Answer:
50 J
Explanation:
The net force acting on the box is given by the algebraic sum of the two forces, so:
The net work done on the box is equal to (assuming the net force is parallel to the displacement of the object)
where
F = 5 N is the net force on the object
d = 10 m is the displacement of the object
Substituting,
Answer:
Average speed will be 48.23 km/h
Explanation:
Let the distance up to hill is = d km
Speed when car goes to hill = 38 km/h
So time required 
Speed when car return from hill = 66 km/h
So time required to return fro hill 
Total time 
Total distance = d+d =2d
So average speed
Answer:
Average velocity v = 21.18 m/s
Average acceleration a = 2 m/s^2
Explanation:
Average speed equals the total distance travelled divided by the total time taken.
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
Average acceleration equals the change in velocity divided by change in time.
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
Where;
v1 and v2 are velocities at time t1 and t2 respectively.
And x1 and x2 are positions at time t1 and t2 respectively.
Given;
t1 = 3.0s
t2 = 20.0s
v1 = 11 m/s
v2 = 45 m/s
x1 = 25 m
x2 = 385 m
Substituting the values;
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
v = (385-25)/(20-3)
v = 21.18 m/s
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
a = (45-11)/(20-3)
a = 2 m/s^2
Answer:
835.29 Hz
Explanation:
When moving towards the source of sound, frequency will be given by
f*=f(vd+v)/v
Where f is the freqiency of the source, vd is the driving speed, v is the speed of sound in air, f* is the inkown frequency when moving forward.
Substituting 800 Hz for f, 340 m/s for v and 15 m/s for vd then
f*=800(15+340)/340=835.29411764704 Hz
Rounded off, the frequency is approximately 835.29 Hz
