First convert 90km/hr to m/s.
Initiate velocity = 0m/s (car was at rest)
Final velocity is 25m/s (90km/hr converted)
25m/s - 0m/s / 8s = 3.125 m/s^s
Therefore the answer is option A (3.13m/s^2)
Answer: Force applied by trampoline = 778.5 N
<em>Note: The question is incomplete.</em>
<em>The complete question is : What force does a trampoline have to apply to a 45.0 kg gymnast to accelerate her straight up at 7.50 m/s^2? note that the answer is independent of the velocity of the gymnast. She can be moving either up or down or be stationary.
</em>
Explanation:
The total required the trampoline by the trampoline = net force accelerating the gymnast upwards + force of gravity on her.
= (m * a) + (m * g)
= m ( a + g)
= 45 kg ( 7.50 * 9.80) m/s²
Force applied by trampoline = 778.5 N
Answer:
<h2>Magnitude of the second charge is
</h2>
Explanation:
According to columbs law;
F = 
F is the attractive or repulsive force between the charges = 12N
q1 and q2 are the charges
let q1 = - 8.0 x 10^-6 C
q2=?
r is the distance between the charges = 0.050m
k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²
On substituting the given values
12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²
Cross multiplying
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.
