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Varvara68
8 days ago
12

Tori experiments with pulleys in physics class. She applies 70 newtons of force to a single pulley to lift a bowling ball. By ad

ding another pulley, she’d get a mechanical advantage of 2. What’s the minimum input force she’ll need to lift the ball if she adds another pulley?
Physics
2 answers:
Softa [2.6K]8 days ago
7 0

Result: 35 N

I hope this is useful!:D <3

-Maggie

Yuliya22 [2.9K]8 days ago
6 0

The minimum force required to lift the ball is 35 N.

Explanation:

The mechanical advantage provided by a single pulley is 1. Given that she exerts 70 N to raise the bowling ball, the weight of the ball (output force) is also 70 N.

Now, introducing another pulley gives a mechanical advantage of 2. Thus,

M.A = (Output Force)/(Input Force)

By substituting the values we find

2=\frac{70}{F_{i}}

F_{i}=\frac{70}{2} = 35 N

This means that an input force of 35 N is required to lift the ball.

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You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.
serg [3222]

Answer:

d = 2021.6 km

Explanation:

This distance problem can be solved using vector analysis; it's best to find each plane's position components before applying the Pythagorean theorem to calculate the separation between them.

For Airplane 1:

Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m = 7607 m

For Plane 2:

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 sin 25 = 8.452 103 m = 8452 m

To determine the distance between the planes using the Pythagorean theorem:

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Now, we perform the calculations:

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 + 9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

7 0
24 days ago
The Hall effect can be used to calculate the charge-carrier number density in a conductor. If a conductor carrying a current of
kicyunya [2911]

Answer:

6.6*10^{27}e/m^3

Explanation:

When calculating Hall voltage, it is crucial to have the current, magnetic field strength, length, area, and number of charge carriers available. The Hall voltage can be expressed using the equation:

V_h = \frac{iB}{neL}

Where:

i= the current

B= the magnetic field strength

L = the length

n = the number of charge carriers

e= charge of an electron

We need to replace values and solve for n:

n= \frac{iB}{V_h e L}

n= \frac{2*1.2}{4.5*10^{-6}*5^10^{-3}*1.6*10^{-19}}

n= 6.6*10^{27}electron.m^{-3}

As a result, the charge carrier density is 6.6*10^{27}e/m^3

5 0
1 month ago
An axle passes through a pulley. Each end of the axle has a string that is tied to a support. A third string is looped many time
Keith_Richards [2877]

Answer:

ΔL = MmRgt / (2m + M)

Explanation:

The system starts from rest, so the change in angular momentum correlates directly to its final angular momentum.

ΔL = L − L₀

ΔL = Iω − 0

ΔL = ½ MR²ω

To determine the angular velocity ω, begin by drawing a free body diagram for both the pulley and the block.

For the block, two forces act: the weight force mg downward and tension force T upward.

For the pulley, three forces are present: weight force Mg down, a reaction force up, and tension force T downward.

For the sum of forces in the -y direction on the block:

∑F = ma

mg − T = ma

T = mg − ma

For the sum of torques on the pulley:

∑τ = Iα

TR = (½ MR²) (a/R)

T = ½ Ma

Substituting gives:

mg − ma = ½ Ma

2mg − 2ma = Ma

2mg = (2m + M) a

a = 2mg / (2m + M)

The angular acceleration of the pulley is:

αR = 2mg / (2m + M)

α = 2mg / (R (2m + M))

Finally, the angular velocity after time t is:

ω = αt + ω₀

ω = 2mg / (R (2m + M)) t + 0

ω = 2mgt / (R (2m + M))

Substituting into the previous equations gives:

ΔL = ½ MR² × 2mgt / (R (2m + M))

ΔL = MmRgt / (2m + M)

3 0
1 month ago
A system expands from a volume of 1.00 l to 2.00 l against a constant external pressure of 1.00 atm. what is the work (w) done b
Keith_Richards [2877]
The amount of work performed by a system at consistent pressure is defined by the following equation:
W=p \Delta V = p (V_f - V_i)
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V_f as the final volume
V_i as the initial volume

Plugging the values given in this case into the formula gives us
W=p (V_f -V_i)=(1.00 atm)(2.00 L-1.00 L)=1.00 L\cdot atm

Considering that 1 atm \cdot L = 101.3 J, the result for the work done becomes
W= 1.00 atm \cdot L = 101.3 J
8 0
19 days ago
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Sav [2826]
E_total = 5.8 x 10⁴ N/C Explanation: To determine the electric field at specified points, we must calculate the vectors individually for each charge and sum them. The electric field caused by each charged conductive sheet can be derived via Gauss's law with the understanding of scalar products between the electric field and relevant surfaces.
4 0
16 days ago
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