Answer:
Explanation:
Average acceleration is found by dividing the change in acceleration by the time.
The shuttle bus has an acceleration of -2.4 meters per square second. It slows from 9.0 meters per second to rest, or 0 meters per second. Therefore:
Substitute the values into the formula.
Solve the numerator.
We want to solve for t, the time. We have to isolate the variable. Let's cross multiply.
t is being multiplied by -2.4. The inverse of multiplication is division, so divide both sides by -2.4
It takes <u>3.75 seconds.</u>
Answer:
Explanation:
Force of friction at car B ( break was applied by car B ) =μ mg = .65 x 2100 X 9.8 = 13377 N .
work done by friction = 13377 x 7.30 = 97652.1 J
If v be the common velocity of both the cars after collision
kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²
= 1800 v²
so , applying work - energy theory ,
1800 v² = 97652.1
v² = 54.25
v = 7.365 m /s
This is the common velocity of both the cars .
To know the speed of car A , we shall apply law of conservation of momentum .Let the speed of car A before collision be v₁ .
So , momentum before collision = momentum after collision of both the cars
1500 x v₁ = ( 1500 + 2100 ) x 7.365
v₁ = 17.676 m /s
= 63.63 mph .
( b )
yes Car A was crossing speed limit by a difference of
63.63 - 35 = 28.63 mph.
Answer:
b
Explanation:
Given:
- The ball is fired at a upward initial speed v_yi = 2*v
- The ball in first experiment was fired at upward initial speed v_yi = v
- The ball in first experiment was as at position behind cart = x_1
Find:
How far behind the cart will the ball land, compared to the distance in the original experiment?
Solution:
- Assuming the ball fired follows a projectile path. We will calculate the time it takes for the ball to reach maximum height y. Using first equation of motion:
v_yf = v_yi + a*t
Where, a = -9.81 m/s^2 acceleration due to gravity
v_y,f = 0 m/s max height for both cases:
For experiment 1 case:
0 = v - 9.81*t_1
t_1 = v / 9.81
For experiment 2 case:
0 = 2*v - 9.81*t_2
t_2 = 2*v / 9.81
The total time for the journey is twice that of t for both cases:
For experiment 1 case:
T_1 = 2*t_1
T_1 = 2*v / 9.81
For experiment 2 case:
T_2 = 2*t_2
T_2 = 4*v / 9.81
- Now use 2nd equation of motion in horizontal direction for both cases:
x = v_xi*T
For experiment 1 case:
x_1 = v_x1*T_1
x_1 = v_x1*2*v / 9.81
For experiment 2 case:
x_2 = v_x2*T_2
x_2 = v_x2*4*v / 9.81
- Now the x component of the velocity for each case depends on the horizontal speed of the cart just before launching the ball. Using conservation of momentum we see that both v_x2 = v_x1 after launch. Since the masses of both ball and cart remains the same.
- Hence; take ratio of two distances x_1 and x_2:
x_2 / x_2 = v_x2*4*v / 9.81 * 9.81 / v_x1*2*v
Simplify:
x_1 / x_2 = 2
- Hence, the amount of distance traveled behind the cart in experiment 2 would be twice that of that in experiment 1.
Answer:
6.97 E 16
Explanation:
Frequency is a function of velocity of light to it's wavelength.
Mathematically written as
F = Velocity / wavelength
Velocity of light = 3 x 10^8
Wavelength =430 nm =430 x 10^-9 m
converting wavelength from nanaometer to meter we divide by 10^9
Frequency = (3 x 10^8)/(430 x10^-9) =6.97 E 16
