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2 years ago

If the applied force is 60 N and the lower mass is 1.0 kg, what is the tension in the string?(frictionless surface)

Physics

1 answer:

Alik [6]2 years ago

3 0

The net force on the larger mass in the horizontal direction (where we take the direction of F to be positive) is

F - T = 2Ma

where T is the magnitude of tension and a is the mass's acceleration.

The net force on the smaller mass in the vertical direction (upward is positive) is

T - Mg = Ma

If M = 1.0 kg and F = 60 N, then we end up with two equations with two unknowns,

60 N - T = (2.0 kg) a

T - (1.0 kg) (9.8 m/s²) = (1.0 kg) a

Eliminate a and solve for T :

(60 N - T) - 2 (T - (1.0 kg) (9.8 m/s²)) = (2.0 kg) a - 2 (1.0 kg) a

60 N - T - 2T + 19.6 N = (2.0 kg) a - (2.0 kg) a

40.4 N - 3T = 0

3T = 40.4 N

T = (40.4 N) / 3

T ≈ 13 N

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aliina [53]

Answer:

$\Delta p=-1.56\ kg-m/s$

Explanation:

It is given that,

Mass of the baseball, m = 0.14 kg

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$u=\sqrt{2gh}$

h = 1.8 m

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$v=\sqrt{2gh'}$

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v = -5.23 m/s

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$\Delta p=m(v-u)$

$\Delta p=0.14(-5.23-5.93)$

$\Delta p=-1.56\ kg-m/s$

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4 years ago

The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second if i

Sergio [31]

Answer with Explanation:

We are given that

Diameter of fighter plane=2.3 m

Radius= $r=\frac{d}{2}=\frac{2.3}{2}=1.15 m$

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Frequency= $\frac{1200}{60}=20 Hz$

1 minute=60 seconds

Angular velocity= $\omega=2\pi f$

Angular velocity= $2\times \frac{22}{7}\times 20=125.7 rad/s$

b.We have to find the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac.

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c.Centripetal acceleration= $\omega^2 r=(125.7)^2(1.15)=18170.56 m/s^2$

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4 years ago

Object A has initial momentum p in the forward direction, but loses momentum at a rate of 0.35 kg·m/s2 until it stops moving. Ac

aliina [53]

Answer: a. net force is acting on object A in the same direction of its motion.

Explanation:

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Therefore, based on the information given in the question, a force is acting on object A in the same direction of its motion.

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Y_Kistochka [10]

Answer:

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