An object is pulled with a 85 N force inclined at 27° along a horizontal surface ABC
where AB is smooth but BC is rough.
a) Compute the mass of the object if the object moves with an acceleration of 5
ms-1 along AB.​

Answers

Answer 1

Answer:

m = 15.15 kg

Explanation:

Newton's Second Law of motion states that when an unbalanced force is applied on a body, an acceleration is produced in it in the direction of force. The component of force along the horizontal direction here, will be given by the Newton's Second Law as:

Fx = ma

F Cosθ = ma

where,

F = Magnitude of Force = 85 N

θ = Angle with horizontal = 27°

m = mass of object = ?

a = acceleration of object = 5 m/s²

Therefore,

85 N Cos 27° = m(5 m/s²)

m = 75.73 N/5 m/s²

m = 15.15 kg


Related Questions

A ball rotates on a string as shown below.



If the string in the image above breaks at point 1, which of the following images best shows the direction the ball will travel?

Question 8 options:

Diagram A

Diagram B

Diagram C

Diagram D

Diagram E

Answers

Answer:

Diagram A

Explanation:

When an object is travelling in a circular path it has an acceleration called centripetal acceleration, even if it is moving with uniform speed. This acceleration is attributed to the object due to change in the direction of velocity at every instant during the motion.

The direction of the velocity of the object is tangent to the circular path at every instant and if the object beaks free of the circular motion, such as breaking of the string, then it will go in the direction tangent to the circle at that instant. Hence, the diagram that best shows the direction ball will travel is:

Diagram A

А A pool of water of refractive index
4/3! is 60cm deep. Find its apparent
depth when viewed vertically through
air.​

Answers

Answer:

Apparent depth = 45 cm

Explanation:

The refractive index of water in a pool, n = 4/3

Real depth, d = 60 cm

We need to find its apparent  depth when viewed vertically through  air.​ The ratio of real depth to the apparent depth is equal to the refractive index of the material. Let the apparent depth is d'. So,

[tex]n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{60}{\dfrac{4}{3}}\\\\d'=45\ cm[/tex]

So, the apparent depth is 45 cm.

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