The concept used to solve this problem is that given in the kinematic equations of motion. From theory we know that the change in velocities of a body is equivalent to twice the distance traveled by acceleration, in other words:
Where,
Final and initial velocity
a = Acceleration
x = Displacement
For the given case, the displacement is equivalent to the height (x = h) and the acceleration is the same gravitational acceleration (a = g). In turn we do not have initial speed therefore
Our values are given as
Replacing we have that,
Therefore the speed with which the liquid sulfur left the volcano is 529.15m/s
The mass of an object stays the same wherever it is, but its weight can change. This happens if the object goes where the gravitational field strength is different from the gravitational field strength on Earth, such as into space or another planet.
Explanation:
the object has constant velocity for 2 seconds and it get a constant accelration (2ms-2)
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
I believe the answer is b) slowly heating the surface
