Ask question

Ask question

All categories

3 years ago

11

If a 1.50 kg mass revolves at the end of a string 0.50 m long, and its tangential speed is 6.0 m/s, calculate the centripetal fo

rce.

2 answers:

Allisa [31]3 years ago

8 0

Answer:

$108\ N$

Explanation:

Mass of object, $m=1.50\ kg.$

Length of string, $r=0.50\ m.$

Tangential speed, $v_t=6.0\ m/s.$

Now, centripetal force F of a object moving in given radius r and mass m moving with velocity v.

Here , object moves along the ends of string. Therefore, radius is equal to length of string.

$F=\dfrac{m \times v_t^2}{r}.$

Putting values of m,v and r in above equation.

We get, $F=\dfrac{1.50\times (6.0)^2}{0.50} \ N=108 \ N.$

Hence, this is the required solution.

rosijanka [135]3 years ago

5 0

You should get about 110 for an answer

You might be interested in

Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

Over [174]

Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

$F_f = \mu m_a g + \mu m_b g$

$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

$F_{net} = ma$

so we have

$F_p - F_f = (m_a + m_b) a$

$9.1 - 3.45 = (10.6 + 7) a$

$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

$F_c = \mu m_b g + m_b a$

$F_c = 0.02(7)(9.8) + 7(0.32)$

$F_c = 3.6 N$

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

$F_p - F_f - F_c = m_b a$

$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

$F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)$

$F_c = 5.5 N$

3 0

2 years ago

What is the potential energy of a 25 kg bicycle resting at the top of a hill 3 m high? (Formula: PE = mgh)

Lisa [10]

Answer:

The answer would be 735J

Explanation:

PE=mgh

=(mass)(force of gravity)(height)

=(25kg)(9.8m/s^2)(3m)

=735J

8 0

3 years ago

Read 2 more answers

Does the rankine degree represent a larger or smaller temperature unit than the kelvin degree

zvonat [6]

The best and most correct answer among the choices provided by the question is the first choice, larger.

Rankine is Fahrenheit + 460 , while Kelvin is Celsius + 273. We all know that Fahrenheit has larger number compared to kelvin , thus rankine is much larger.
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.

8 0

2 years ago

A 7.5 cm object is 14 cm from a concave lens, which has a focal length of –7 cm. Its image is 5 cm in front of the lens. What is

Rama09 [41]

Given conditions:
height of object = 7.5cmdistance of object from mirror = 14 cmfocus length = -7 cmimage distance = ?
Using mirror formula:
1/(focus length) = 1/(object distance) + 1/(image distance)
or, -1/7 = 1/14 + 1/(image distance)
or, image distance = -4.66cm (the image formed is a virtual image)

Also, magnification of image is:
image height /height of object = - image distance /object distance
or, image height = - image distance / object distance * height of object
or, image height = -(-4.66) / 14 * 7.5 = 2.49 = 3(nearest whole number)

3 0

2 years ago

Read 2 more answers

In trampoline competitions, a good jump is one that lasts about 1.8 seconds. (A) How high can an athlete who stays in the air 1.

KonstantinChe [14]

Answer:

3.97305 m

Explanation:

a = Acceleration due to gravity = 9.81 m/s²

If a jump lasts for 1.8 seconds this means that from the moment when the person leaves the ground till the person touches the ground again it takes 1.8 seconds. So, maximum height reached will be at half the time of the jump i.e., 0.9 seconds.

u = Initial velocity = 0

Equation of motion

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0t+\frac{1}{2}9.81\times 0.9^2\\\Rightarrow s=3.97305\ m$

So, height of the jump is 3.97305 m.

4 0

3 years ago