kinetic energy is given as
KE = (0.5) m v²
given that : v = speed of the bottle in each case = 4 m/s
when m = 0.125 kg
KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J
when m = 0.250 kg
KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J
when m = 0.375 kg
KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J
when m = 0.0.500 kg
KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J
Answer:
3.5 cm
Explanation:
mass, m = 50 kg
diameter = 1 mm
radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m
L = 11.2 m
Y = 2 x 10^11 Pa
Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3
= 7.85 x 10^-7 m^2
Let the wire is stretch by ΔL.
The formula for Young's modulus is given by
ΔL = 0.035 m = 3.5 cm
Thus, the length of the wire stretch by 3.5 cm.
Answer: 6.48m/s
Explanation:
First, we know that Impulse = change in momentum
Initial velocity, u = 19.8m/s
Let,
Velocity after first collision = x m/s
Velocity after second collision = y m/s
Also, we know that
Impulse = m(v - u). But then, the question said, the guard rail delivered a "resistive" impulse. Thus, our impulse would be m(u - v).
5700 = 1500(19.8 - x)
5700 = 29700 - 1500x
1500x = 29700 - 5700
1500x = 24000
x = 24000/1500
x = 16m/s
Also, at the second guard rail. impulse = ft, so that
Impulse = 79000 * 0.12
Impulse = 9480
This makes us have
Impulse = m(x - y)
9480 = 1500(16 -y)
9480 = 24000 - 1500y
1500y = 24000 - 9480
1500y = 14520
y = 14520 / 1500
y = 9.68
Then, the velocity decreases by 3.2, so that the final velocity of the car is
9.68 - 3.2 = 6.48m/s
Complete Question:
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.
Answer:
m = 0.001 M
For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/
Time before projectile hits wall
= 88.2 m / 29.4 m/s = 3 seconds
Vertical velocity of projectile after three seconds
= 3*9.8 = 29.4 m/s
Horizontal velocity of projectile after three seconds, assuming no air resistance
= 29.4 m/s (given)
Conclusion:
velocity of projectile when it hits the wall
= < 29.4, -29.4> m/s
= sqrt(29.4^2+29.4^2) m/s east-bound at 45 degrees below horizontal
= 41.58 m/s east-bound at 45 degrees below horizontal.
