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antiseptic1488 [7]
3 years ago
9

30 POINTS

Physics
1 answer:
S_A_V [24]3 years ago
8 0

Answer:

13.4s

Explanation:

t = usinx/g. (X= angle , u= initial velocity)

= (147×sin63)/9.8

=13.4

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A man of mass 85 kg runs up a flight of stairs of height 4.6 m in a time period
seraphim [82]

Explanation:

a) Power = work / time = force × distance / time

P = Fd/t

P = (85 kg × 9.8 m/s²) (4.6 m) / (12 s)

P ≈ 319 W

b) P = Fd/t

0.70 (319 W) = (m × 9.8 m/s²) (4.6 m) / (9.6 s)

m = 47.6 kg

7 0
3 years ago
Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot
almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0 (Ec. 1)

\Sigma F_{y} = N - W = 0 (Ec. 2)

Where:

P - Pushing force, measured in newtons.

T - Tension, measured in newtons.

\mu_{k} - Coefficient of kinetic friction, dimensionless.

N - Normal force, measured in newtons.

W - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

P+T -\mu_{k}\cdot W = 0

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

\mu_{k} =\frac{P+T}{m\cdot g}

If we know that P = 400\,N, T = 290\,N, m = 300\,kg and g = 9.807\,\frac{m}{s^{2}}, then:

\mu_{k} = \frac{400\,N+290\,N}{(300\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\mu_{k} = 0.235

The kinetic coefficient of friction of the crate is 0.235.

5 0
3 years ago
A 50 kg pitcher throws a baseball with a mass of 0. 15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is
dsp73

The velocity of the pitcher at the given mass is 0.1 m/s.

The given parameters:

  • <em>Mass of the pitcher, m₁ = 50 kg</em>
  • <em>Mass of the baseball, m₂ = 0.15 kg</em>
  • <em>Velocity of the ball, u₂ = 35 m/s</em>

<em />

Let the velocity of the pitcher = u₁

Apply the principle of conservation of linear momentum to determine the velocity of the pitcher as shown below;

m₁u₁ = m₂u₂

u_1 = \frac{m_2 u_2}{m_1} \\\\u_1 = \frac{0.15 \times 35}{50} \\\\u_1 = 0.105 \ m/s\\\\u_1 \approx 0.1 \ m/s

Thus, the velocity of the pitcher at the given mass is 0.1 m/s.

Learn more about conservation of linear momentum here: brainly.com/question/13589460

4 0
2 years ago
Suppose the initial kinetic energy and final potential energy in an experiment are both zero. What can you conclude?
Kisachek [45]

Initially, the experiment has only potential energy (since total energy is the sum of kinetic and potential energy). And at the end, the experment has only kinetic energy.

5 0
3 years ago
Can someone help me its really hard to do d is stuff​
Tasya [4]

Answer:

direction, speed

means the object is staying still, 0

newtons, N

the sum of all the forces acting on an object

Explanation:

3 0
3 years ago
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