All categories

3 years ago

PLZZ ANSWER THE QUESTION

Physics

1 answer:

Alex_Xolod [135]3 years ago

6 0

Answer:

I think that bicycle B is<u><em> C 6km/min </em></u>

hope this helps <3

You might be interested in

A 215-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a

Misha Larkins [42]

Answer:

303.9481875 N

Explanation:

t = Time taken = 2 seconds

F = Force

r = Radius = 1.5 m

I = Moment of Inertia

$\alpha$ = Angular Acceleration

Torque

$\tau=F\times r$

$\tau=I\times \alpha$

$\\\Rightarrow F\times r=I\times \alpha\\\Rightarrow F=\frac{I\times \alpha}{r}$

Angular velocity

$\omega=rev/s\times 2\pi\\\Rightarrow \omega=0.6\times 2\pi\\\Rightarrow \omega=3.76991\ rad/s$

Angular acceleration

$\alpha=\frac{\omega}{t}\\\Rightarrow \alpha=\frac{3.76991}{2}\\\Rightarrow \alpha=1.88495\ rad/s^2$

$I=\frac{1}{2}mr^2\\\Rightarrow I=\frac{1}{2}215\times 1.5^2\\\Rightarrow I=241.875\ kgm^2$

$F=\frac{I\times \alpha}{r}\\\Rightarrow F=\frac{241.875\times 1.88495}{1.5}\\\Rightarrow F=303.9481875\ N$

The magnitude of the force to stop the merry-go-round is 303.9481875 N

3 0

3 years ago

Given the following lens combination:

natulia [17]

Given:

Lens.........diameter ...fl#

eyepiece...2cm............5

objective...40cm........15

focal length of eyepiece = 2*5 = 10cm

focal length of objective = 40*15 = 600cm

magnification = FL obj / FL eyp = 600/10 = 60x

7 0

3 years ago

Read 2 more answers

Objects 1 and 2 attract each other with a electrostatic force of 18.0 units. If the charge

xxMikexx [17]

Answer:

new force is 6 times of the initial force.

Explanation:

Let the charges on two objects is q₁ and q₂. The electric force between charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

Objects 1 and 2 attract each other with a electrostatic force of 18.0 units

If the charge of Object 1 is doubled and the charge of object 2 is tripled, it means, $q_1'=2q_1$ and $q_2'=3q_2$ . New force is given by :

$F'=\dfrac{kq_1'q_2'}{r^2}\\\\F'=\dfrac{k(2q_1)(3q_2)}{r^2}\\\\F'=6\dfrac{kq_1q_2}{r^2}\\\\F'=6F$

So, the new electrostatic force between objects will become 6 times of the initial force.

5 0

3 years ago

What would you do if someone came up atta nowhere and stole your kneecaps, what do you do?

arsen [322]

Answer:

peat them with my dislocated legs

3 0

3 years ago

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor

const2013 [10]

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is $\tau = 34.3 \ N\cdot m$