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3 years ago

12

Thing 1 and Thing 2 push with a force of 2000 N to move a rather small 500 kg elephant 30 m across the kitchen floor. How much w

ork did they do?

1 answer:

Amanda [17]3 years ago

3 0

Answer:

60000 J

Explanation:

Assuming the force is applied parallel to the displacement of the elephant, the work done to move it across the floor is

$W=Fd$

where

F = 2000 N is the force applied

d = 30 m is the displacement of the elephant

Substituting the numbers into the formula, we find

$W=(2000 N)(30 m)=60,000 J$

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Answer:

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Explanation:

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2 years ago

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The alarm at a fire station rings and a 87.5-kg fireman, starting from rest, slides down a pole to the floor below (a distance o

blsea [12.9K]

Answer:

$F_f=840N$

Explanation:

From the question we are told that

Weight of fireman $W_f= 87.5kg$

Pole distance $D=4.10m$

Final speed is $V_f 1.75m/s$

Generally the equation for velocity is mathematically represented as

$v^2 = v_0^2 + 2 a d$

Therefore Acceleration a

$a'= v^2 / 2 d$

$a'= 0.21m/s^2$

Generally the equation for Frictional force $F_f$ is mathematically given as

$F_f=m*a$

$F_f=m*(g-0.21)$

$F_f=87.5*(9.81-0.21)$

Therefore

$F_f=840N$

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2 years ago

A 10n falling object encounters 4n of air resistance. what is the net force on the object?

Margaret [11]

It would be 6n down.

5 0

2 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

5 0

3 years ago

Newer aircraft jet catapult systems use magnets instead of steam. The launch still takes 1.19 seconds, but the acceleration is a

Westkost [7]

Answer:

33.6 m

Explanation:

Given:

v₀ = 0 m/s

a = 47.41 m/s²

t = 1.19 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (1.19 s) + ½ (47.41 m/s²) (1.19 s)²

Δx = 33.6 m

8 0

2 years ago