2 years ago

What is the quantity of work done when a crane lifts a 100-n block from 2 m above the ground to 6 m above the ground?.

1 answer:

7 0

Answer: by definition work = Force * distanceForce = 100Ndistance = 6 - 2 = 4work = 100 * 4 = 400 J or CStill stuck? Get 1-on-1 help from an expert tutor now.

Explanation:

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What speed would a proton need to achieve in order to circle Earth 1790 km above the magnetic equator, where the Earth's mag- ne

irinina [24]

Answer:

The velocity is 31.25 m/s and direction is toward west.

Explanation:

Given that,

Distance $h= 1790 km = 1.790\times10^{6}\ m$

Magnetic field $B=4\times10^{-8}\ T$

Mass of proton $m=1.673\times10^{-21}\ Kg$

Radius of earth $R =6.38\times10^{6}\ m$

Radius of orbit $r=R+h$

$r=6.38\times10^{6}+1.790\times10^{6}$

$r=8170000\ m$

We need to calculate the speed

Using formula of magnetic field

$Bvq=\dfrac{mv^2}{r}$

$v=\dfrac{Bqr}{m}$

Put the value into the formula

$v=\dfrac{4\times10^{-8}\times1.6\times10^{-19}\times8170000}{1.673\times10^{-21}}$

$v=31.25\ m/s$

Hence, The velocity is 31.25 m/s and direction is toward west.

6 0

3 years ago

Right answer gets brainlist

ira [324]

Numberrrr 1 Inertiaaaaa

3 0

2 years ago

A 40 kg skier starts at the top of a 12-meter high slope at the bottom she is traveling 10 m/s how much energy does she lose fri

Iteru [2.4K]

Potential energy at top:
PE = mgh
PE = 40 x 9.81 x 12
P.E = 4,708.8 J

Kinetic energy at bottom:
KE = 1/2 mv²
KE = 1/2 x 40 x 10²
K.E = 2,000 J

P.E = K.E + Frictional losses
Frictional losses = 4708 - 2000
Frictional losses = 2708 J

The answer is D.

7 0

3 years ago

Read 2 more answers

2. A proton and an electron are sitting 0.5m away from each other. What is the

Brrunno [24]

Answer:we can calculate the answer by the following formula;

Explanation: