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Hunter-Best [27]
2 years ago
7

1. A block is pulled to the right at constant velocity by a 20N force acting at 30o above the horizontal. If the coefficient of

sliding friction is 0.5, what is the weight of the block?
explain briefly?
Physics
2 answers:
DiKsa [7]2 years ago
8 0

Answer:

44.6 N

Explanation:

Draw a free body diagram of the block.  There are four forces on the block:

Weight force mg pulling down,

Normal force N pushing up,

Friction force Nμ pushing left,

and applied force F pulling right 30° above horizontal.

Sum of forces in the y direction:

∑F = ma

N + F sin 30° − mg = 0

N = mg − F sin 30°

Sum of forces in the x direction:

∑F = ma

F cos 30° − Nμ = 0

F cos 30° = Nμ

N = F cos 30° / μ

Substitute:

mg − F sin 30° = F cos 30° / μ

mg = F sin 30° + (F cos 30° / μ)

Plug in values:

mg = 20 N sin 30° + (20 N cos 30° / 0.5)

mg = 44.6 N

Free_Kalibri [48]2 years ago
5 0

Answer:

24.64N

Explanation:

The force inclined at 30° exerted on the block has two components the horizontal and vertical components .

The vertical component force is ;

FsinA; where A is angle of inclination.

20sin30° = 10N

The horizontal component is ;

20cos30° = 17.32N

Now the normal reaction is a vertical upward force acting opposite to the weight.

Note also that for the body to slide it has to overcome a limiting Frictional force and this force must be 17.32N since it's this horizontal component that makes the body slide.

We can then go on to compute the normal reaction from the coefficient of friction formula given as;

Limiting Frictional Force/ Normal reaction

By change of subject formula;

Normal reaction = Limiting frictional force /coefficient of friction

By substituting the equivalent values of limiting Frictional force and coefficient of friction, we have:

17.32N/0.5 = 34.64N

Now since the object didn't fall downwards it means the sum of upward force is the same as the sum of down ward force.

The downward forces are the weight of the object and the vertical component of the applied force when the normal reaction is the upward force.

Expressed mathematically, we have:

34.64= 10 + weight of object

Hence weight of object = 34.64-10= 24.64N

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A 52 kg child on a swing is travelling at 6 m/s . What is his gravitational potential energy if he has 1000 J of the mechanical
DiKsa [7]

Answer:

The correct answer is "64 J".

Explanation:

The given values are:

Mass,

m = 52 kg

Velocity,

v = 6 m/s

Mechanical energy,

= 1000 J

Now,

The gravitational potential energy will be:

⇒ P.E=1000-\frac{1}{2}mv^2

           =1000-\frac{1}{2}\times 52\times (6)^2

           =1000-26\times 36

           =1000-936

           =64 \ J

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Read 2 more answers
Cohort studies allow the investigator to examine multiple outcomes and multiple exposures. Consider the following three exposure
mojhsa [17]

Answer:

See the explanation for the answer.

Explanation:

                          <u>Cohort study for smoking</u>

                      <u>Outcomes that were examined</u>

1. People at risk of lung cancer due to smoking

2. Influence of outdoor air pollution in lung cancer

3. Variation in tobacco smoking and heavy tobacco smoking was observed.

4. Occupational factors and housing factor were observed.

                       

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2. Observing value in summer.

3. Correlating vitamin D value with weather and age

4. 25-hydroxy vitamin D was observed.

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                      <u>Outcomes that were examined</u>

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3 0
3 years ago
Two Physics quick help
Free_Kalibri [48]

Answer:

21.3 V, 1.2 A

Explanation:

1.

These resistors are in series, so the net resistance is:

R = R₁ + R₂ + R₃

R = 20 + 30 + 45

R = 95

So the current is:

V = IR

45 = I (95)

I = 9/19

So the voltage drop across R₃ is:

V = IR

V = (9/19) (45)

V ≈ 21.3 V

2.

First, we need to find the equivalent resistance of R₂ and R₃, which are in parallel:

1/R₂₃ = 1/R₂ + 1/R₃

1/R₂₃ = 1/10 + 1/10

R₂₃ = 5

Now we find the overall resistance by adding the resistors in series:

R = R₁ + R₂₃ + R₄

R = 10 + 5 + 10

R = 25

So the current through R₁ is:

V = IR

30 = I (25)

I = 1.2 A

7 0
3 years ago
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