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Ede4ka [16]
3 years ago
15

PLEASE ANSWER FAST In which of the following situations is the greatest amount of work accomplished? 1. A boy lifts a 2-newton b

ox 0.8 meters. 2. A boy lifts a 5-newton box 0.8 meters. 3.A boy lifts a 8-newton box 0.2 meters. 4.A boy lifts a 10-newton box 0.2 meters.
Physics
1 answer:
GuDViN [60]3 years ago
3 0

Explanation:

Work done is given by the product of force and displacement.

Case 1,

1. A boy lifts a 2-newton box 0.8 meters.

W = 2 N × 0.8 m = 1.6 J

2. A boy lifts a 5-newton box 0.8 meters.

W = 5 N × 0.8 m = 4 J

3. A boy lifts a 8-newton box 0.2 meters.

W = 8 N × 0.2 m = 1.6 J

4. A boy lifts a 10-newton box 0.2 meters.

W = 10 N × 0.2 m = 2 J

Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.

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A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 ×
HACTEHA [7]

Answer:

f_r = 150.47 N

Explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

\dfrac{mv^2}{r}= N sin \theta + f_r cos \theta

∑ F_y = 0

0 = N cos \theta - f_r sin \theta - mg

N = \dfrac{f_rsin \theta + mg}{cos \theta}

\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta

              = f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta

        f_r = \dfrac{\dfrac{mv^2}{r}- mg tan\theta}{sin\theta tan \theta + cos \theta}

         f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}

f_r = 150.47 N

8 0
3 years ago
A point charge with charge q1 = 4.00 μC is held stationary at the origin. A second point charge with charge q2 = -4.40 μC moves
Bezzdna [24]

Answer:

W=0.94J

Explanation:

Electrostatic potential energy is the energy that results from the position of a charge in an electric field. Therefore, the work done to move a charge from point 1 to point 2 will be the change in electrostatic potential energy between point 1 and point 2.

This energy is given by:

U=\frac{K\left |q_1 \right |\left |q_2 \right |}{r}\\

So, the work done to move the chargue is:

W=U_1-U_2\\W=\frac{K\left |q_1 \right |\left |q_2 \right |}{r_1}-\frac{K\left |q_1 \right |\left |q_2 \right |}{r_2}\\r_1=\sqrt{((0.155 m)^2+0 m)^2}=0.115m\\r_2=\sqrt{((0.245 m)^2+(0.270 m)^2}=0.365m\\W=K\left |q_1 \right |\left |q_2 \right |(\frac{1}{r_1}-\frac{1}{r_2})\\W=8.99*10^9\frac{Nm^2}{c^2}(4.00*10^{-6}C)(4.40*10^{-6}C)(\frac{1}{0.115m}-\frac{1}{0.365})\\W=0.94J

The work is positive since the potential energy in 1 is greater than 2.

5 0
4 years ago
A 60 kg runner accelerates during a race at 3m/s2. what force is exerted on the earth with each step
marusya05 [52]

Answer:

180 Newton(N)

Explanation:

force =mass *acceleration

=60 * 3

=180 kgm/s^2

=180 N

6 0
3 years ago
What formula is used to calculate resistance using the color code?
jolli1 [7]
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bazaltina [42]

Answer:

if one bulb burns out the other bulbs will also turn off because they are connected to each other.

8 0
3 years ago
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