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2 years ago

Joe leans against a lamppost with a force of 200 N. The lamppost does not move. How much work does he do?

Physics

1 answer:

hoa [83]2 years ago

6 0

Answer:

No work

Explanation:

He is not moving the lamppost as the lamppost cannot move to begin with.

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A bus accelerates from 5.75 m/s at a rate of 1.25 m/s/s for 3.50

marysya [2.9K]

Answer:

10.125 meters?

Explanation:

Im taking 5.75m/s + 1.25 m/s/s (3.5) = my answer.

In those 3.5 seconds it travels 4.375.

I added that to 5.75 to get 10.125m

3 0

3 years ago

Bosons & Fermions A new type of quantum object has been discovered. When a large collection of these are cooled to almost ab

Lena [83]

Answer:

Only 2,3,4 are true

Explanation:

Bosons Particles are particles that condense to the same state. Bosons particle have integral spin like 0 , $\hbar$ , $2$\hbar$ , $3$\hbar$$ , etc. Bosons particles always have asymmetric wave function and there is exchange of particles.

1) It does not obey Fermi_ Dirac statistics

2) It obeys Bose-Einstein statistics

3) The object can have intrinsic spin $2$\hbar$

4) Yes the Bosons particle is always symmetric with exchange of particles

5) No Bosons particle are symmetric and not asymmetric

5 0

3 years ago

Death Star has a diameter of 160,000m and a mass of 5.1e17kg. Millennium Falcon has a mass of 1.36e6kg (data from Wookieepedia)

lions [1.4K]

Answer:

7229 N

Explanation:

The gravitational force between the Death Star and the Millenium Falcon is given by:

$F=G\frac{mM}{R^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$M=5.1\cdot 10^{17} kg$ is the mass of the Death Star

$m=1.36\cdot 10^6 kg$ is the mass of the Millennium Falcon

$R=\frac{160,000 m}{2}=80,000 m$ is the radius of the Death Star

Substituting numbers into the equation, we find the force

$F=(6.67\cdot 10^{-11})\frac{(1.36\cdot 10^6 kg)(5.1\cdot 10^{17} kg)}{(80,000 m)^2}=7229 N$

5 0

3 years ago

A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500

defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

$M = \frac{\mu_0 N_1 N_2A_1}{l_1}$

Where,

$\mu =$ permeability of free space

$N_1 =$ Number of turns in solenoid 1

$N_2 =$ Number of turns in solenoid 2

$A_1=$ Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

$\mu_0 = 4\pi *10^{-7}H/m$

$N_1 = 500$

$N_2 = 40$

$A = 7.5*10^{-4}m^2$

$l = 0.5m$

Substituting,

$M = \frac{\mu_0 N_1 N_2A_2}{l_1}$

$M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}$

$M = 3.77*10^{-4}H$

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0

3 years ago

The weight of a hydraulic barber's chair with a client is 2100 N. When the barber steps on the input piston with a force of 44 N

guajiro [1.7K]

Answer:

$\frac{r_1}{r_2}=6.9$

Explanation:

According to Pascal's Law, the pressure transmitted from input pedal to the output plunger must be same:

$P_1 = P_2\\\\\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\\\frac{F_1}{F_2}=\frac{A_1}{A_2}\\\\\frac{F_1}{F_2}=\frac{\pi r_1^2}{\pi r_2^2}\\\\\frac{F_1}{F_2}=\frac{r_1^2}{r_2^2}$

where,

F₁ = Load lifted by output plunger = 2100 N

F₂ = Force applied on input piston = 44 N

r₁ = radius of output plunger

r₂ = radius of input piston

Therefore,

$\frac{r_1^2}{r_2^2}=\frac{2100\ N}{44\ N}\\\\\frac{r_1}{r_2}=\sqrt{\frac{2100\ N}{44\ N}} \\\\\frac{r_1}{r_2}=6.9$

4 0

2 years ago