Wearing rubber or stay away from water or/ and a conductor
Answer:
35%
Explanation:
The car's engine gives off 65% thermal energy
So only 35 % is converted into mechanical energy .
input heat = Q₁ = 100
output heat = Q₂ = 65
Work output = Q₁ - Q₂ = W
W = 100 - 65 = 35
Efficiency = W / Q₁ X 100
= (35/ 100) X 100
= 35%.
Answer:
h' = 55.3 m
Explanation:
First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:
s = vt
where,
s = horizontal distance between arrow and orange = 60 m
v = initial horizontal speed of the arrow = v₀ Cos θ
θ = launch angle = 30°
v₀ = launch speed = 35 m/s
Therefore,
60 m = (35 m/s)Cos 30° t
t = 60 m/30.31 m/s
t = 1.98 s
Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:
h = Vi t + (1/2)gt²
where,
Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°
Vi = 17.5 m/s
Therefore,
h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²
h = 34.6 m + 19.2 m
h = 53.8 m
since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:
h' = h + y
h' = 53.8 m + 1.5 m
<u>h' = 55.3 m</u>
Answer:
Between 0 and 1 seconds (B)
Explanation:
The velocity of the car over time is represented by the line graphed here
the steeper the line, the greater change in velocity that occurred in a given time frame.
The steepest portion of the line is between 0-1 seconds, which means that the greatest rate of change occurred between 0-1 seconds.
(acceleration is the rate of change)
To solve this problem we will apply the concept of centripetal acceleration. This type of acceleration is described as the product between the square of the angular velocity and the turning radius. Mathematically the expression can be expressed as
Here,
Angular velocity
r = Radius
Our values are given as,
Replacing,
Therefore the electron's centripetal acceleration is 
