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andreyandreev [35.5K]
3 years ago
7

Acording to Newton’s second law of motion if a rigid body of unchanging mass is observed accelerating what must be happening

Physics
2 answers:
blondinia [14]3 years ago
6 0

A force is being applied to the mass.

Kitty [74]3 years ago
5 0

The net force on the object can't be zero.

The forces acting on it are not balanced.

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A mass m = 75 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.2 m and finally
dalvyx [7]

Answer:

The velocity is 19.39 m/s

Solution:

As per the question:

Mass, m = 75 kg

Radius, R = 19.2 m

Now,

When the mass is at the top position in the loop, then the necessary centrifugal force is to keep the mass on the path is provided by the gravitational force acting downwards.

F_{C} = F_{G}

\frac{mv^{2}}{R} = mg

where

v = velocity

g = acceleration due to gravity

v = \sqrt{2gR} = \sqrt{2\times 9.8\times 19.2} = 19.39\ m/s

4 0
3 years ago
Which would take more force to stop in 10 seconds: an 8.0-kilogram ball rolling in a straight line at a speed of 0.2 m/sec or a
gayaneshka [121]
I use the impulse momentum formula.
the 4.0 kilogram ball requires more force to stop

6 0
3 years ago
Read 2 more answers
I need help in this pls i really need a answer
IrinaK [193]

Answer:

In terms of distance, average speed is 20 km/h

In terms of displacement, average speed is 0 km/h.

Explanation:

Total distance:

= (40.0 - 10.0) + (20.0 - 10.0) + (40.0 - 20.0) \\  = 30.0 + 10.0 + 20.0 \\  = 60.0 \: km

Total time is 3.0 hours

but:

average \: speed =  \frac{total \: distance}{total \: time}  \\

In terms of distance.

substitute:

average \: speed =  \frac{60.0}{3.0}  \\  \\  = 20 \:  {kmh}^{ - 1}

displacement = ( - 30.0) + 10 .0+ 20.0 \\  = 0

In terms of displacement:

speed =  \frac{0}{3}  \\  \\  = 0 \:  {kmh}^{ - 1}

3 0
3 years ago
You could use newton’s second law to calculate the force applied to an object if you knew the objects mass and its _____.
olga2289 [7]

Answer:

You could use newton’s second law to calculate the force applied to an object if you knew the objects mass and its <u>acceleration.</u>

Explanation:

By, Newtons second law, the force applied on an object directly varies with the acceleration caused and the mass of the object.

This is given by :

F=m\ a

Where F represents force applied on the object , m represents mass of the object and a represents the acceleration.

In order to calculate force applied on object we require the mass of the object and its acceleration. The force can be calculated by finding the product of mass and acceleration of the object.

4 0
3 years ago
The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by
Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, \mu = 0.2

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, f_{y} = 40sin \theta

\sum f(y) = 0

N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N

\sum f(x) = 0

40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}

v^{2}  = u^{2} + 2as\\u = 0 m/s\\v^{2}  =  2 * 0.731 * 5\\v^{2}  = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s

Power, P = Fvcos \theta

P = 40 *2.704 cos60\\P = 54.074 W

7 0
3 years ago
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