Answer:
Explanation:
To solve this problem, we can use the following suvat equation:
where
is the vertical displacement of the frog
is the initial vertical velocity
t is the time
a is the acceleration
We have chosen this formula because apart from , all the other quantities are known. In fact:
is the vertical displacement
t = 2 s is the total time of flight
is the acceleration due to gravity (negative because it is downward)
Therefore, solving for , we find the initial velocity of the frog:
Answer: 8.1 x 10^24
Explanation:
I(t) = (0.6 A) e^(-t/6 hr)
I'll leave out units for neatness: I(t) = 0.6e^(-t/6)
If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).
For neatness let k = 1/(6x3600) = 4.63x10^-5, then:
I(t) = 0.6e^(-kt)
Providing t is in seconds, total charge Q in coulombs is
Q= ∫ I(t).dt evaluated from t=0 to t=∞.
Q = ∫(0.6e^(-kt)
= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.
= -(0.6/k)[e^-∞ - e^-0]
= -0.6/k[0 - 1]
= 0.6/k
= 0.6/(4.63x10^-5)
= 12958 C
Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.
Answer:
The torque on the child is now the same, τ.
Explanation:
- It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
- In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
- The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.
- When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:
- When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:
- Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:
τ = 3/4*m*r² * (2α) = 3/2*m*r²
same result than in (2), so the torque remains the same.
The braking force is -400 N
Explanation:
We can solve this problem by using the impulse theorem, which states that the impulse applied on the ferry (the product of force and time) is equal to its change in momentum:
where in this problem, we have:
F is the force applied by the brakes
is the time interval
m = 13,000 kg is the mass of the ferry
u = 2.0 m/s is the initial velocity
v = 0 is the final velocity
And solving for F, we find the force applied by the brakes:
where the negative sign indicates that the direction is backward.
Learn more about impulse:
brainly.com/question/9484203
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The question is incomplete. Here is the entire question.
A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?
Answer: Δx = - 42m
Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.
For this "type" of motion, displacement (Δx) can be determined by:
is the initial velocity
a is acceleration and can be positive or negative, according to the referential.
For Referential, let's assume rightward is positive.
Calculating displacement:
= - 42
Displacement of the boat for t=6.0s interval is = - 42m, i.e., 42 m to the left.
