Answer:
To convert inches to centimeters, use an easy formula and multiply the length by the conversion ratio.
Since one inch is equal to 2.54 centimeters, this is the inches to cm formula to conver
Explanation:
the independent variable is what you're testing or changing in an experiment, so the answer is the temperature of the ball when its dropped.
i hope that helped <3
Answer:
Angular acceleration, ![\alpha =9.49\ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D9.49%5C%20rad%2Fs%5E2)
Explanation:
It is given that,
Mass of the solid sphere, m = 245 g = 0.245 kg
Diameter of the sphere, d = 4.3 cm = 0.043 m
Radius, r = 0.0215 m
Force acting at a point, F = 0.02 N
Let
is its angular acceleration. The relation between the angular acceleration and the torque is given by :
![\tau=I\times \alpha](https://tex.z-dn.net/?f=%5Ctau%3DI%5Ctimes%20%5Calpha)
I is the moment of inertia of the solid sphere
For a solid sphere, ![I=\dfrac{2}{5}mr^2](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7B2%7D%7B5%7Dmr%5E2)
![\alpha =\dfrac{\tau}{I}](https://tex.z-dn.net/?f=%5Calpha%20%3D%5Cdfrac%7B%5Ctau%7D%7BI%7D)
![\alpha =\dfrac{F.r}{(2/5)mr^2}](https://tex.z-dn.net/?f=%5Calpha%20%3D%5Cdfrac%7BF.r%7D%7B%282%2F5%29mr%5E2%7D)
![\alpha =\dfrac{5F}{2mr}](https://tex.z-dn.net/?f=%5Calpha%20%3D%5Cdfrac%7B5F%7D%7B2mr%7D)
![\alpha =\dfrac{5\times 0.02}{2\times 0.245\times 0.0215}](https://tex.z-dn.net/?f=%5Calpha%20%3D%5Cdfrac%7B5%5Ctimes%200.02%7D%7B2%5Ctimes%200.245%5Ctimes%200.0215%7D)
![\alpha =9.49\ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D9.49%5C%20rad%2Fs%5E2)
So, its angular acceleration is
. Hence, this is the required solution.
A contact force since you are making physical contact with the dog
Mark my brainliest please
Answer
given,
mass of the package = 12 kg
slides down distance = 2 m
angle of inclination = 53.0°
coefficient of kinetic friction = 0.4
a) work done on the package by friction is
W_f = -μk R d
= -μk (mg cos 53°)(2.0)
=-(0.4)(8.0 )(9.8)(cos 53°)(2.0)
= -37.75 J
b)
work done on the package by gravity is
W_g = m (g sin 53°) d
= (8.0 )(9.8 )(sin 53°)(2.0 )
=125.23 J
c)
the work done on the package by the normal force is
W_n = 0
d)
the net work done on the package is
W = -37.75 + 125.23 + 0
W = 87.84 J