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
If the velocity is decreased by half, then Kinetic energy will reduced to it's 4 times,
= 1.9 * 10³ / 4
= 4.75 * 10²
In short, Your Answer would be Option B
Hope this helps!
Answer:
Explanation:
Let equal mass of Ne and Kr be m gm
no of moles of Ne and Kr will be m / 20 and m / 84 ( atomic weight of Ne and Kr is 20 and 84 )
Let the pressure and volume of both the gases be P and V respectively .
The temperature of Ne be T₁ and temperature of Kr be T₂.
For Ne
PV = (m / 20) x R T₁
For Kr
PV = (m / 84) x R T₂
T₁ / T₂ = 84 / 20
We know that
average KE of an atom of mono atomic gas = 3 / 2 x k T
k is boltzmann constant and T is temperature .
KEKr/KENe = T₂ / T₁
= 20 / 84
Answer:
Part 1)
Part 2)
Part 3)
Part 4)
Since torque on right side is more so here it will turn and slip over it
Explanation:
As we know that the block A is placed at distance
d = 50 cm from the hinge at 70 cm mark
So torque due to weight of A is given as
the block B is placed at distance
d = 30 cm from the hinge at 70 cm mark
So torque due to weight of B is given as
Now torque due to weight of the scale is given as
now torque on left side of scale is given as
Torque on right Side is given as
Since torque on right side is more so here it will turn and slip over it
It is trajectory acceleration. A friction track is a device to study motion in low friction environments, I believe. Does this help?
