All categories

2 years ago

A uniform thin circular rubber band of mass M and spring constant k has an original radius R?

Physics

1 answer:

Gemiola [76]2 years ago

6 0

Answer: Yes

Explanation:

A ruber band can be described by his mas, M, the spring constant K, and his rest radius, wich is the radius of the circular rubber band when there is no force applied to it, so yes, a uniform thin circular rubber band of mass M and spring constant k has an original radius R

You might be interested in

An object is located 13.5 cm in front of a convex mirror, the image being 7.05 cm behind the mirror. A second object, twice as t

goldfiish [28.3K]

Answer:

Second object is located at 42.03 cm in front of mirror

Explanation:

In this question we have given,

object distance from convex mirror ,u=-13.5cm

Image distance from convex mirror,v=7.05cm

let focal length of convex mirror be f

we have to find the distance of second object from convex mirror

we know that u, v and f are related by following formula

$\frac{1}{f} =\frac{1}{v}+ \frac{1}{u}$ .............(1)

put values of u and v in equation (1)

we got,

$\frac{1}{f} =\frac{1}{7.05}+ \frac{1}{-13.5}$

$\frac{1}{f}=\frac{13.5-7.05}{13.5\times 7.05}$

$\frac{1}{f}=\frac{6.45}{13.5\times 7.05}\\f=13.5\times 1.09\\f=14.75$

we have given that

second object is twice as tall as the first object

and image height of both objects are same

it means

$o_{2}=2o_{1}\\i_{1}=i_{2}$ .............(2)

we know that

$\frac{v}{u}=\frac{i}{o}\\i=\frac{o\times v}{u}$

therefore,

$i_{1}=\frac{o_{1}\times v}{u}$ .................(3)

put values of v and u in equation 3

$i_{1}=-\frac{o_{1}\times 7.05}{13.5}$

$i_{1}=-0.52o_{1}$

therefore from equation 2

$i_{2}=-0.52o_{1}$

we know that

$i_{2}=\frac{o_{2}\times V}{U}$ .................(4)

put value of $i_{2}$ and $o_{2}$ in equation 4

$-.52o_{1}=\frac{2o_{1}\times V}{U}$

$U=\frac{2o_{1}\times V}{-.52o_{1}} \\U=-3.85V$

we know that U,V and f are related by following formula

$\frac{1}{f} =\frac{1}{V}+ \frac{1}{U}$ .............(5)

put values of f and U in equation 5

we got

$\frac{1}{14.75} =\frac{1}{V}- \frac{1}{3.85V}$

$\frac{1}{14.75} =\frac{2.85}{3.85V}$

$\frac{1}{14.75} =\frac{2.85}{3.85V}\\V=\frac{2.85\times 14.75}{3.85}\\V=10.91 cm$

Therefore,

U=-10.91\times 3.85

U=-42.03 cm

Second object is located at 42.03 cm in front of mirror

4 0

2 years ago

You hold a piece of wood in one hand and a piece of iron in the other. both pieces have the same volume, and you hold them fully

aivan3 [116]

You hold a piece of wood in one hand and a piece of iron in the other. Both pieces have the same volume, and you hold them fully under water at the same depth. At the moment you let go of them, which one experiences the greater buoyancy force?<span>
</span>

5 0

2 years ago

A car drives off a cliff next to a river at a speed of 30 m/s and lands on the bank on theother side. The road above the cliff i

dezoksy [38]

Answer:1.301 s

Explanation:

Given

Initial Velocity(u)=30 m/s

Height of cliff=8.3 m

Time taken to cover 8.3 m

$h=ut+\frac{at^2}{2}$

here Initial vertical velocity is 0

$8.3=\frac{gt^2}{2}$

$t^2=1.69$

$t=1.301 s$

Horizontal distance

$R=u\times t$

$R=30\times 1.301=39.04 m$

7 0

2 years ago

89. An electron is moving in a straight line with a velocity of 4.0×105 m/s. It enters a region 5.0 cm long where it undergoes a

ddd [48]

Explanation:

Given that,

Initial speed of the electron, $u=4\times 10^5\ m/s$

Distance, s = 5 cm = 0.05 cm

Acceleration of the electron, $a=6\times 10^{12}\ m/s^2$

(a) Let v is the electron's velocity when it emerges from this region. It can be calculated as :

$v^2=u^2+2as$

$v^2=(4\times 10^5)^2+2\times 6\times 10^{12}\times 0.05$

v = 871779.788 m/s

$v=8.71\times 10^5\ m/s$

(b) Let t is the time for which the electron take to cross the region. It can be calculated as:

$t=\dfrac{v-u}{a}$

$t=\dfrac{8.71\times 10^5-4\times 10^5}{6\times 10^{12}}$

$t=7.85\times 10^{-8}\ s$

Hence, this is the required solution.

4 0

2 years ago

A professor's office door is 0.89 m wide, 2.0 m high, and 4.0 cm thick; has a mass of 25 kg ; and pivots on frictionless hinges.

taurus [48]

In order to answer this question ... strange as it may seem ...
we only need one of those measurements that you gave us
that describe the door.

The door is hanging on frictionless hinges, and there's a torque
being applied to it that's trying to close it. All we need to do is apply
an equal torque in the opposite direction, and the door doesn't move.

Obviously, in order for our force to have the most effect, we want
to hold the door at the outer edge, farthest from the hinges. That
distance from the hinges is the width of the door ... 0.89 m.

We need to come up with 4.9 N-m of torque,
applied against the mechanical door-closer.

Torque is (force) x (distance from the hinge).

4.9 N-m = (force) x (0.89 m)

Divide each side by 0.89m: Force = (4.9 N-m) / (0.89 m)

= 5.506 N .

7 0

2 years ago