Answer:
Work done is equal to 125.44 J
Explanation:
We have given mass of the student m = 80 kg
Distance moved d = 4 m
Acceleration due to gravity 
Coefficient of kinetic friction 
So normal force exerted by student 
We know that work done is equal to multiplication of force and distance
So work done 
Answer:
Explanation:
This problem bothers on the energy stored in a spring in relation to conservation of energy
Given data
Mass of block m =200g
To kg= 200/1000= 0.2kg
Spring constant k = 1.4kN/m
=1400N/m
Compression x= 10cm
In meter x=10/100 = 0.1m
Using energy considerations or energy conservation principles
The potential energy stored in the spring equals the kinetic energy with which the block move away from the spring
Potential Energy stored in spring
P.E=1/2kx^2
Kinetic energy of the block
K.E =1/mv^2
Where v = velocity of the block
K.E=P.E (energy consideration)
1/2kx^2=1/mv^2
Kx^2= mv^2
Solving for v we have
v^2= (kx^2)/m
v^2= (1400*0.1^2)/0.2
v^2= (14)/0.2
v^2= 70
v= √70
v= 8.36m/s
a. Distance moved if the ramp exerts no force on the block
Is
S= v^2/2gsinθ
Assuming g= 9. 81m/s^2
S= (8.36)^2/2*9.81*sin60
S= 69.88/19.62*0.866
S= 69.88/16.99
S= 4.11m
Answer:
F=mg
F= 0.153kg x 9.8 m/s^2= 1.5 N
42) The sailboat travels east with velocity
, while the current moves south with speed
. Since the two velocities are perpendicular to each other, he resultant velocity will be given by the Pytagorean theorem:
and the direction is in between the two original directions, therefore south-east. So, the correct answer is
D) 42 mph southeast
43) Since the light moves by uniform motion, we can calculate the distance corresponding to one light year by using the basic relationship between velocity, space and time. In fact, we know the velocity:
and the time is one year, corresponding to:
therefore, the distance corresponding to one light year is:
Therefore, the correct answer is D.
44) For the purpose of the problem, we can assume that the light travels instantaneously from the flash to us (because the distances involved are very small), so the time between the flash and the thunder corresponds to the time it took for the sound to travel to us.
The speed of sound is
And since the time between the flash and the thunder is t=3 s, the distance is
Therefore, the correct answer is A) 3/5 mile.
