a.
The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:
where F is the magnitude of the force, theta is the angle between them and d is the distance.
The problen gives the following data:
The magnitude of the force 750 N.
The angle between the force and the displacement which is 25°
The distance, 26 m.
Plugging this in the formula we have:
Therefore the work done is 17673 J.
b)
The power is given by:
the problem states that the time it takes is 6 s. Then:
Therefore the power is 2945.5 W
Answer:it takes approximately 148.8 seconds to achieve. The average person in a free-fall will hit the ground going at 9.66 m/s from the top of the Empire State Building.
Explanation:
Answer:
14m/s
Explanation:
Given parameters:
Radius of the curve = 50m
Centripetal acceleration = 3.92m/s²
Unknown:
Speed needed to keep the car on the curve = ?
Solution:
The centripetal acceleration is the inwardly directly acceleration needed to keep a body along a curved path.
It is given as;
a = 
a is the centripetal acceleration
v is the speed
r is the radius
Now insert the parameters and find v;
v² = ar
v² = 3.92 x 50 = 196
v = √196 = 14m/s
Answer:
1) The force Christian can exert on his bicycle before picking up the the cargo is 529.74 N
2) The force Christian can exert on his bicycle after picking up the the cargo is 647.46 N
Therefore, Christian has to exert more force on his bike after picking up the cargo
Explanation:
The given parameters are;
The mass of Christian and his bicycle = 54 kg
The mass of the cargo = 12 kg
1) The force Christian can exert on his bicycle before picking up the the cargo = Mass of Christian and his bicycle × Acceleration due to gravity
∴ The force Christian can exert on his bicycle before picking up the the cargo = 54 kg × 9.81 m/s² = 529.74 N
2) The force Christian can exert on his bicycle after picking up the the cargo = (54 + 12) kg × 9.81 m/s² = 647.46 N
Therefore, Christian has to exert more force on his bike after picking up the cargo.
Answer:
a. 1.027 x 10^7 m/s b. 3600 V c. 0 V and d. 1.08 MeV
Explanation:
a. KE =1/2 (MV^2) where the M is mass of electron
b. E = V/d
c. V= 0 V (momentarily the pd changes to zero)
d KE= 300*3600 v = 1.08 MeV
