Answer:
d. less than 20m/s
Explanation:
To the 2nd car, the first car is travelling 10m/s east and 10m/s south. So the total velocity of the first car with respect to the 2nd car is
[tex]\sqrt{10^2 + 10^2} =10\sqrt{2}=14.14m/s
As 14.14m/s is less than 20m/s. d is the correct selection for this question.
Answer:
0
Explanation:
Assuming your graph and question match the attachment, the average speed is 0. The bug ends up where it started, so its displacement is zero.
average speed = displacement/time = 0/(8 s)
average speed = 0
Answer:
acceleration = 2.4525 m/s²
Explanation:
Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²
Tension in the rope = T
Sol: m2 > m1
i) for downward motion of m2:
m2 a = m2 g - T
5 a = 5 × 9.81 m/s² - T
⇒ T = 49.05 m/s² - 5 a Eqn (a)
ii) for upward motion of m1
m a = T - m1 g
3 a = T - 3 × 9.8 m/s²
⇒ T = 3 a + 29.43 m/s² Eqn (b)
Equating Eqn (a) and(b)
49.05 m/s² - 5 a = T = 3 a + 29.43 m/s²
49.05 m/s² - 29.43 m/s² = 3 a + 5 a
19.62 m/s² = 8 a
⇒ a = 2.4525 m/s²
Answer:
height is 69.68 m
Explanation:
given data
before it hits the ground = 46 % of entire distance
to find out
the height
solution
we know here acceleration and displacement that is
d = (0.5)gt² ..............1
here d is distance and g is acceleration and t is time
so when object falling it will be
h = 4.9 t² ....................2
and in 1st part of question
we have (100% - 46% ) = 54 %
so falling objects will be there
0.54 h = 4.9 (t-1)² ...................3
so
now we have 2 equation with unknown
we equate both equation
1st equation already solve for h
substitute h in the second equation and find t
0.54 × 4.9 t² = 4.9 (t-1)²
t = 0.576 s and 3.771 s
we use here 3.771 s because 0.576 s is useless displacement in the last second before it hits the ground is 46 % of the entire distance it falls
so take t = 3.771 s
then h from equation 2
h = 4.9 t²
h = 4.9 (3.771)²
h = 69.68 m
so height is 69.68 m
Answer:
clockwise
Explanation:
According to the law given by Lenz, known as the Lenz law, it is said that a current induced in the circuit which is due to the change in the magnetic field and is so directed so as to oppose the change in the flux and to apply a force in the opposite direction if the force.
Here, as the magnetic field is directed out of the screen, the current flows in the direction which is clockwise in the loop and it opposes the increasing magnetic field.
The clockwise induced current will produce magnetic field in to the screen.