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lidiya [134]
3 years ago
15

A 17 kg box experiences an applied force of +175 N and a force of friction of -125 N. While experiencing these unbalanced forces

, the box travels a total of 15.0 meters in 5.25 seconds.
Calculate the power developed by the unbalanced forces to move the box.



A
44.1 watts
B
143 watts
C
263 watts
D
750 watts
Physics
1 answer:
Flura [38]3 years ago
7 0

Answer: (b)

Explanation:

Given

Mass of box m=17\ kg

Applied force F=175\ N

Friction force f=-125\ N

Box travels a distance of s=15\ m

time taken t=5.25\ s

Net unbalanced force F_{net}=175-125=50\ N

Work done by the unbalanced force

\Rightarrow W=F_{un}\cdot s\\\Rightarrow W=50\times 15\\\Rightarrow W=750\ N

Power developed by unbalanced force

\Rightarrow P=\dfrac{W}{t}\\\\\Rightarrow P=\dfrac{750}{5.25}\\\\\Rightarrow P=142.85\approx 1423\ W

Thus, option (b) is correct.

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Answer: because  ν = velocity/λ where ν and λ are the frequency and wavelegth of the wave.

Explanation: In order to explain this problem we have to consider the relationship between frequency and wavelengths which  are related by the velocity of the wave as follows ν*λ=v where ν and λ are the frequency and wavelegth of the wave. These parameters have an inverse proportionality.

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7 0
3 years ago
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I’m not sure how to solve this
spayn [35]

Answer:

Option 10. 169.118 J/KgºC

Explanation:

From the question given above, the following data were obtained:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1.61 KJ

Mass of metal bar = 476 g

Specific heat capacity (C) of metal bar =?

Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:

1 kJ = 1000 J

Therefore,

1.61 KJ = 1.61 KJ × 1000 J / 1 kJ

1.61 KJ = 1610 J

Next, we shall convert 476 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

476 g = 476 g × 1 Kg / 1000 g

476 g = 0.476 Kg

Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1610 J

Mass of metal bar = 0.476 Kg

Specific heat capacity (C) of metal bar =?

Q = MCΔT

1610 = 0.476 × C × 20

1610 = 9.52 × C

Divide both side by 9.52

C = 1610 / 9.52

C = 169.118 J/KgºC

Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC

6 0
2 years ago
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Answer:

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If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a
PolarNik [594]

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

b) the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s w_o = 12 rad/s

Angular Velocity at time 0.15s w_f = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( w_f - w_o ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

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