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Artyom0805 [142]
3 years ago
10

A gymnast of mass 62.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t

hat the rope does not stretch. Use the value 9.81m/s2 for the acceleration of gravity. Calculate the tension T in the rope if the gymnast hangs motionless on the rope.Calculate the tension T in the rope if the gymnast climbs the rope at a constant rate.Calculate the tension T in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 1.20m/s2 .Calculate the tension T in the rope if the gymnast slides down the rope with a downward acceleration of magnitude 1.20m/s2 .*all answers in Newtons
Physics
1 answer:
MrRissso [65]3 years ago
5 0

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

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Answer:

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Explanation:

Given that,

Mass of a ball, m = 1.5 kg

Initial velocity of the ball, u = 7 m/s

We need to find the maximum height reached by the ball. Let it is be h. Using the conservation of energy to find it such that,

mgh=\dfrac{1}{2}mu^2\\\\h=\dfrac{u^2}{2g}

Put all the values,

h=\dfrac{7^2}{2\times 9.8}\\\\=2.5\ m

So, it will reach to a height of 2.5 m.

8 0
3 years ago
A circular curve of radius 150 m is banked at an angle of 15 degrees. A 750-kg car negotiates the curve at 85.0 km/h without ski
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Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.

Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:

Pcos 15°-N=0

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from the first we obtain N, the normal force

N=750Kg*9.8* cos (15°)= 7.1 *10^3 N

Then to calculate the frictional force (f) we can use the second equation

f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r

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A constant 20 N force is applied to a 7 kg box to push it along the ground. How
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Answer:

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Explanation:

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6 0
3 years ago
A zero-order reaction has a constant rate of 2.30×10−4 M/s. If after 80.0 seconds the concentration has dropped to 1.50×10−2 M,
dolphi86 [110]

Answer:

Initial concentration of the reactant = 3.34 × 10^(-2)M

Explanation:

Rate of reaction = 2.30×10−4 M/s,

Time of reaction = 80s

Final concentration = 1.50×10−2 M

Initial concentration = Rate of reaction × Time of reaction + Final concentration

= 2.30×10−4 M/s × 80s + 1.50×10−2 M = 3.34 × 10^(-2)M

Initial concentration = 3.34 × 10^(-2)M

6 0
2 years ago
Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m
Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC

In words, the value of q₃ must be 5.3nC.

7 0
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