All categories

3 years ago

An object is pushed with a force of 400 N for 20 meters; what is the work done on the object? Solve using W=F*d

Physics

1 answer:

nika2105 [10]3 years ago

5 0

Answer:

W = Fd = 400(20) = 8000 J

Explanation:

You might be interested in

A cold glass is left sitting outside on a hot day. Soon, water droplets form on the outside of the glass. Describe the events th

ikadub [295]

Evaporation must happen

8 0

3 years ago

The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole huma

gayaneshka [121]

Explanation:

Below is an attachment containing the solution.

6 0

3 years ago

A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t

kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

$\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W$

where we have used the values given by the information of the problem for N B and A.

the emf is given by:

$emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV$

hope this helps!!

5 0

3 years ago

A car slows from 27 m/s to 5 m/s with a constant acceleration for 6.87 s. What is the car’s acceleration?

Kobotan [32]

In this case, the movement is uniformly delayed (the final rapidity is less than the initial rapidity), therefore, the value of the acceleration will be negative.

1. The following equation is used:

a = (Vf-Vo)/ t

a: acceleration (m/s2)

Vf: final rapidity (m/s)

Vo: initial rapidity (m/s)

t: time (s)

2. Substituting the values in the equation:

a = (5 m/s- 27 m/s)/6.87 s

3. The car's acceleration is:

a= -3.20 m/ s^2

5 0

3 years ago

A ball is dropped from a building taking 3sec to fall to the ground. Calculate:

GenaCL600 [577]

Answer:

Vf = 29.4 m/s

h = 44.1 m

Explanation:

Data:

Initial Velocity (Vo) = 0 m/s
Gravity (g) = 9.8 m/s²
Time (t) = 3 s
Final Velocity (Vf) = ?
Height (h) = ?

==================================================================

Final Velocity

Use formula:

Vf = g * t

Replace:

Vf = 9.8 m/s² * 3s

Multiply:

Vf = 29.4 m/s

==================================================================

Height

Use formula:

$\boxed{h=\frac{g*(t)^{2}}{2}}$

Replace:

$\boxed{h=\frac{9.8\frac{m}{s^{2}}*(3s)^{2}}{2}}$

Multiply time squared:

$\boxed{h=\frac{9.8\frac{m}{s^{2}}*9s^{2}}{2}}$

Simplify the s², and multiply in the numerator:

$\boxed{h=\frac{88.2m}{2}}$

It divides:

$\boxed{h=44.1\ m}$

What is the velocity when falling to the ground?

The final velocity is 29.4 meters per seconds.

How high is the building?

The height of the building is 44.1 meters.

3 0

3 years ago