Answer:
30Km/h
Explanation:
acceleration is the change of speed in a given time so when we substract the accelerations we can know how much the car goes per an hour
Answer:
Given:
mass of the ball m = 0.144 kg
velocity v = 38 m/s
now, change in momentum
P = m v- ( - mv)
= 2 mv
=2 x (0.144) x (38)
= 10.944 kg-m/s
Impulse J= F. Δt
change in momentum is equal to impulse
J = 10.944 kg-m/s
we know force is equal to change in momentum per unit time
F = 13.68 x 10³ N
F = 13.68 kN
The cart is at rest, so it is in equilibrium and there is no net force acting on it. The only forces acting on the cart are its weight (magnitude <em>w</em>), the normal force (mag. <em>n</em>), and the friction force (maximum mag. <em>f</em> ).
In the horizontal direction, we have
<em>n</em> cos(120º) + <em>f</em> cos(30º) = 0
-1/2 <em>n</em> + √3/2 <em>f</em> = 0
<em>n</em> = √3 <em>f</em>
and in the vertical,
<em>n</em> sin(120º) + <em>f</em> sin(30º) + (-<em>w</em>) = 0
<em>n</em> sin(120º) + <em>f</em> sin(30º) = (50 kg) (9.80 m/s²)
√3/2 <em>n</em> + 1/2 <em>f</em> = 490 N
Substitute <em>n</em> = √3 <em>f</em> and solve for <em>f</em> :
√3/2 (√3 <em>f </em>) + 1/2 <em>f</em> = 490 N
2 <em>f</em> = 490 N
<em>f</em> = 245 N
(pointed up the incline)
Answer:
v=5.86 m/s
Explanation:
Given that,
Length of the string, l = 0.8 m
Maximum tension tolerated by the string, F = 15 N
Mass of the ball, m = 0.35 kg
We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :
v is the maximum speed
Hence, the maximum speed of the ball is 5.86 m/s.
Answer:
3.44 W/m²
1.134 J
Explanation:
E₀ = Intensity of electric field = 50.9 V/m
I = Intensity of electromagnetic wave
Intensity of electromagnetic wave is given as
I = (0.5) ε₀ E₀² c
I = (0.5) (8.85 x 10⁻¹²) (50.9)² (3 x 10⁸)
I = 3.44 W/m²
A = Area = 0.0277 m²
t = time interval = 11.9 s
Amount of energy is given as
U = I A t
U = (3.44) (0.0277) (11.9)
U = 1.134 J
