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ivanzaharov [21]
3 years ago
15

A force of 100 newtons is applied to a box at an angle of 36º with the horizontal. If the mass of the box is 25 kilograms, what

is the horizontal acceleration of the box?
Physics
1 answer:
Furkat [3]3 years ago
5 0
The horizontal acceleration of the box is 3.24 m/s^2.

First you find the horizontal component of the force via cosine 36 degrees.

Then with cosine 36 find the horizontal force by multiplying it with 100 N.

Then divide the horizontal force by the mass -- 25 kgs.  F = ma
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Inertia is defined as the property of matter by which causes it to resist changes in its state of motion such as changes in velocity. From the given options above, the option that has the greatest inertia would be option B. A jet airliner. 
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What is meant by the statement,the linear expansivity of copper is 0.000017k
Aliun [14]

Answer:

The change in length per unit length per degree rise in temperature of copper is 0.000017k

Explanation:

Given that :

The linear expansivity of copper is 0.000017k. This simply means that ; for a given copper length, the length of such copper will increase by 0.000017k for every degree rose in temperature of the copper rod.

Therefore, the change in length per unit length per degree rise in temperature (k) is 0.000017

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one group of atoms on the periodic tables are known as "noble gases". why have they been given this name?​
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3 years ago
A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.10 104 rad/s to an angular speed of 3.14
Anestetic [448]

Answer:

3.63 s

Explanation:

We can solve the problem by using the equivalent SUVAT equations for the angular motion.

To find the angular acceleration, we can use the following equation:

\omega_f^2 - \omega_i ^2 =2 \alpha \theta

where

\omega_f = 3.14\cdot 10^4 rad/s is the final angular speed

\omega_i = 1.10 \cdot 10^4 rad/s is the initial angular speed

\theta= 2.00 \cdot 10^4 rad is the angular distance covered

\alpha is the angular acceleration

Re-arranging the formula, we can find \alpha:

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}=\frac{(3.14\cdot 10^4 rad/s)^2-(1.10\cdot 10^4 rad/s)^2}{2(2.00\cdot 10^4 rad)}=2.16\cdot 10^4 rad/s^2

Now we want to know the time the bit takes starting from rest to reach a speed of \omega_f=7.85\cdot 10^4 rad/s. So, we can use the following equation:

\alpha = \frac{\omega_f-\omega_i}{t}

where:

\alpha=2.16\cdot 10^4 rad/s^2 is the angular acceleration

\omega_f = 7.85\cdot 10^4 rad/s is the final speed

\omega_i = 0 is the initial speed

t is the time

Re-arranging the equation, we can find the time:

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{7.85\cdot 10^4 rad/s-0}{2.16\cdot 10^4 rad/s^2}=3.63 s

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Answer:

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Explanation:

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