Answer:
12.5J
Explanation:
Given parameters:
Mass of bucket = 1.7kg
Height = 75cm = 0.75m
Unknown;
Work done on the bucket by the person = ?
Solution:
To solve this problem, we use the work done equation;
Work done = force x distance = mgh
m is the mass
g is the acceleration due to gravity
h is the height
Now, insert parameters and solve ;
Work done = 1.7 x 9.8 x 0.75 = 12.5J
Answer:
v ’= v + v₀
a system can be another vehicle moving in the opposite direction.
Explanation:
In an inertial reference frame the speed of the vehicle is given by the Galileo transformational
v ’= v - v₀
where v 'is the speed with respect to the mobile system, which moves with constant speed, v is the speed with respect to the fixed system and vo is the speed of the mobile system.
The vehicle's speedometer measures the harvest of a fixed system on earth, in this system v decreases, for a system where v 'increases it has to be a system in which the mobile system moves in the negative direction of the x axis, whereby the transformation ratio is
v ’= v + v₀
Such a system can be another vehicle moving in the opposite direction.
Answer:
is 3 and 2
Explanation: the firth one is 3 and the 2
Answer:
32000 N
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 40 m/s
Distance (s) = 10 m
Final velocity (v) = 0 m/s
Mass (m) of car = 400 Kg
Force (F) =?
Next, we shall determine the acceleration of the the car. This can be obtained as follow:
Initial velocity (u) = 40 m/s
Distance (s) = 10 m
Final velocity (v) = 0 m/s
Acceleration (a) =?
v² = u² + 2as
0² = 40² + (2 × a × 10)
0 = 1600 + 20a
Collect like terms
0 – 1600 = 20a
–1600 = 20a
Divide both side by –1600
a = –1600 / 20
a = –80 m/s²
The negative sign indicate that the car is decelerating i.e coming to rest.
Finally, we shall determine the force needed to stop the car. This can be obtained as follow:
Mass (m) of car = 400 Kg
Acceleration (a) = –80 m/s²
Force (F) =?
F = ma
F = 400 × –80
F = – 32000 N
NOTE: The negative sign indicate that the force is in opposite direction to the motion of the car.