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satela [25.4K]
3 years ago
12

What force is necessary to accelerate a 70 kg object at a rate of 4.2 m/s squares

Physics
1 answer:
Vladimir79 [104]3 years ago
6 0
F = ma
F = 70*4.2
F = 294 N
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HELPP!!
lisov135 [29]

Answer:

1.2 MPS

Explanation:

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3 years ago
An object is moving to the left with a constant speed. What can be concluded about the forces acting upon the object?
MariettaO [177]

Answer: sum of forces is zero

Explanation: acording to the newton first law you can say that the sum of the forces acting over the object is zero no matter the direction of movement

good luck

mario

8 0
3 years ago
A 1.0-kg cue ball traveling at 15 m/s strikes a stationary billiard ball of mass 1.5 kg. After the collision, the cue ball remai
Igoryamba

Answer:

Other ball's velocity is 10 m/s

Explanation:

We can use conservation of momentum:

P_{1i} +P_{2i} =P_{1i} +P_{2i}\\1 * 15 + 1.5 * 0 = 1 * 0 + 1.5 * v\\15 = 1.5 * v\\v=\frac{15}{1.5} \,\frac{m}{s} \\v=10 \,\frac{m}{s}

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3 years ago
A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be
FrozenT [24]

Answer:

\omega=0.31\frac{rad}{s}

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

a_c=\frac{v^2}{r}(1)

Here, r is the radius and v is the tangential speed, which is given by:

v=\omega r(2)

Here \omega is the angular velocity, we replace (2) in (1):

a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r

Recall that r=\frac{d}{2}=\frac{200m}{2}=100m.

Solving for \omega:

\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}

3 0
4 years ago
Electrons and protons travel from the Sun to the Earth at a typical velocity of 3.83 ✕ 105 m/s in the positive x-direction. Thou
Leona [35]

Answer:

F=2.84*10^{-26}N  & -y direction

F=2.84*10^{-26}N & +y direction

Explanation:

From the question we are told that:

Speed of electron V_e=3.83 * 10^5 m/s +x direction

Earths magnetic field B_e=3.04 * 10^-^8 +z direction

a)

Generally the equation for magnetic force F_m is mathematically given by

F=q(V_e*B_e)

where

q=1.6*10^{-19}c\\\=i*\=z=-\=j

F=1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)

F=1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)

F=-2.84*10^{-26}N \=j

Magnitude & Direction

F=2.84*10^{-26}N  -y direction

b)

Generally the equation for magnitude and direction of the magnetic force on an electron. is mathematically given by

\=F'=-1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)

\=F'=-2.84*10^{-26}N \=j

Magnitude & Direction

F=2.84*10^{-26}N & +y direction

5 0
3 years ago
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