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Andre45 [30]
3 years ago
9

Which object will have the greatest acceleration if acted upon by a net force of 100N?

Physics
1 answer:
VashaNatasha [74]3 years ago
5 0

Answer:

The object with the greatest acceleration is the one with the mass of 5 kg. Option A

Explanation:

The conclusion is easy for getting because we can use all the given options in the equation of newton for the mass, the force, and the acceleration. We can all the results in the attached image.

Note: We can realize that for a given force, while the mass of the object is less it will be more easy to accelerate it.

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Answer:

Answer:B

Explanation:

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The exact speed of an object in a specific instant is?
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In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i
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a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is 2.7 rad/s^2

Explanation:

a)

In a uniform circular motion, the centripetal acceleration is given by

a_c=\omega^2 r

where:

\omega is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

\omega=1.7 rad/s is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity (g=9.8 m/s^2), so:

a_c=3.2 g = 3.2(9.8)=31.4 m/s^2

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m

b)

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

a=4.4 g

So, 4.4 times the acceleration due to gravity.

The total acceleration is the resultant of the centripetal acceleration (a_c) and the tangential acceleration (a_t):

a=\sqrt{a_c^2+a_t^2}

We know that:

a = 4.4g

a_c = 3.2 g

So, we can find the tangential acceleration:

a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2

The angular acceleration is related to the tangential acceleration by

\alpha = \frac{a_t}{r}

where r = 10.9 m is the length of the centrifuge. Substituting,

\alpha = \frac{29.6}{10.9}=2.7 rad/s^2

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Two long, parallel wires separated by 2.00 cm carry currents in opposite directions. The current in one wire is 1.75 A, and the
Natasha_Volkova [10]

The force per unit length between the two wires is 6.0\cdot 10^{-5} N/m

Explanation:

The magnitude of the force per unit length exerted between two current-carrying wires is given by

\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}

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I_1, I_2 are the currents in the two wires

r is the separation between the two wires

For the wires in this problem, we have

I_1 = 1.75 A

I_2 = 3.45 A

r = 2.00 cm = 0.02 m

Substituting into the equation, we find

\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(1.75)(3.45)}{2\pi (0.02)}=6.0\cdot 10^{-5} N/m

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