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FrozenT [24]
3 years ago
5

An object of mass 10 kg is at rest when a net force of 20 N acts on it in a direction east. The acceleration produced on the obj

ect is?
a) 2 m/s^2 in the direction east b) 2 m/s^2 in the direction west c) 0.5 m/s^2 in the direction east d) 0.5 m/s^2 in the direction west e) 200 m/s^2 in the direction east
Physics
1 answer:
torisob [31]3 years ago
6 0
It's a since a = F / m
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A thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with
NNADVOKAT [17]

Answer: 1.59 sec

Explanation:

The kinetic energy contained in the rotation of the cylinder is 1/2 m v^2

The kinetic energy of translation is also 1/2 m v^2

so the total energy is m v^2

The force applied is mg sin (theta)

= m x 9.8 x 1/2

= 4.9 m

Now equate

F x d = m v^2

4.9 m x 3.1 = m v^2

v^2 = 4.9 x 3.1

v = sqrt(4.9 x 3.1) = 3.9 m/s

Acceleration.

V^2 = 2 a s

4.9 x 3.1 = 2 x a x 3.1

4.9 = 2 a

a = 2.45 m/s^2

Time

T = v/a

= 3.9/2.45

= 1.59 sec

6 0
4 years ago
Two weights are connected by a very light cord that passes over an 80.0Nfrictionless pulley of radius 0.300m. The pulley is a so
Citrus2011 [14]

Answer:

The force does the ceiling exert on the hook is 269.59 N

Explanation:

Applying the second Newton law:

F = m*a

From the attached diagram, the net force in object 1 is:

m_{1} a=T_{1} -W_{1}

In object 2:

m_{2} a=W_{2} -T_{2}

Adding the two equations:

m_{2} a+m_{1} a=T_{1} -W_{1} +W_{2} -T_{2} \\m_{1} =\frac{W_{1} }{g} \\m_{2} =\frac{W_{2} }{g} \\Replacing\\T_{2}-T_{1}=W_{2}   -W_{1} -(\frac{W_{1} }{g} +\frac{W_{2} }{g} )a  (eq. 1)

The torque:

\tau =I\alpha

Where

I = moment of inertia

α = angular acceleration

If the linear acceleration is

a=r\alpha \\\alpha =\frac{a}{r} \\I=\frac{1}{2} mr^{2} \\\tau =\frac{mra}{2}

Torque due the tension is equal:

\tau =r(T_{2} -T_{1} )

Substituting torque, mass, in equation 1, the expression respect the acceleration is:

a=\frac{g*(W_{2}-W_{1})}{W_{1}+W_{2} +\frac{W}{2} }

Where

W₁ = 75 N

W₂ = 125 N

W = 80 N

a=\frac{9.8*(125-75)}{75+125+\frac{80}{2} } =2.04m/s^{2}

The net force is:

F_{n} =F-W-T_{1} -T_{2}\\0=F-W-W_{1} (\frac{a}{g} +1)-W_{2} (1-\frac{a}{g})\\F=W+W_{1} +W_{2} +\frac{a}{g} (W_{1} -W_{2} )\\F=80+75+125+\frac{2.04}{9.8} (75-125)\\F=269.59N

4 0
3 years ago
What are the layers of the earth
AleksAgata [21]

Answer:

CRUST, UPPER MANTLE, LOWER MANTLE, INNER CORE, OUTER CORE,

Explanation:

how do u not know this

8 0
3 years ago
PLEASE HELP ME on this question.
VladimirAG [237]
The second one is the answer
4 0
3 years ago
Help me please asap :(
mixer [17]
17. a
18. c
19. a
20. c
21. c
8 0
3 years ago
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