(u) = 20 m/s
(v) = 0 m/s
<span> (t) = 4 s
</span>
<span>0 = 20 + a(4)
</span><span>4 x a = -20
</span>
so, the answer is <span>-5 m/s^2. or -5 meter per second</span>
The final velocity of the block A will be 2.5 m/sec. The principal of the momentum conversation is used in the given problem.
<h3>What is the law of conservation of momentum?</h3>
According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
In a given concern, mass m₁ is M, mass m₂ is 3M. Initial speed for the mass m₁ and m₂ will be u₁=5 and u₂=0 m/s respectively,
According to the law of conservation of momentum
Momentum before collision =Momentum after collision
m₁u₁+m₂u₂=(m₁+m₂)v
M×5+3M×0=[M+3M]v
The final velocity is found as;
V=51.25 m/s
The velocity of block A is found as;

Hence, the final velocity of the block A will be 2.5 m/sec.
To learn more about the law of conservation of momentum, refer;
brainly.com/question/1113396
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Answer:
(a) 1.3 x 10^6 Hz
(b) 76.73 cm
Explanation:
(a)
the formula for the frequency is given by
f = B q / 2 π m
where, B be the strength of magnetic field, q be the charge on one electron, m is the mass of one electron.
B = 46.7 micro tesla = 46.7 x 10^-6 T
q = 1.6 x 10^-19 C
m = 9.1 x 10^-31 kg
f = (46.7 x 10^-6 x 1.6 x 10^-19) / (2 x 3.14 x 9.1 x 10^-31) = 1.3 x 10^6 Hz
(b) K = 114 eV = 114 x 1.6 x 10^-19 J = 182.4 x 10^-19 J
K = 1/2 mv^2
182.4 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2
v = 6.3 x 10^6 m/s
r = m v / B q
Where, r be the radius of circular path
r = (9.1 x 10^-31 x 6.3 x 10^6) / (46.7 x 10^-6 x 1.6 x 10^-19)
r = 0.7673 m = 76.73 cm
Answer:
The net power needed to change the speed of the vehicle is 275,000 W
Explanation:
Given;
mass of the sport vehicle, m = 1600 kg
initial velocity of the vehicle, u = 15 m/s
final velocity of the vehicle, v = 40 m/s
time of motion, t = 4 s
The force needed to change the speed of the sport vehicle;

The net power needed to change the speed of the vehicle is calculated as;
![P_{net} = \frac{1}{2} F[u + v]\\\\P_{net} = \frac{1}{2} \times 10,000[15 + 40]\\\\P_{net} = 275,000 \ W](https://tex.z-dn.net/?f=P_%7Bnet%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20F%5Bu%20%2B%20v%5D%5C%5C%5C%5CP_%7Bnet%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%2010%2C000%5B15%20%2B%2040%5D%5C%5C%5C%5CP_%7Bnet%7D%20%3D%20275%2C000%20%5C%20W)
Answer:
The manufacturer of a 9V dry-cell flashlight battery says that the battery will deliver 20 mA for 80 continuous hours. During that time the voltage will drop from 9V to 6V. Assume the drop in voltage is linear with time. How much energy does the battery deliver in this 80 h interval?
Explanation: