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>
Answer:
Explanation:
Let the velocity of projectile be v and angle of throw be θ.
The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m
considering its vertical displacement
h = - ut +1/2 g t²
100 = - vsinθ x 5 + .5 x 9.8 x 5²
5vsinθ = 222.5
vsinθ = 44.5
It covers 160 horizontally in 5 s
vcosθ x 5 = 160
v cosθ = 32
squaring and adding
v²sin²θ +v² cos²θ = 44.4² + 32²
v² = 1971.36 + 1024
v = 54.73 m /s
<span>Antimony I am pretty sure is one. </span>
Answer:
The instantaneous velocity is the specific rate of change of position (or displacement) with respect to time at a single point (x,t) , while average velocity is the average rate of change of position (or displacement) with respect to time over an interval.Average velocity : Average velocity of a body is defined as the change in position or displacement (Δx) divided by time interval (Δt) in which that displacement occurs.
Instantaneous velocity : The instantaneous velocity of a body is the velocity of the body at any instant of time or at any point of its path .
velocity can be positive , negative or zero.
By studying speed and velocity we come to the result that at any time interval average speed of an object is equal or more than the average but instantaneous speed is equal to instantaneous velocity.
Answer:
0.249
Explanation:
Perihelion = 4.43 x 10^9 km
Aphelion = 7.37 x 10^9 km
Let e be the eccentricity and a be the length of semi major axis.
The relation between the semi major axis, perihelion and aphelion s given by
Semi major axis = half of sum of perihelion and aphelion
a = 5.9 x 10^9 km
The relation between the perihelion, semi major axis and the eccentricity is given by
Perihelion = a (1 - e)
4.43 x 10^9 = 5.9 x 10^9 (1 - e)
0.751 = 1 - e
e = 0.249
