Answer:
<em>The work done by the car is 363 kJ</em>
Explanation:
Work : Work is said to be done when a Force moves an object through a certain distance. Work and Energy are interchangeable because they have the same unit. The unit of work is Joules (J).
Mathematically work done can be expressed as,
E = W = 1/2mv²
W = 1/2mv²................................ Equation 1
Where E = Energy, W = work done, m = mass of the car, v = velocity of the car
<em>Given: m=1500 kg, v=22 m/s</em>
<em>Substituting these values into equation 1</em>
<em>W = 1/2(1500)(22)²</em>
<em>W = 750 × 484</em>
<em>W = 363000 J</em>
<em>W = 363 kJ</em>
<em>Thus the work done by the car is 363 kJ</em>
Answer:
98.13m
Explanation:
Complete question
Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building
CHECK THE ATTACHMENT
From the figure, using trigonometry
Tan(θ ) = opposite/adjacent
Where Angle (θ )= 63°
Opposite= X = height of the building
Adjacent= 50 m
Then substitute the values we have
Tan(63)= X/50
1.9626= X/50
X= 1.9626 × 50
X= 98.13m
Hence, the height of the building is 98.13m
<span>The ball clears by 11.79 meters
Let's first determine the horizontal and vertical velocities of the ball.
h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s
v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s
Now determine how many seconds it will take for the ball to get to the goal.
t = 36.0 m / 15.04 m/s = 2.394 s
The height the ball will be at time T is
h = vT - 1/2 A T^2
where
h = height of ball
v = initial vertical velocity
T = time
A = acceleration due to gravity
So plugging into the formula the known values
h = vT - 1/2 A T^2
h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2
h = 42.92 m - 4.9 m/s^2 * 5.731 s^2
h = 42.92 m - 28.0819 m
h = 14.84 m
Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
Answer:
11.4 m/s
Explanation:
The expression for the Centripetal acceleration is :
Where, a is the accleration
v is the velocity around circumference of circle
R is radius of circle
In the given question,
a = g = Acceleration due to gravity as the car is at top = 
v = ?
R = 13.2 m
So,
<u>v = 11.4 m/s</u>
