I had to look for the options and here is my answer:
Based on the actual options attached to this question, the statements that are considered true about electromagnetic induction are the following:
-It is possible to induce a current in a closed loop of wire without the aid of a power supply or battery.
-It is possible to induce a current in a closed loop of wire by changing the strength of a magnetic field enclosed by the wire.
-It is possible to induce a current in a closed loop of wire by change the orientation of a magnetic field enclosed by the wire.
-It is possible to induce a current in a closed loop of wire located in a uniform magnetic field by either increasing or decreasing the <span>area enclosed by the loop.</span>
Answer:
strong winds that blow for a long time over a great distance
weak winds that blow for short periods of time with a short fetch
Explanation:
When the winds are weak and blow for short periods, we experience the smallest ocean waves but when there are strong winds over a longer duration, the largest ocean waves are seen. Therefore, the conditions to produce the smallest and largest ocean waves are strong winds that blow for a long time over a great distance and weak winds that blow for short periods of time with a short fetch.
Given:
m(mass of the car)=1134 Kg
u(Initial velocity)=83Km/HR=23m/s
s(distance traveled by the car)=98m
v(final velocity)=0(as it is given the car stops).
Now we know,
v=u+at
Where v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
0=23+at
at=-23
Also
s=ut+1/2(at^2)
s is the distance covered by the car
u is the initial velocity
t is the time necessary for the car to cover a particular distance.
a is the acceleration
Now substituting these values we get
98=23t-1/2(23t)
98=23t-11.5t
11.5t=98
t=8.52secs
Now we have already derived
at=-23
ax8.52=-23
a=-23/8.52
a=-2.75 m/s^2
F=mxa
Where F is the force acting on the car.
m is the mass of the car.
a is the acceleration.
F=1134 x-2.75
F=-3119N
Answer:
a) 30.58°
b) 367.55 N
Explanation:
Given:
Diameter of the specimen, D = 63mm = 0.063 m
Height of the specimen, H = 25 mm
Shear force at failure, 
Shear stress at failure, 
Area = 
or
Area = 
thus,
Shear stress at failure, 
a)
............(1)
where,
C' = cohesion
for sand C' = 0
∅' = angle of friction
σ' = Normal stress
on substituting the values we get,

or

or

b) Shear force required at the time of failure with normal stress 200 KN/m² can be calculated by using the equation 1
on substituting the values, we get
or

shear force will be = 118.185 × area
or
shear force will be = 118.185 × 0.00311 = 0.36755 KN = 367.55 KN
If they have self motivation or others motivation, they will show their full potential.