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2 years ago

If you apply 100.0 N of force to lift an object with a single, fixed pulley, then what is the resistive force?

1 answer:

6 0

<span>If you apply 100.0 N of force to lift an object with a single, fixed pulley, then the resistive force is also equivalent to 100 Newtons of force. Since the weight of the object was not mentioned, it is assumed that it has already been taken into account in the 100 N value of force. This follows Newton's law of motion of equal action and reaction.</span>

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The air around a pool and the water in the pool receive equal amounts of energy from the sun. Why does the air experience a grea

labwork [276]

Answer:

Explanation:

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1 year ago

A skier is moving down a snowy hill with an acceleration of 0.40 m/s2. The angle of the slope is 5.0∘ to the horizontal. What is

kirill115 [55]

Answer:

1.25377 m/s²

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

$\mu$ = Coefficient of friction

$\theta$ = Slope

From Newton's second law

$mgsin\theta-f=ma\\\Rightarrow mgsin\theta-\mu mgcos\theta=ma\\\Rightarrow \mu=\frac{gsin\theta-a}{gcos\theta}\\\Rightarrow \mu=\frac{9.81\times sin5-0.4}{9.81\times cos5}\\\Rightarrow \mu=0.04655$

Applying $\mu$ to the above equation and $\theta=10^{\circ}$

$mgsin\theta-\mu mgcos\theta=ma\\\Rightarrow a=gsin\theta-\mu gcos\theta\\\Rightarrow a=9.81\times sin10-0.04655\times 9.81\times cos10\\\Rightarrow a=1.25377\ m/s^2$

The acceleration of the same skier when she is moving down a hill is 1.25377 m/s²

3 0

2 years ago

Choose the correct statement of Kirchhoff's voltage law.

frutty [35]

Answer:

A) If one travels around a closed path adding the voltages for which one enters the negative reference and subtracting the voltages for which one enters the positive reference, the total is zero.

Explanation:

Kirchhoff's voltage law deals with the conservation of energy when the current flows in a closed-loop path.

It states that the algebraic sum of the voltages around any closed loop in a circuit is always zero.

In other words, the algebraic sum of all the potential differences through a loop must be equal to zero.

3 0

2 years ago

A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve

Troyanec [42]

The velocity of tennis racket after collision is 14.96m/s

<u>Explanation:</u>

Given-

Mass, m = 0.311kg

u1 = 30.3m/s

m2 = 0.057kg

u2 = 19.2m/s

Since m2 is moving in opposite direction, u2 = -19.2m/s

Velocity of m1 after collision = ?

Let the velocity of m1 after collision be v

After collision the momentum is conserved.

Therefore,

m1u1 - m2u2 = m1v1 + m2v2

$v1 = (\frac{m1-m2}{m1+m2})u1 + (\frac{2m2}{m1+m2})u2$

$v1 = (\frac{0.311-0.057}{0.311+0.057})30.3 + (\frac{2 X 0.057}{0.311 + 0.057}) X-19.2\\\\v1 = (\frac{0.254}{0.368} )30.3 + (\frac{0.114}{0.368}) X -19.2\\ \\v1 = 20.91 - 5.95\\\\v1 = 14.96$

Therefore, the velocity of tennis racket after collision is 14.96m/s

7 0

2 years ago

A child of mass m is at the edge of a merry-go-round of diameter d. When the merry-go-round is rotating with angular acceleratio

dem82 [27]

Answer:

The torque on the child is now the same, τ.

Explanation:

It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.
When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:

$I_{to} = I_{d} + m*r^{2} = m*\frac{r^{2}}{2} + m*r^{2} = \frac{3}{2}* m*r^{2} (1)$

So, τ = 3/2*m*r²*α (2)

When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:

$I_{t} = I_{d} + m*\frac{r^{2}}{4} = m*\frac{r^{2}}{2} + m*\frac{r^{2}}{4} = \frac{3}{4}* m*r^{2} (3)$

Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:

τ = 3/4*m*r² * (2α) = 3/2*m*r²

same result than in (2), so the torque remains the same.

7 0

2 years ago