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jarptica [38.1K]
3 years ago
12

A 225 kg block is pulled by two horizontal forces. The first force is 178 N at a 41.7-degree angle and the second is 259 N at a

108-degree angle. What is the x and y component of the total force acting on the block?
Physics
1 answer:
yawa3891 [41]3 years ago
7 0

Answer:

52.9 N, 364.7 N

Explanation:

First of all, we need to resolve both forces along the x- and y- direction. We have:

- Force A (178 N)

A_x = (178 N)(cos 41.7^{\circ})=132.9 N\\A_y = (178 N)(sin 41.7^{\circ})=118.4 N

- Force B (259 N)

B_x = (259 N)(cos 108^{\circ})=-80.0 N\\B_y = (259 N)(sin 108^{\circ})=246.3 N

So the x- and y- component of the total force acting on the block are:

R_x = A_x + B_x = 132.9 N - 80.0 N =52.9 N\\R_y = A_y + B_y = 118.4 N +246.3 N = 364.7 N

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The circuit breaker will tripped. Details below

<h3>What is power? </h3>

This is defined as the rate in which energy is consumed. Electrical power is expressed mathematically as:

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The following data were obtained from the question:

  • Power of heater = 1.50×10³ W
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Total power = power of heat + power of toaster + power of grill

Total power = 1.50×10³ + 750 + 1×10³

Total power = 3250 W

<h3>How to determine whether or not the circuit breaker will tripped</h3>

To determine if the circuit breaker will tripped or not, we shall determine the current used by the appliances. Detail below:

  • Total power (P) = 3250 W
  • Voltage (V)= 120 V
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P = IV

Divide both side by V

I = P / V

I = 3250 / 120

I = 27.08 A

Since the current used by the three appliances (i.e 27.08 A) is greater than the current the circuit breaker (i.e 25 A), can with stand,thus, the circuit breaker will tripped

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Answer:

Distance covered is 37.63 ft

Solution:

As per the question:

Circumference of the bowling ball, C = 27 in

Weight of the bowling ball, w = 14.8 lb

Radius of gyration, k = 3.43 in

Velocity of release of the ball, v' = 23 ft/s = 276 in/s

Coefficient of friction, \mu = 0.19

Now,

We know that the circumference of the circle is given by:

C = 2\pi R

27 = 2\pi R

R = \frac{27}{2\pi } = 4.29\ in

Also, in case of rolling, we know that:

Angular velocity, \omega = \frac{v}{R}

v = \omega R

Now, applying the conservation of angular momentum along the floor:

Initial angular momentum = Final angular momentum

m\omega' = m\omega + I\omega

Moment of Inertia, I = \frac{2}{5}mR^{2} = mk^{2}

mv'R = mvR + mk^{2}\times \frac{v}{R}

v'R = vR + k^{2}\times \frac{v}{R}

276\times 4.29 = 4.29v + 3.43^{2}\times \frac{276}{4.29}

1184.04 - 756.903 = 4.29v

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Now,

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Also, acceleration of the ball can be computed as:

\mu mg = - ma

a = - \mu g = - 0.19\times 386.09 = - 73.357\ in/s^{2}

Now, the distance, d covered by the ball before rolling without slipping:

v^{2} = v'^{2} + 2ad

99.56^{2} = 276^{2} + 2\times (- 73.357)d

99.56^{2} - 276^{2} = 2\times (- 73.357)d

d = 451.65 in = 37.53 ft

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