Answer:
Explanation:
Let the velocity of observer the driver be V₀ and the velocity of police car the source be. Let V be the velocity of sound in air.
The apparent frequency f is given by the following relation
F is real frequency.
Now F = 2600,
Putting these values in the equation
f = 3146 Hz.
b ) After the police passes , the sign of velocity of observer and source get reversed.
f = 2175 Hz
c ) When the police car is also going northward and trying to overtaking the driver the frequency is calculated as follows .
f = 2799 Hz
When the policy car already crosses the drives the frequency will be as follows
f = 2445 Hz
Answer:
3.43 x 10⁵ N/m
Explanation:
M = mass of the car alone = 2000 kg
m = mass of the four people combined = 315 kg
x = compression of the spring when there is car alone
x' = compression of the spring when there is car and four people = x + 0.009
k = spring constant
When there is car, the weight of the car is balanced by the spring force and the force equation is given as
k x = Mg
k x = (2000)(9.8)
k x = 19600 eq-1
When there is car and four people, the weight of the car and four people is balanced by the spring force and the force equation is given as
k x' = (M + m) g
k (x + 0.009) = (2000 + 315) (9.8)
k x + 0.009 k = (2000 + 315) (9.8)
using eq-1
19600 + 0.009 k = 22687
k = 3.43 x 10⁵ N/m
Alpha particles are released by proton rich unstable nuclei. It has no electrons to balance the protons making it positively charged particles moving at very high speeds. Gamma articles are electromagnetic waves of short wavelength and high frequency. It has a speed same as the speed of light. Beta particles are released by neutron rich unstable nuclei. It has a speed of less than the gamma particles.
Answer:
Explanation:
The forces acting on the crates when the train starts stopping are their weights, the normal force from the train, the static frictional force and the fictional force that is produced by the deceleration of the train. As the gravitational force, this fictional force is equal to the mass of the crates multiplied by the magnitude of the acceleration of the train. So, the equations of motion of the crates will be:
Since the static frictional force is , we get:
So we have a limit to the acceleration of the train. Now, we have to know the distance traveled by the train when it is stopping. Then, we use the kinematic formula:
Now we solve for the acceleration to combine this equation to the inequality we got before:
And solve for x:
Since we are looking for the minimum value for x, we consider the case in which that inequality becomes an equation:
Before we finish, we have to convert the unities of the initial velocity to meters per second:
Finally, we plug in the known values to get :
It means that the train can be stopped at a minimum distance of 36.2m at constant acceleration without causing the crates slide over the floor.