Answer:
![f_k = 310N](https://tex.z-dn.net/?f=f_k%20%3D%20310N)
the answer is A.
Explanation:
Using the laws of newton:
∑F = ma
where ∑F is the sumatory of forces acting in the system, m the mass and a the acelertion of the system.
Then, if the block is moving with constant velocity, its aceleration is equal to 0, so:
∑F = m(0)
∑F = 0
It means that:
F -
= 0
where F is the force applied and
is the friction force. Replacing the value of F, we get:
310N -
= 0
Finally, solving for
:
![f_k = 310N](https://tex.z-dn.net/?f=f_k%20%3D%20310N)
I'm not sure what "60 degree horizontal" means.
I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith.
Now, I'll answer the question that I have invented.
When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is S cos(60) = 0.5 S ,
and the vertical component is S sin(60) = S√3/2 = 0.866 S . (rounded)
-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.
-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change.
-- So at the top of its trajectory, its KE is 0.25 of what it had originally.
That's E/4 .
Answer:
(a) 490 N on earth
(b) 80 N on earth
(c) 45.4545 kg on earth
(d) 270.27 kg on moon
Explanation:
We have given 1 kg = 9.8 N = 2.2 lbs on earth
And 1 kg = 1.6 N = 0.37 lbs on moon
(a) We have given mass of the person m = 50 kg
As it is given that 1 kg = 9.8 N
So 50 kg = 50×9.8 =490 N
(b) Mass of the person on moon = 50 kg
As it is given that on moon 1 kg = 1.6 N
So 50 kg = 50×1.6 = 80 N
(c) We have given that weight of the person on the earth = 100 lbs
As it is given that 1 kg = 2.2 lbs on earth
So 100 lbs = 45.4545 kg
(d) We have given weight of the person on moon = 100 lbs
As it is given that 1 kg = 0.37 lbs
So 100 lbs ![\frac{100}{0.37}=270.27kg](https://tex.z-dn.net/?f=%5Cfrac%7B100%7D%7B0.37%7D%3D270.27kg)
Answer:
5.4 J.
Explanation:
Given,
mass of the object, m = 2 Kg
initial speed, u = 5 m/s
mass of another object,m' = 3 kg
initial speed of another orbit,u' = 2 m/s
KE lost after collusion = ?
Final velocity of the system
Using conservation of momentum
m u + m'u' = (m + m') V
2 x 5 + 3 x 2 = ( 2 + 3 )V
16 = 5 V
V = 3.2 m/s
Initial KE = ![\dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dmu%5E2%20%2B%20%5Cdfrac%7B1%7D%7B2%7Dm%27u%27%5E2)
= ![\dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%202%5Ctimes%205%5E2%20%2B%20%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%203%20%5Ctimes%202%5E2)
= 31 J
Final KE = ![\dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%20%28m%2Bm%27%29V%5E2%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%205%20%5Ctimes%203.2%5E2%20%3D%2025.6%20J)
Loss in KE = 31 J - 25.6 J = 5.4 J.
Answer:
Final velocity of the car will be -9.28 m/sec
Explanation:
We have given that the car starts from the rest so initial velocity of the car u = 0 m /sec
Acceleration of the car
in negative direction so acceleration will be ![a=-1.6m/sec^2](https://tex.z-dn.net/?f=a%3D-1.6m%2Fsec%5E2)
From first equation of motion we know that
v = u+at
So ![v=0+(-1.6)\times 5.8=-9.28m/sec](https://tex.z-dn.net/?f=v%3D0%2B%28-1.6%29%5Ctimes%205.8%3D-9.28m%2Fsec)
So final velocity will be -9.28 m/sec