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yan [13]
3 years ago
10

Find the magnitude of the torque produced by a 4.5 N force applied to a door at a perpendicular distance of 0.26 m from the hing

e. Answer in units of N · m.
Physics
1 answer:
Alex787 [66]3 years ago
3 0

Answer:

The required torque is 1.17 N-m.

Explanation:

The given data :-

The magnitude of force ( F ) = 4.5 N.

The length of arm ( r ) = 0.26 m.

Here given that force is applied at perpendicular means ( ∅ ) = 90°.

The torque ( T ) is given by

T = F * r * sin∅

T = 4.5 * 0.26 * sin 90°

T = 4.5 * 0.26 * 1

T = 1.17 N-m.

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Answer:

The answer is "5.53 \ \frac{m}{s}"

Explanation:

apply the formula for calculating the average velocity to the relative air

V_{PG} =V_{PA}+V_{AG}

\Rightarrow  V_{PA} = V_{PG} -V_{AG}

Given value:

V_{AG} = -2 \ \frac{m}{s}

V_{PG} =3.53

\Rightarrow  V_{PA} = 3.53 - (-2) \\\\\Rightarrow  V_{PA} = 3.53 +2 \\\\\Rightarrow  V_{PA} = 5.53  \\\\

The final answer is "5.53 \ \frac{m}{s}" in the south-east direction.

4 0
2 years ago
A force of 6600 N is exerted on a piston that has an area of 0.010 m2
sveticcg [70]

Answer:

Choice A: approximately 0.015\; \rm m^2, assuming that the two pistons are connected via some confined liquid to form a simple machine.

Explanation:

Assume that the two pistons are connected via some liquid that is confined. Pressure from the first piston:

\displaystyle P_1 = \frac{F_1}{A_1} = \frac{6.600\times 10^3\; \rm N}{1.0\times 10^{-2}\; \rm m^{2}} = 6.6\times 10^{5}\; \rm N \cdot m^{-2}.

By Pascal's Principle, because the first piston exerted a pressure of 6.6\times 10^{5}\; \rm N \cdot m^{-2} on the liquid, the liquid will now exert the same amount of pressure on the walls of the container.

Assume that the second piston is part of that wall. The pressure on the second piston will also be 6.6\times 10^{5}\; \rm N \cdot m^{-2}. In other words:

P_2 = P_1 = 6.6\times 10^{5}\; \rm N \cdot m^{-2}.

To achieve a force of 9.900 \times 10^3\; \rm N, the surface area of the second piston should be:

\displaystyle A_2 = \frac{F_2}{P_2} = \frac{9.900\times 10^{3}\; \rm N}{6.6\times 10^5\; \rm N \cdot m^{-2}} \approx 0.015\; \rm m^{2}.

4 0
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What does the zigzag line in the circuit diagram represent? A. a source of electrical energy B. an electrical conductor C. an el
lora16 [44]
C. an electrical resistor 
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A very light ping-pong ball moving east at a speed of 4 m/s collides with a very heavy stationary bowling ball. The Ping-Pong ba
KengaRu [80]

Answer:

They experience the same magnitude impulse

Explanation:

We have a ping-pong ball colliding with a stationary bowling ball. According to the law of conservation of momentum, we have that the total momentum before and after the collision must be conserved:

where is the initial momentum of the ping-poll ball

is the initial momentum of the bowling ball (which is zero, since the ball is stationary)

is the final momentum of the ping-poll ball

is the final momentum of the bowling ball

We can re-arrange the equation as follows or

which means (1) so the magnitude of the change in momentum of the ping-pong ball is equal to the magnitude of the change in momentum of the bowling ball.

However, we also know that the magnitude of the impulse on an object is equal to the change of momentum of the object:

(2) therefore, (1)+(2) tells us that the ping-pong ball and the bowling ball experiences the same magnitude impulse:

3 0
3 years ago
The force of gravity on a 2-kg rock is twice as great as that on a 1-kg rock. why then doesn't the heavier rock fall faster?
katrin2010 [14]
It takes twice the force to produce the same acceleration in the 2kg rock. 
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