Answer:
B.The force of friction between the block and surface will decrease.
Explanation:
The force of friction is given by
where 
is the coefficient of friction and 
is the normal force.
When the student pulls on the block with force 
at an angle 
, the normal force on the block becomes
and hence the frictional force becomes
.
Now, as we increase , 
increases which as a result decreases the normal force 
, which also means the frictional force decreases; Hence choice B stands true.
<em>P.S: Choice D is tempting but incorrect since the weight </em>
<em> is independent of the external forces on the block. </em>
A haha snake e enajene. skskskksksks and
Answer:
Vf=3
Explanation:
you must first write your data
data before impact
M1=1000 M2=5000
V1=0 m/s V2 =0m/s
data after impact
M1=1000 M2=5000
V1=15m/s V2=?
M1V1 +M2V2=M1V1 +M2V2f
(1000)(0)+(5000)(0)=(1000)(15)+(5000)Vf
0=15000+5000Vf
- 15000÷5000=5000Vf÷5000
Vf= -3
Vf =3
Answer:
41.74 m/s
Explanation:
The energy used to draw the bowstring = the kinetic energy of the arrow.
Fd = 1/2mv²................................ Equation 1
Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.
make v the subject of the equation
v = √(2Fd/m)...................... Equation 2
Given: F = 201 N, m = 0.3 kg, d = 1.3 m.
Substitute into equation 2
v = √(2×201×1.3/0.3)
v = √(1742)
v = 41.74 m/s.
Hence the arrow leave the bow with a speed of 41.74 m/s
Answer with Explanation:
We are given that mass of block=0.0600 kg
Initial speed of block=0.63 m/s
Distance of block from the hole when the block is revolved=0.47 m
Final speed=3.29 m/s
Distance of block from the hole when the block is revolved=
a.We have to find the tension in the cord in the original situation when the block has speed =
Because tension is equal to centripetal force
Substitute the values
b.
c.Work don=Final K.E-Initial K.E
