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Mamont248 [21]
3 years ago
6

A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to acc

elerate across the ice. If drag and the friction from the ice apply a force of 5 N on the boy, what is the magnitude of the net force acting on him? (a) 70 N (b) 370 N (c) 80 N (d) 380 N
Physics
2 answers:
JulsSmile [24]3 years ago
4 0

Answer:

Choice a. 70\; \rm N, assuming that the skating rink is level.

Explanation:

<h3>Net force in the horizontal direction</h3>

There are two horizontal forces acting on the boy:

  • The pull of his friend, and
  • Frictions.

The boy should be moving in the direction of the pull of his friend. The frictions on this boy should oppose that motion. Therefore, the frictions on the boy would be in the opposite direction of the pull of his friend.

The net force in the horizontal direction should then be the difference between the pull of the friend, and the friction on this boy.

\text{Net force, horizontal} = 75\; \rm N - 5\; \rm N = 70\; \rm N.

<h3>Net force in the vertical direction</h3>

The net force on this boy should be zero in the vertical direction. Consider Newton's Second Law of motion. The net force on an object is proportional to its acceleration. In this question, the net force on this boy in the vertical direction should be proportional to the vertical acceleration of this boy.

However, because (by assumption) the ice rink is level, the boy has no motion in the vertical direction. His vertical acceleration will be zero. As a result, the net force on him should also be zero in the vertical direction.

<h3>Net force</h3>

Therefore, the (combined) net force on this boy would be:

\sqrt{(70\; \rm N)^2 + (0\; \rm N)^2} = 70\; \rm N.

S_A_V [24]3 years ago
4 0
The answer is a hope it helps
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Citrus2011 [14]

Answer:

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3 years ago
A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 430 n. the ropes,
ICE Princess25 [194]
Refer to the diagram shown below.

For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂                (1)

For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392        (2)

Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N

Answer:
T₂ = 339.06 N
T₃ = 276.57 N

7 0
3 years ago
A simple pendulum has length of 820mm. Calculate the frequency (g = 9.8 ms -2)<br>​
Vadim26 [7]

Answer:

\huge\boxed{\sf f=0.55 \ Hz}

Explanation:

<u>Given Data:</u>

Length = l = 820 mm = 0.82 m

Acceleration due to gravity = g = 9.8 ms⁻²

<u>Required:</u>

Frequency = f = ?

<u>Formula:</u>

\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} }

<u>Solution:</u>

\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}

7 0
2 years ago
A hot drink in a room is at a temperature of 20 C
IRISSAK [1]

Answer: the drink cooled down to the same temperature as the room it was in.

Explanation: due to the change and temperature of the room, it cause the heat particles to exchange, causing it cool

8 0
3 years ago
A milk truck carries milk with density 64.6 lb/ft3 in a horizontal cylindrical tank with diameter 12 ft. (a) Find the force exer
7nadin3 [17]

Answer:

F = 351×10³lb

Explanation:

Given the density

ρg = 64.6lb/ft³

Diameter d = 12ft

The tank is horizontally cylindrical. The vertical distance from the top to the bottom of the tank is h = 12ft

The pressure in the tank is

P = ρgh = 64.6 × 12 = 775.2lb/ft²

The force exerted on one end of the tank is therefore F = PA = 775.2 × πd² = 775.2π×12²

F = 351×10³lb.

4 0
3 years ago
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