Answer:
350 N
Explanation:
From the question given above, the following data were obtained:
Mass (m) of Go Kart = 35 kg
Initial velocity (u) = 12 m/s
Distance (s) = 7.2 m
Force (F) =?
Next, we shall determine the acceleration of the Go Kart. This can be obtained as follow:
Initial velocity (u) = 12 m/s
Distance (s) = 7.2 m
Final velocity (v) = 0 m/s
Acceleration (a) =.?
v² = u² + 2as
0² = 12² + (2 × a × 7.2)
0 = 144 + 14.4a
Collect like terms
0 – 144 = 14.4a
– 144 = 14.4a
Divide both side by 14.4
a = – 144 / 14.4
a = – 10 m/s²
The negative sign indicate that the Go Kart is decelerating when the brake was applied.
Finally, we shall determine the force the Go Kart have when the student locked the brake. This can be obtained as follow:
Mass (m) of Go Kart = 35 kg
Acceleration (a) = 10 m/s
Force (F) =?
F = ma
F = 35 × 10
F = 350 N
Thus, the Go Kart has a force of 350 N when the student locked the brake.
Answer:4.04 m
Explanation:
mass of athlete
initial speed
To get the maximum height h of Athlete we conserve energy i.e.
Kinetic Energy of Athlete=Potential energy gained by athlete
(b)Speed at half of maximum height
Considering v be the velocity at half of maximum height
conserving Energy we can write
(as )
thus
So Athlete interact with the gravitational Field of Earth
Answer:
The work required to cause this volume change is 2 x 10⁶ J
Explanation:
Given;
constant pressure of the gas, P = 100 kPa = 100,000 Pa
change in volume of the gas, ΔV = 20 m³
The work required to cause this volume change is calculated as;
W = PΔV
Substitute the given values and solve for the required work (W).
W = (100,000)(20)
W = 2 x 10⁶ J
Therefore, the work required to cause this volume change is 2 x 10⁶ J
Answer:
Explanation:
Let the initial speed be u . final speed = 0
v² = u² - 2gh
v is final speed , u is initial speed , h is height .
0 = u² - 2g x 134
u² = 2 x 9.8 x 134
u = 51.25 m /s .
Answer:
30 m/s
Explanation:
Applying,
v = u+at................ Equation 1
Where v = final speed of the ball, u = initial speed of the ball, a = acceleration, t = time.
From the question,
Given: u = 0 m/s (stationary), a = 600 m/s², t = 0.05 s
Substitute these values into equation 5
v = 0+(600×0.05)
v = 30 m/s
Hence the speed at which the ball leaves the player's boot is 30 m/s