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Goshia [24]
3 years ago
6

A car travels at a constant speed of 25 m/s. Find the power supplied by the engine if it can supply a maximum force of 18,000 N​

Physics
1 answer:
mina [271]3 years ago
8 0

Answer:

720

Explanation:

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If you hold a 50 kilogram barbell above your head for 3 seconds and
aalyn [17]

Answer:

c

if you calculate the net force you get 490 N

3 0
3 years ago
as an aid in understanding this problem. The drawing shows a positively charged particle entering a 0.61-T magnetic field. The p
miv72 [106K]

Answer:

E = 420.9 N/C

Explanation:

According to the given condition:

Net\ Force = 2(Magnetic\ Force)\\Electric\ Force - Magnetic\ Force = 2(Magnetic\ Force)\\Electric\ Force = 3(Magnetic\ Force)\\qE = 3qvBSin\theta\\E = 3vBSin\theta

where,

E = Magnitude of Electric Field = ?

v = speed of charge = 230 m/s

B = Magnitude of Magnetic Field = 0.61 T

θ = Angle between speed and magnetic field = 90°

Therefore,

E = (3)(230\ m/s)(0.61\ T)Sin90^o

<u>E = 420.9 N/C</u>

3 0
3 years ago
Please Help Me!!!!!!!​
Elden [556K]

Answer:

B

Explanation:

5 0
3 years ago
The pic says everything, thank you !
Alenkasestr [34]

Answer:

1/5 or .2

Explanation:

you put miles over seconds 3/15 then simplify to 1/5 to get miles per second

4 0
3 years ago
two blocks are held together with a compressed spring between them on the surface of a slippery table .one block has three times
Marina CMI [18]

Explanation:

The initial kinetic energy KE_0 for both blocks is zero. Let m_1= m and m_2 =3m. So using the conservation law of linear momentum, we can write

0 = m_1v_1 - m_2v_2

or

v_2 = \dfrac{m_1}{m_2}v_1 = \dfrac{m}{3m}v_1 = \dfrac{1}{3}v_1

The final kinetic energies for the two masses are

KE_1 = \frac{1}{2}m_1v_1^2 = \frac{1}{2}mv_1^2

KE_2 = \frac{1}{2}m_2v_2^2 = \frac{1}{2}(3m)(\frac{1}{3}v_1)^2 = \frac{1}{2}m(\frac{1}{3}v_1^2)

Therefore, the ratio of their kinetic energies is

\dfrac{\Delta KE_2}{\Delta KE_1}  = \dfrac{\frac{1}{2}(\frac{1}{3}v_1^2)}{\frac{1}{2}v_1^2} = \dfrac{1}{3}

4 0
3 years ago
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