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Andrews [41]
2 years ago
6

Help me, this is my last question pls.

Physics
2 answers:
Cloud [144]2 years ago
6 0

If resultant force on the body is 0 the acceleration will also be 0.​

<h3>What is acceleration?</h3>

The term "acceleration" refers to the change in velocity with time. We must also recall that force is the product of mass and acceleration. If that is so, we can write; F = ma.

Now, we are told that the force on the body is zero so making the acceleration the subject of the formula; a = 0/mand a = 0.

Hence,  if resultant force on the body is 0 the acceleration will also be 0.​

Learn more about acceleration: brainly.com/question/2437624

Alex777 [14]2 years ago
4 0

Answer:

it means that the acceleration vanishes we can say that from newton first law of motion if an object experiences zero resultant force then its force wont change and so the object will have zero acceleration

Explanation:

n say that

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A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimumcoefficient of static friction
zhenek [66]

Answer:

c. 0.816

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Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

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The frictional force is given as:

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f=\mu N

As there is no vertical motion, therefore, N=mg. So,

f=\mu mg

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816

Therefore, option (c) is correct.

7 0
3 years ago
In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr
spin [16.1K]

Answer:

a) v_p=9.35m/s

Explanation:

From the question we are told that:

Open cart of mass   M_o=50.0 kg

Speed of cart   V=5.00m/s

Mass of package   M_p=15.0kg

Speed of package at end of chute V_c=3.00m/s

Angle of inclination   \angle =37

Distance of chute from bottom of cart   d_x=4.00m

a)

Generally the equation for work energy theory is mathematically given by

  \frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2

Therefore

  \frac{1}{2}u^2+gh=\frac{1}{2}v_p^2

  v_p=\sqrt{2(\frac{1}{2}u^2+gh)}

  v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}

  v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}

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4 0
2 years ago
A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of
sladkih [1.3K]

Answer:

The magnitude of the resultant of the magnetic field is 4.11\times10^{-5}\ T

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Given that,

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Using formula of magnetic field

B'=\dfrac{\mu_{0}I}{2\pi r}

Where, r = distance

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Using formula of resultant

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B''=\sqrt{(3.7\times10^{-5})^2+(1.8\times10^{-5})^2}

B''=4.11\times10^{-5}\ T

Hence, The magnitude of the resultant of the magnetic field is 4.11\times10^{-5}\ T

6 0
3 years ago
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