5.47 m
The bullet undergoes a non-elastic collision with the block of wood and momentum is conserved. The initial momentum is 0.029 kg * 510 m/s = 14.79 kg*m/s. The combined mass of the block and bullet is 1.40 kg * 0.029 kg = 1.429 kg. Since momentum is conserved, the velocity of both combined will then be 14.79 kg*m/s / 1.429 kg = 10.34989503 m/s.
With a local gravitational acceleration of 9.8 m/s^2, it will take 10.34989503 m/s / 9.8 m/s^2 = 1.056111738 s for their upward velocity to drop to 0, just prior to descending.
The equation for distance under constant acceleration is
d = 0.5 A T^2
so
d = 0.5 * 9.8 m/s^2 * (1.056111738 s)^2
d = 4.9 m/s^2 * 1.115372003 s^2
d = 5.465322814 m
Rounding to 3 significant figures gives a height of 5.47 meters.
Answer:
v = -1.8t+36
20 seconds
360 m
40 seconds
36 m/s
The object speed will increase when it is coming down from its highest height.
Explanation:
Differentiating with respect to time we get
a) Velocity of the object after t seconds is v = -1.8t+36
At the highest point v will be 0
b) The object will reach the highest point after 20 seconds
c) Highest point the object will reach is 360 m
d) Time taken to strike the ground would be 20+20 = 40 seconds
Acceleration will be taken as positive because the object is going down. Hence, the sign changes. 2 is multiplied because the expression is given in the form of
e) The velocity with which the object strikes the ground will be 36 m/s
f) The speed will increase when the object has gone up and for 20 seconds and falls down for 20 seconds. The object speed will increase when it is coming down from its highest height.
Despite current has a magnitude and a direction, like vectors, it is a scalar because it doesn't obey laws of vector addition. For instance, if we consider a junction of
in a circuit, and two currents entering this junction, we know that the resultant current is just the algebraic sum of the two currents, not the vector sum, so it is not a vector quantity.
Answer:
(C) 40m/s
Explanation:
Given;
spring constant of the catapult, k = 10,000 N/m
compression of the spring, x = 0.5 m
mass of the launched object, m = 1.56 kg
Apply the principle of conservation of energy;
Elastic potential energy of the catapult = kinetic energy of the target launched.
¹/₂kx² = ¹/₂mv²
where;
v is the target's velocity as it leaves the catapult
kx² = mv²
v² = kx² / m
v² = (10000 x 0.5²) / (1.56)
v² = 1602.56
v = √1602.56
v = 40.03 m/s
v ≅ 40 m/s
Therefore, the target's velocity as it leaves the spring is 40 m/s
Answer:Twice of given mass
Explanation:
Given
Two Particles of Equal mass placed at the base of an equilateral Triangle
let mass of two equal masses be m and third mass be m'
Taking one of the masses at origin
Therefore co-ordinates of first mass be (0,0)
Co-ordinates of other equal mass is (a,0)
if a is the length of triangle
co-ordinates of final mass
Given its center of mass is at midway between base and third vertex therefore