Answer:
1.25377 m/s²
Explanation:
m = Mass of person
g = Acceleration due to gravity = 9.81 m/s²
= Coefficient of friction
= Slope
From Newton's second law
Applying to the above equation and
The acceleration of the same skier when she is moving down a hill is 1.25377 m/s²
Given that,
Distance =30 m
speed = 0.5c
(A). We need to find the bell and siren simultaneous events for a passenger seated in the car
According to given data
The distance travelled by the light to reach either side of the rocket train car is same.
So, The two events are simultaneous and the bell and siren are the simultaneous events for a passenger seated in the car.
(B). We need to calculate time interval between the events
Using formula of time dilation
.....(I)
Where, = proper time
= time interval between the events
The time interval between the events measured in a reference frame
The proper time in this case is
For the second interval,
Put the value of in the equation (I)
Put the value in the equation
Negative sign shows the siren rings before the bell ring.
Hence, (A). Yes, the bell and siren are simultaneous events.
(B). The siren sounds before the bell rings.
Answer:
291.598 N-m
291.6 N-m
Explanation:
Let's first take a look at the free bodily diagrammatic representation.
The first diagram will aid us in answering question (a), so as the second diagram will facilitate effective understanding when solving for question (b).
Let's first determine our angle θ from the diagram
To find angle θ ; we have :
tan θ =
tan θ =
tan θ = 1.333
θ = tan⁻¹ (1.333)
θ = 53.13°
Now, to determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A.
We have:
where Force(F) = Force in the cord AC = 1350 N and θ = 53.13° ; we have:
Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A = 291.598 N-m
b) From the second diagram, taking the moment at point B ,
we have:
where Force(F) = 1350 N and θ = 53.13° ; we have:
Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point C = 291.6 N-m
Answer:
Explanation:
It is given that,
Speed of one quoll around a curve, v = 3.2 m/s (maximum speed)
Radius of the curve, r = 1.4 m
On the curve, the centripetal force is balanced by the frictional force such that the coefficient of frictional is given by :
So, the coefficient of static friction between the quoll's feet and the ground in this trial is 0.74. Hence, this is the required solution.