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Tpy6a [65]
2 years ago
15

11. A box with a mass of 1.0 kg is resting on a horizontal surface and the coefficient of friction between the block and the sur

face is 0.20. It is accelerated by attaching a 1.5 kg mass, as shown in the diagram. Assume that the pulley is frictionless and that the cord has negligible mass.
a. What is the acceleration of the box? What is the tension?
Physics
2 answers:
Volgvan2 years ago
6 0

Answer:

A force is applied acting to the right. Assume that friction is negligible. Ikg 0.5N. For each question, one or more features of the system.

Explanation:hope this helps

AlladinOne [14]2 years ago
5 0

Answer:

We know that the force pulling the box in the positive x direction has a magnitude of m g sin 30 . Using Newtons Second Law, F = ma , we just need to solve for a :

ma=mgsin30

a=gsin30

=(10m/s2)(0.500)

=5m/s2

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A ball with a momentum of 16 kg.M/s strikes a ball at rest. What is the total momentum of both the balls after the collision.
melomori [17]

Answer:

Total momentum = 16 Kgm/s

Explanation:

Let the momentum of the two balls be A and B respectively.

Momentum A = 16 kgm/s

Momentum B = 0 kgm/s (since the ball is at rest).

Total momentum = A + B

Total momentum = 16 + 0

Total momentum = 16 Kgm/s

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

Momentum = mass * velocity

8 0
3 years ago
Read 2 more answers
You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo
jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

\Delta E_k=W=W_g-W_p

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J

hence, the change in Er is about 1.52J times the initial rotational energy

5 0
3 years ago
Read 2 more answers
Explain the costruction and working of windmill
fiasKO [112]
A windmill produces wind without electricity. wind blows on a windmill and it and that’s how it produces wind. You can put a windmill in anywhere outdoors (ex. Deserts or plains)
—————————————————————


Hope this helped you:)
6 0
3 years ago
Scientists hypothesize that there are no dinosaurs alive today because
Wittaler [7]
Its c after the meteorite hit earth it created a different environment worldwide that animals weren't able to adapt to due to its harsh conditions 
8 0
3 years ago
Read 2 more answers
Two vectors of magnitudes 30 units and 70 units are added to each other. What are possible results of this addition? (section 3.
Yanka [14]

Answer:

the correct answer is option C which is 50 units.

Explanation:

given,

two vector of magnitude = 30 units and of 70 units

to calculate resultants vector = \sqrt{a^2+b^2+2 a b cos\theta}

cos θ value varies from -1 to 1

so, resultant vector

=\sqrt{a^2+b^2-2 a b cos\theta}\ to\ \sqrt{a^2+b^2+ 2 a b cos\theta}

a = 30 units    and  b = 70 units

= \sqrt{30^2+70^2-2\times 30\times 70}\ to\ \sqrt{30^2+70^2+2\times 30\times 70}

=   40 units to 100 units

hence, the correct answer is option C which is 50 units.

                       

4 0
3 years ago
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