Answer:
30.96 m
Explanation:
If the particle has a lifetime of 129 ns as measured by observer A, and has a speed of 0.8c as measured by observer A, the distance between the markers will be:
d = v * Δt
v = 0.8*c = 0.8 * 3e8 = 2.4e8
Δt = ζ = 129 ns = 1.29e-7 s
d = 2.4e8 * 1.29e-7 = 30.96 m
This is the distance as measured by observer A.
Answer:
a) T = 608.22 N
b) T = 608.22 N
c) T = 682.62 N
d) T = 533.82 N
Explanation:
Given that the mass of gymnast is m = 62.0 kg
Acceleration due to gravity is g = 9.81 m/s²
Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.
So;
To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;
T = mg
= (62.0 kg)(9.81 m/s²)
= 608.22 N
When the gymnast climbs the rope at a constant rate tension in the string is
= (62.0 kg)(9.81 m/s²)
= 608.22 N
When the gymnast climbs up the rope with an upward acceleration of magnitude
a = 1.2 m/s²
the tension in the string is T - mg = ma (Since acceleration a is upwards)
T = ma + mg
= m (a + g )
= (62.0 kg)(9.81 m/s² + 1.2 m/s²)
= (62.0 kg) (11.01 m/s²)
= 682.62 N
When the gymnast climbs up the rope with an downward acceleration of magnitude
a = 1.2 m/s² the tension in the string is mg - T = ma (Since acceleration a is downwards)
T = mg - ma
= m (g - a )
= (62.0 kg)(9.81 m/s² - 1.2 m/s²)
= (62.0 kg)(8.61 m/s²)
= 533.82 N
Answer:
Distance, d = 778.05 m
Explanation:
Given that,
Force acting on the car, F = 981 N
Mass of the car, m = 1550 kg
Initial speed of the car, v = 25 mi/h = 11.17 m/s
We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:
Let d is the distance covered by car. Using second equation of motion as :
So, the car will cover a distance of 778.05 meters.
Answer:As the temperature increases, the helium in the ballon expands.
Explanation:
I took the quiz already
To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.
Planet gravitational force
Distance between planet and star
Gravitational force is
Applying the new distance,
Replacing with the previous force,
Replacing our values
Therefore the magnitude of the force on the star due to the planet is 
