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Otrada [13]
3 years ago
6

What is the force needed to move an 225 kg car a distance of 150 meters

Physics
2 answers:
Kobotan [32]3 years ago
4 0

Physics

mass = 225 kg

weight = m × a = 2250 N

distance = 150 m

force : ______?

Answers :

W = F × s

W = 2250 × 150

W = 337500 ✅

jekas [21]3 years ago
4 0
Depends on how fast you want it to go I would say 225kg it should take a while though
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Answer:

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Explanation:

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The cochlea nerve is also called the auditory nerve.

7 0
3 years ago
Help please!!!
TiliK225 [7]

Answer:

Cu 8.92

Explanation:

The formula for density is mass/volume. If you were to divide 62.44 by 7, you would get 8.92. Since copper is the only metal in this table that has a density of 8.92, that is the answer.

7 0
3 years ago
The acceleration of a particle traveling along a straight line isa=14s1/2m/s2, wheresis in meters. Ifv= 0,s= 1 m whent= 0, deter
Yuliya22 [10]

Answer:

0.78m/s

Explanation:

We are given that

Acceleration=a=\frac{1}{4}s^{\frac{1}{2}}m/s^2

v=0, s=1 when t=0

We have to find the particle's velocity at s=2m

We know that

a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=\frac{dv}{ds}v

vdv=ads

\int_{0}^{v} vdv=\int_{1}^{s}0.25s^{\frac{1}{2}}ds

\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2})^{s}_{1}

By using formula:\int x^ndx=\frac{x^{n+1}}{n+1}+C

\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2}}-1

Substitute s=2

\frac{v^2}{2}=\frac{0.50}{3}((2^{1.5})-1)

\frac{v^2}{2}=\frac{0.50}{3}\times 1.83

v^2=2\times 0.305=0.61

v=\sqrt{0.61}=0.78m/s

Hence, the velocity of particle at s=2m=0.78m/s

3 0
3 years ago
A 40 kg dog is sitting on top of a hillside and has a potential energy of 1,568 J. What is the height of the hillside? (Formula:
Tpy6a [65]

Answer:

The answer is 4.0 m

Explanation:

6 0
3 years ago
Read 2 more answers
A solenoid has a radius Rs = 14.0 cm, length L = 3.50 m, and Ns = 6500 turns. The current in the solenoid decreases at the rate
babymother [125]

Answer:

E = 58.7 V/m

Explanation:

As we know that flux linked with the coil is given as

\phi = NBA

here we have

A = \pi R_s^2

B = \mu_o N i/L

now we have

\phi = N(\mu_o N i/L)(\pi R_s^2)

now the induced EMF is rate of change in magnetic flux

EMF = \frac{d\phi}{dt} = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L

now for induced electric field in the coil is linked with the EMF as

\int E. dL = EMF

E(2\pi r_c) = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L

E = \frac{\mu_o N^2 R_s^2 \frac{di}{dt}}{2 r_c L}

E = \frac{(4\pi \times 10^{-7})(6500^2)(0.14^2)(79)}{2(0.20)(3.50)}

E = 58.7 V/m

3 0
3 years ago
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