3 trams must be added
Explanation:
In this problem, there are 12 trams along the ring road, spaced at regular intervals.
Calling L the length of the ring road, this means that the space between two consecutive trams is
(1)
In this problem, we want to add n trams such that the interval between the trams will decrease by 1/5; therefore the distance will become
And the number of trams will become
So eq.(1) will become
(2)
And substituting eq.(1) into eq.(2), we find:
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Answer:
2666 kg
0.11567 m/s²
Explanation:
m = Mass of boat
a = Acceleration of boat
From Newton's second law
Force
Force on the first boat is 333.25 N
Hence, mass of the second boat is 2666 kg
Combined mass = 2666+215 = 2881 kg
The acceleration on the combined mass is 0.11567 m/s²
Answer:
The energy of the system is 15 J.
Explanation:
Given that,
Energy E = 2.5 J
Amplitude = 10 cm
We need to calculate the spring constant
Using formula of mechanical energy of the system
Put the value into the formula
If the block is replaced by a block with twice the mass of the original block
Amplitude = 6 cm
We need to calculate the energy
Using formula of mechanical energy
Put the value into the formula
Hence, The energy of the system is 15 J.
Answer:
They had the same speed.
Explanation:
It won't be velocity, because velocity is a vector quantity. Speed is scalar.
Answer:
the magnitude of the electric field is 1.25 N/C
Explanation:
The induced emf in the cube ε = LB.v where B = magnitude of electric field = 5 T , L = length of side of cube = 1 cm = 0.01 m and v = velocity of cube = 1 m/s
ε = LB.v = 0.01 m × 5 T × 1 m/s = 0.05 V
Also, induced emf in the cube ε = ∫E.ds around the loop of the cube where E = electric field in metal cube
ε = ∫E.ds
ε = Eds since E is always parallel to the side of the cube
= E∫ds ∫ds = 4L since we have 4 sides
= E(4L)
= 4EL
So,4EL = 0.05 V
E = 0.05 V/4L
= 0.05 V/(4 × 0.01 m)
= 0.05 V/0.04 m
= 1.25 V/m
= 1.25 N/C
So, the magnitude of the electric field is 1.25 N/C
