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Montano1993 [528]
3 years ago
6

A 1-kg collar (located at point (2,2) from the origin) is pulled along a vertical, frictionless bar with a force of 10 N applied

along string A at the point (5,8). Calculate the acceleration of the collar in the direction of the beam.
Physics
1 answer:
faltersainse [42]3 years ago
3 0

Answer:

The acceleration of the collar is 10 m/s²

Explanation:

Given;

mass of the collar, m = 1 kg

applied force on the bar, F = 10 N

The acceleration of the collar can be calculated by applying Newton's second law of motion;

F = ma

where;

F is the applied force

m is mass of the object

a is the acceleration

a = F / m

a = 10 / 1

a = 10 m/s²

Therefore, the acceleration of the collar is 10 m/s²

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8 0
3 years ago
[1] The assembly starts from rest and reaches an angular speed of 150 rev/min under the action of a 20-N force T applied to the
ExtremeBDS [4]

Answer:

t = 5.89 s

Explanation:

To calculate the time, we need the radius of the pulley and the radius of the sphere which was not given in the question.

Let us assume that the radius of the pulley (r_p) = 0.4 m

Let the radius of the sphere (r) = 0.5 m

w = angular speed = 150 rev/min = (150 × 2π / 60) rad/s = 15.708 rad/s

Tension (T) = 20 N

mass (m) = 3 kg each

\int\limits^0_t {Tr_p} \, dt=H_2-H_1\\( Tr_p)t=4rm(rw)\\( Tr_p)t=4r^2mw

t = \frac{4r^2mw}{Tr_P}

Substituting values:

t = \frac{4r^2mw}{Tr_P}= \frac{4*(0.5)^2*3*15.708}{20*0.4}=5.89s

7 0
3 years ago
Read 2 more answers
The boiling point for liquid nitrogen at atmospheric pressure is 77k. Is the temperature of an open container of liquid nitrogen
algol13
Answer:
It would be lower than because, if the boiling point of that element is 77 Kelvin degrees then if it isn’t at boiling point it would automatically be cooler than that. Even if it’s at its original state. The normal temperature of Liquid Nitrogen is really cold -320.8 degrees.

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8 0
3 years ago
A car has a mass of 1000 kg. What is the acceleration produced by a force of 2000 N?
EastWind [94]
F=ma
a=F/m
a=2000/1000
a=2 m/s^2
6 0
2 years ago
Problem: Suppose that the rate of flow of water through a fire hose is 21.4 kg/s and the stream of water from the hose moves at
Sloan [31]

Answer:

The force exerted is 318.86 N

Explanation:

The force exerted by such a stream is calculated by multiplying the mass flow rate of water by the velocity of the stream of water.

mass flow rate = 21.4 kg/s

velocity = 14.9 m/s

Force exerted = 21.4 kg/s × 14.9 m/s = 318.86 kgm/s^2 = 318.86 N

6 0
3 years ago
Read 2 more answers
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