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nasty-shy [4]
2 years ago
9

A 5.00 kg crate is on a 21.0° hill.

Physics
1 answer:
WARRIOR [948]2 years ago
4 0

Answer:

17.56 N

Explanation:

Given that,

Mass of a crate, m = 5 kg

It is on a 21.0° hill

We need to find the y component of the weight.

y component = mgsinθ

Put all the values,

y-component = 5×9.8×sin(21)

= 49×sin(21)

= 17.56 N

So, the y-component of the weight is 17.56 N.

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50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the bui
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Answer:

98.13m

Explanation:

Complete question

Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building

CHECK THE ATTACHMENT

From the figure, using trigonometry

Tan(θ ) = opposite/adjacent

Where Angle (θ )= 63°

Opposite= X = height of the building

Adjacent= 50 m

Then substitute the values we have

Tan(63)= X/50

1.9626= X/50

X= 1.9626 × 50

X= 98.13m

Hence, the height of the building is 98.13m

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2 years ago
A sports car accelerates from 0 to 25 meters per second in 4 seconds. What is its acceleration?
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6.25 ms²

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Four solid plastic cylinders all have radius 2.41 cm and length 5.94 cm. Find the charge of each cylinder given the following ad
Paladinen [302]

Answer:

Check explanation

Explanation:

QUICK NOTE: THE QUESTION IS NOT COMPLETE. Although it is not, we can make assumptions, since we only need values for the UNIFORM CHARGE DENSITY.

SO, LET US BEGIN;

To solve this question we are to use the equation (1) below;

Charge,Q = uniform charge density,p × Total area of the cylinder,A ------------------------------------------------------------------------(1).

From the question, we are given radius, R to be 2.41 cm and length, L to be 5.94 cm.

Step one: calculate for the total area of the cylinder, A.

Total area of the cylinder, A= area of the top surface + area of the buttom + area of the curved surface of the cylinder.

Hence, total area of the cylinder,A is;

==> πR^2 + πR^2 + 2πRL. -------------------------------------------------------------------------(2).

Then, total area of the cylinder,A is;

==> (L + R)2πR.

Step two: find the charge of each cylinder.

===> For the first cylinder; we have the uniform charge density to be 35 nC/m^2.

Therefore, the combination of equation (1) and (3) gives;

Charge Q= p × (L + R)2πR...----------------------------(4)

Hence, Q= 35 × [(5.94 + 2.41) 2× 3.143 × 5.94].= 10912.615 coulumb.

====> For the second cylinder, we have a uniform charge density of 50 nC/m^2.

Using equation (4), charge,Q= 15,589.45 Coulumb

=====> For THE third cylinder, the uniform charge density is 600, we make use of equation (4);

Charge,Q= 600×311.789.

Charge,Q= 187,073.4 coulumb.

====> For THE fourth cylinder, the uniform charge density is 750 nC/m^2.., we make use of equation (4);

Charge,Q= 233,841.75 coulumb.

7 0
3 years ago
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