when object goes under acceleration
c).its velocity always increases
<h3><u>Additional</u><u> </u><u>information</u><u>:</u><u>-</u></h3>
★ Acceleration: Rate of increase in velocity.
★ Velocity: Distance travelled by a body per unit time in given direction is called velocity .
Answer:
The car would travel after applying brakes is, d = 14.53 m
Explanation:
Given that,
The time taken to apply brakes fully is, t = 0.5 s
The velocity of the car, v = 29.06 m/s
The distance traveled by the car in 0.5 s, d = ?
The relation between the velocity, displacement, and time is given by the formula
d = v x t m
Substituting the values in the above equation,
d = 29.06 m/s x 0.5 s
= 14.53 m
Therefore, the car would travel after applying brakes is, d = 14.53 m
In order to get the propoerty of work you need to use the following formula
<span>work = force times distance
</span>replacing data you will get:
W = (1.500) (3)
W = 4.500 NM
The answer should be in NM. So it will be 4500 NM againts the force of gravity
Answer:
94
Explanation:
f = 2.57 x 10^13 Hz
E = 10 eV = 10 x 1.6 x 10^-19 J = 1.6 x 10^-18 J
Energy of each photon = h f
Where, h is Plank's constant
Energy of each photon = 6.63 x 10^-34 x 2.57 x 10^13 = 1.7 x 10^-20 J
Number of photons = Total energy / energy of one photon
N = (1.6 x 10^-18) / (1.7 x 10^-20) = 94.11 = 94
Answer:
4.5m/s
Explanation:
Linear speed (v) = 42.5m/s
Distance(x) = 16.5m
θ= 49.0 rad
radius (r) = 3.67 cm
= 0.0367m
The time taken to travel = t
Recall that speed = distance / time
Time = distance / speed
t = x/v
t = 16.5/42.5
t = 0.4 secs
tangential velocity is proportional to the radius and angular velocity ω
Vt = rω
Angular velocity (ω) = θ/t
ω = 49/0.4
ω = 122.5 rad/s
Vt = rω
Vt = 0.0367 * 122.5
Vt =4.5 m/s
