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marysya [2.9K]
3 years ago
6

A ball moving at positive 3.0 m per s along a table rolls off a table and lands on the ground 2.0 m away. How high was the table

?
Question 3 options:

A 3.3 m

B 1.1 m

C 2.2 m

D 3.0 m
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
3 0

consider the motion along the horizontal direction :

v₀ = initial velocity in horizontal direction as the ball rolls off the table = 3.0 m/s

X = horizontal displacement of the ball = 2.0 m

a = acceleration along the horizontal direction = 0 m/s²

t = time taken to land = ?

using the kinematics equation

X = v₀ t + (0.5) a t²

2.0 = 3.0 t + (0.5) (0) t²

t = 2/3


consider the motion of the ball along the vertical direction

v₀ = initial velocity in vertical direction as the ball rolls off the table = 0 m/s

Y = vertical displacement of the ball = height of the table = h

a = acceleration along the vertical direction = 9.8 m/s²

t = time taken to land = 2/3

using the kinematics equation

Y = v₀ t + (0.5) a t²

h = 0 t + (0.5) (9.8) (2/3)²

h = 2.2 m


C 2.2 m

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Explanation:

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Then, divide 20.93 m/s by 5.0s

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Answer:

a) T = 608.22 N

b) T = 608.22 N

c) T = 682.62 N

d) T = 533.82 N

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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

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= 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

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the tension in the string is  T - mg = ma (Since acceleration a is upwards)

T = ma + mg

= m (a + g )

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= (62.0 kg) (11.01 m/s²)

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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

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= (62.0 kg)(8.61 m/s²)

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8 0
3 years ago
Urgent!
Masja [62]

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m_1 = 1.90 kg

density of iron block is

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now the volume of the iron piece is given as

V = \frac{m}{\rho}

V = \frac{1.90}{7860} = 2.42* 10^{-4} m^3

Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as

F_b = \rho_L V g

here we know that

\rho_L = density of liquid = 916 kg/m^3

F_b = 916* 2.42 * 10^{-4} * 9.8

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F_s + 2.17 = 1.90* 9.8

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