5

A sumo wrestler is 330 pounds. How much force does the earth exert in him? Define Mass Define Weight

Physics
1 answer:
Oksana_A [137]3 years ago
5 0

Answer:

a large matters of body with no shape is mass

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A sphere of radius R has total charge Q. The volume charge density (C/m3) within the sphere is rho(r)=C/r2, where C is a constan
san4es73 [151]

Answer: C = Q/4πR

Explanation:

Volume(V) of a sphere = 4πr^3

Charge within a small volume 'dV' is given by:

dq = ρ(r)dV

ρ(r) = C/r^2

Volume(V) of a sphere = 4/3(πr^3)

dV/dr = (4/3)×3πr^2

dV = 4πr^2dr

Therefore,

dq = ρ(r)dV ; dq =ρ(r)4πr^2dr

dq = C/r^2[4πr^2dr]

dq = 4Cπdr

FOR TOTAL CHANGE 'Q', we integrate dq

∫dq = ∫4Cπdr at r = R and r = 0

∫4Cπdr = 4Cπr

Q = 4Cπ(R - 0)

Q = 4CπR - 0

Q = 4CπR

C = Q/4πR

The value of C in terms of Q and R is [Q/4πR]

7 0
3 years ago
A 1200 kg truck is moving to the right at the speed of 30 m/s. It hit another identical truck that was at rest. The two trucks s
Marrrta [24]

Answer:

ok wht is the question

Explanation:

4 0
3 years ago
A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction
Vaselesa [24]

Answer:

H=1020.12m

Explanation:

From a balance of energy:

\frac{m*Vo^2}{2} -mg*H=-Ff*d   where H is the height it reached, d is the distance it traveled along the ramp and Ff = μk*N.

The relation between H and d is given by:

H = d*sin(30)   Replace this into our previous equation:

\frac{m*Vo^2}{2} -mg*d*sin(30)=-\mu_k*N*d

From a sum of forces:

N -mg*cos(30) = 0    =>  N = mg*cos(30)   Replacing this:

\frac{m*Vo^2}{2} -mg*d*sin(30)=-\mu_k*mg*cos(30)*d   Now we can solve for d:

d = 2040.23m

Thus H = 1020.12m

6 0
3 years ago
Read 2 more answers
calculate the work done in kilo joules in lifting a mass of 20kg at steady velocity through a vertical height of 20m
bulgar [2K]

Answer:

\huge\boxed{\sf Work\ done = 4 kJ}

Explanation:

Since work done is in the form of potential energy, we will use the formula of potential energy here.

We know that,

<h3>P.E. = mgh </h3>

Where,

m = mass = 20 kg

g = acceleration due to gravity = 10 m/s²

h = vertical height = 20 m

So,

<h3>Work done = mgh</h3>

Work done = (20)(10)(20)

Work done = 4000 joules

Work done = 4 kJ

\rule[225]{225}{2}

5 0
2 years ago
When you kick a soccer ball across a field it will eventually come to a stop. Does this prove or disapprove newton's first law o
jek_recluse [69]
I think its D. I am pretty sure 
6 0
3 years ago
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