After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm . The explorer finds that
1 answer:
Answer:
12.4 m/s²
Explanation:
L = length of the simple pendulum = 53 cm = 0.53 m
n = Number of full swing cycles = 99.0
t = Total time taken = 128 s
T = Time period of the pendulum
g = magnitude of gravitational acceleration on the planet
Time period of the pendulum is given as
T = 1.3 sec
Time period of the pendulum is also given as
g = 12.4 m/s²
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Answer:
Twice as fast
Explanation:
Solution:-
- The mass of less massive cart = m
- The mass of Massive cart = 2m
- The velocity of less massive cart = u
- The velocity of massive cart = v
- We will consider the system of two carts to be isolated and there is no external applied force on the system. This conditions validates the conservation of linear momentum to be applied on the isolated system.
- Each cart with its respective velocity are directed at each other. And meet up with head on collision and comes to rest immediately after the collision.
- The conservation of linear momentum states that the momentum of the system before ( P_i ) and after the collision ( P_f ) remains the same.
- Since the carts comes to a stop after collision then the linear momentum after the collision ( P_f = 0 ). Therefore, we have:
- The linear momentum of a particle ( cart ) is the product of its mass and velocity as follows:
m*u - 2*m*v = 0
Where,
( u ) and ( v ) are opposing velocity vectors in 1-dimension.
- Evaluate the velcoity ( u ) of the less massive cart in terms of the speed ( v ) of more massive cart as follows:
m*u = 2*m*v
u = 2*v
Answer: The velocity of less massive cart must be twice the speed of more massive cart for the system conditions to hold true i.e ( they both come to a stop after collision ).
Answer:
Explanation:
To find the magnitude of the magnetic field, you use the following formula for the calculation of the magnetic field generated by a current in a wire:
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
I: current = 6.0 A
r: distance to the wire in which magnetic field is measured
In this case, you have four wires at corners of a square of length 9.0cm = 0.09m
You calculate the magnetic field in one corner. Then, you have to sum the contribution of all magnetic field generated by the other three wires, in the other corners. Furthermore, you have to take into account the direction of such magnetic fields. The direction of the magnetic field is given by the right-hand side rule.
If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i) and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:
I1 = I2 = I3 = 6.0A
r1 = 0.09m
r2 = 0.09m
Then you have:
