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:
H = 45 m
Explanation:
First we find the launch velocity of the ball by using the following formula:
v₀ = √(v₀ₓ² + v₀y²)
where,
v₀ = launching velocity = ?
v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s
v₀y = Vertical Component of Launch Velocity = 30 m/s
Therefore,
v₀ = √[(15 m/s)² + (30 m/s)²]
v₀ = 33.54 m/s
Now, we find the launch angle of the ball by using the following formula:
θ = tan⁻¹ (v₀y/v₀ₓ)
θ = tan⁻¹ (30/15)
θ = tan⁻¹ (2)
θ = 63.43°
Now, the maximum height attained by the ball is given by the formula:
H = (v₀² Sin² θ)/2g
H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)
<u>H = 45 m</u>
Answer:
(4xy+5ab)(4xy-5ab)
Explanation:
16
-25
4^2 is 16 and 5^2 is 25,
Also, (x-a)(x+a) = x^2-a^2
So, this factorized is:
(4xy+5ab)(4xy-5ab)
Hope this helps!
The answer is distressing
