Answer:
kinetic energy is 2, 4, 50 thirties, 2.45 into 10, raise to three Julie.
Explanation:
In the given question, we are told that what is the kinetic energy of mass M equals 0.1 kg bullet traveling at a velocity Velocity is given and 700 m/s. So we know that kinetic energy mm-hmm k equals one half m v squared. So this will be mass is given 0.1 and velocity is 700 so 700 square this is one half 0.1 in two 49 double zero, double zero. This is one-half into 49 double zero. So kinetic energy is 2, 4, 50 thirties, 2.45 into 10, raise to three Julie. This is kinetic energy. Thank you.
Answer:
g=9.64m/s^2.
Explanation:
Gravitational field strength (in other words, gravitational acceleration) is given as follows:g=GMR2g=R2GMwhere G=6.674×10−11m3kg⋅s2G=6.674×10−11kg⋅s2m3 is the gravitational constant, M=5.972×1024kgM=5.972×1024kg is the mass of the Earth, and R=6.371×106m+0.06×106m=6.431×106mR=6.371×106m+0.06×106m=6.431×106m is the distance from the center of the Earth to the required point above the surface (radius plus 60 km).
The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.
Answer:
400ft. 32ft/s -32ft/s
Explanation:
In reality the gravitational acceleration is 9.81 so the quadratic coefficient of the function should be 9.81/2
Anyway for the sake of assumtion let us takes=160t-16t^2
ds/dt=160-32t=0
t=160/32= 5 seconds.
s=160*160/32-16*(160/32)^2= 400 mts
s=384 mts
160t-16t^2=384
i.e
16t^2-160t+384=0
t^2-10t+24=0
(t-6)(t-4)=0
t=[4,6]
we have to take t=4 because it is all the up i.e <5
velocity =v=ds/dt=160-32t
v=160-32*4=32 ft/sec still going up
for all the way down take t=6 whuch is >5
v=160-6*32=-32 ft/sec (falling down!!!)
Answer:
Part a)
Part b)
Part c)
Part d)
Net force on a closed loop in uniform magnetic field is always ZERO
Explanation:
As we know that force on a current carrying wire is given as
now we have
Part a)
current in side 166 cm and magnetic field is parallel
so we have
here we know that L and B is parallel to each other so
Part b)
For 68.1 cm length wire we have
here we know that
so we have
Part c)
For 151 cm length wire we have
here we know that
so we have
Part d)
Net force on a closed loop in uniform magnetic field is always ZERO
