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Makovka662 [10]
3 years ago
15

A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo

ttom is 8.00 rad/s, what is the tension in the string when the object is at the bottom of the circle?
Physics
1 answer:
Talja [164]3 years ago
5 0

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

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Oduvanchick [21]

Answer:

f = 8 N

Explanation:

Data provided in the question

Radius of the pulley  = r = 0.05 m

Moment of inertia = (I) = 0.2 kg.m^{2}

Angular acceleration = ∝ = 2 rad/sec

Based on the above information

As we know that

Torque is

= force \times  radius

= f \times r

And,

Torque is also

= moment\ of\ inertia \times angular\ acceleration

= I \times \alpha

So,

We can say that

f \times r = I \times \alpha

f \times 0.05 = 0.2 \times 2

0.05f = 0.4

f = 8 N

We simply applied the above formulas

8 0
3 years ago
A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long a
IrinaK [193]

The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

<h3>What's the expression of range of a projectile motion?</h3>
  • Range = U²× sin(2θ)/g
  • U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
  • U=√{Range×g/sin(2θ)}
  • Here, range= 2.20m, = 36.5°
  • U= √{2.20×9.8/sin(73)}

U= √{2.20×9.8/sin(73)} = 22.5m/s

<h3>What's the expression of time of flight in projectile motion?</h3>
  • Time of flight= (2×U×sinθ)/g
  • So, T= (2×22.5×sin36.5°)/9.8

= 2.73 s

Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

Learn more about the range and time period of projectile motion here:

brainly.com/question/24136952

#SPJ1

4 0
2 years ago
A car accelerate from a stop to 27 m/sec in 5 seconds. What was the car's acceleration?
Furkat [3]
The acceleration should 5.4 m/s^2
4 0
3 years ago
A 87.0 kg astronaut is working on the engines of a spaceship that is drifting through space with a constant velocity. The astron
Ket [755]

Answer:

259.62521 seconds

Explanation:

m_1 = Mass of astronaut = 87 kg

m_2 = Mass of wrench = 0.57 kg

v_1 = Velocity of astronaut

v_2 = Velocity of wrench = 22.4 m/s

Here, the linear momentum is conserved

m_1v_1=m_2v_2\\\Rightarrow v_1=\frac{m_2v_2}{m_1}\\\Rightarrow v_1=\frac{0.57\times 22.4}{87}\\\Rightarrow v_1=0.14675\ m/s

Time = Distance / Speed

Time=\frac{38.1}{0.14675}=259.62521\ s

The time taken to reach the ship is 259.62521 seconds

4 0
3 years ago
The reflection of a sound wave from a wal is called
WINSTONCH [101]
I think its an echo?
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3 years ago
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