By
vector addition.
In fact, velocity is a vector, with a magnitude intensity, a direction and a verse, so we can't simply do an algebraic sum of the two (or more velocities).
First we need to decompose each velocity on both x- and y-axis (if we are on a 2D-plane), then we should do the algebraic sum of all the components on the x- axis and of all the components on the y-axis, to find the resultants on x- and y-axis. And finally, the magnitude of the resultant will be given by
where Rx and Rx are the resultants on x- and y-axis. The direction of the resultant will be given by
where
is its direction with respect to the x-axis.
Answer:
a) 37.70 m/s
b)710.6 m/s²
Explanation:
Given that ;
Mass of object = 2 kg
Radius of the motion = 2m
Frequency of motion = 3 rev/s
The formula to apply is;
v= 2πrf where v is linear speed
v = 2×π×2×3 =12π = 37.70 m/s
Centripetal acceleration is given as;
a= 4×π²×r×f²
a= 4×π²×2×3²
a=710.6 m/s²
Answer
given,
mass of the shell = 87 g = 0.087 Kg
speed of the muzzle = 853 m/s
mass of the helicopter = 4410 kg
A burst of 176 shell fired in 2.93 s
resulting average force = ?
momentum of the shell = m v
= 0.087 x 853
= 74.21 kgm/s
momentum of 176 shell is = 176 p
= 176 x 74.21
= 13060.96
momentum of helicopter = - 13060.96 kgm/s
amount of speed reduce a = 
a= 
a = 2.96 m/s²
velocity = \dfrac{2.96}{2.93}
v = 1.01 m/s
The final temperature of the block is 
Explanation:
The amount of thermal energy Q supplied to a substance is related to the increase in temperature of the substance, 
, according to the equation
where:
m is the mass of the substance
is the specific heat capacity of the substance
In this problem, we have:
m = 1.2 kg is the mass of the block
is the amount of energy supplied to the block
is the specific heat capacity of the block
Solving for , we find the increase in temperature:
And since the initial temperature was
The final temperature will be
Learn more about specific heat capacity:
brainly.com/question/3032746
brainly.com/question/4759369
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(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
(b) The net work done on the crate while it is on the rough surface is 23.7 J.
(c) The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
<h3>Magnitude of net force on the crate</h3>
F(net) = F - μFf
F(net) = 280 - 0.351(92 x 9.8)
F(net) = -36.46 N
<h3>Net work done on the crate</h3>
W = F(net) x L
W = -36.46 x 0.65
W = - 23.7 J
<h3>Acceleration of the crate</h3>
a = F(net)/m
a = -36.46/92
a = - 0.396 m/s²
<h3>Speed of the crate</h3>
v² = u² + 2as
v² = 0.845² + 2(-0.396)(0.65)
v² = 0.199
v = √0.199
v = 0.45 m/s
Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
The net work done on the crate while it is on the rough surface is 23.7 J.
The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
Learn more about work done here: brainly.com/question/8119756
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