Answer:
B. 7.5 m/s^2
Explanation:
To find acceleration you need to subtract the final velocity by the starting velocity then divide that by the time
a= v-v/t
a= 60-0/8
a= 60/8
a=7.5 m/s^2
Answer:
A (2066,6 N)
Explanation:
Use the Work formula
62.000J = F . 30
62.000/30 = 2066,6 N
The amout of time it took to move the rock doesn´t matter at all.
It is called a distraction variable, We don´t need it to solve the problem it is there just to confuse.
Answer:
The magnetic force points in the positive z-direction, which corresponds to the upward direction.
Option 2 is correct, the force points in the upwards direction.
Explanation:
The magnetic force on any charge is given as the cross product of qv and B
F = qv × B
where q = charge on the ball thrown = +q (Since it is positively charged)
v = velocity of the charged ball = (+vî) (velocity is in the eastern direction)
B = Magnetic field = (+Bj) (Magnetic field is in the northern direction; pointing forward)
F = qv × B = (+qvî) × (Bj)
F =
| î j k |
| qv 0 0|
| 0 B 0
F = i(0 - 0) - j(0 - 0) + k(qvB - 0)
F = (qvB)k N
The force is in the z-direction.
We could also use the right hand rule; if we point the index finger east (direction of the velocity), the middle finger northwards (direction of the magnetic field), the thumb points in the upward direction (direction of the magnetic force). Hence, the magnetic force is acting upwards, in the positive z-direction too.
Hope this Helps!!!
Answer:
The answer is 100J.
Explanation:
In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. In this question, the mass is equals to 50kg and the velocity is 2m/s
Now,
25kg×4m/s^2 = 100kgm/s^2 or 100J
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
