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Minchanka [31]
3 years ago
9

Two 0.60-kilogram objects are connected by a thread that passes over a light, frictionless pulley. The objects are initially hel

d at rest. If a third object with a mass of 0.30 kilogram is added on top of one of the 0.60-kilogram objects and the objects are released, the magnitude of the acceleration of the 0.30-kilogram object is most nearly:______
Physics
1 answer:
Tasya [4]3 years ago
5 0

Answer:

2 m/s²

Explanation:

From the given information:

The first mass m_1 = 0.6 kg

The second mass m_2 = 0.3 kg

The magnitude for the acceleration of 0.3 kg is:

a = net force/ effective mass

Mathematically, it can be computed as follows:

a = \dfrac{F}{m}

a = \dfrac{(m_2 +m_1 -m_1) }{(m_2+m_1+m_1)}(g)

a = \dfrac{0.3 +0.6 -0.6}{(0.3 +0.6+0.6)}(9.8)

a ≅ 2 m/s²

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Answer:

just put the molecules and atoms where they belong

Explanation:

4 0
2 years ago
A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.
Mama L [17]

Answer: f_{r} = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

W_{x} = horizontal component of the weight = mgsinФ

W_{y} = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - f_{r} = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8m/s^{2}

f_{r} = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

v^{2} = u^{2} + 2aS

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}

we slot in a into the equation below to get frictional force

mgsinФ - f_{r} = ma

3 * 9.8 * sin 37 - f_{r} = 3* 0.4

17.9633 - f_{r} =  1.2

f_{r} = 17.9633 - 1.2

f_{r} = 16.49N

4 0
3 years ago
How would I workout question 4?? Please help
Agata [3.3K]

Answer:

550 kg

Explanation:

mass = E / gh

= 33000/60

=550

plzzz......

mark it as a brilliant answer

3 0
3 years ago
If velocity is decreasing,then acceleration:
ziro4ka [17]
If velocity is decreasing, then acceleration is in the direction
opposite to the velocity.

If the object is moving in the direction that you call 'positive',
then acceleration is negative.
5 0
3 years ago
1) A 1,600 kilogram car is also traveling in a straight line. Its momentum is 32,500 kg*m/s. What is the
BaLLatris [955]

Answer:

v = 20.31 m/s

Explanation:

p = mv -> v = p/m = 32,500 kg*m/s / 1,600 kg = 20.31 m/s

4 0
3 years ago
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