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Gnom [1K]
3 years ago
13

Girl who’s mass is 52kg, experienced a net force of 1800N at bottom of a roller coster loop during her school physics field to t

he local amusement park, determine Sophia’s acceleration at location.
Physics
1 answer:
Andreas93 [3]3 years ago
3 0

Answer:

34.62m/s^2

Explanation:

Force = mass x acceleration

Given

Force = 1800N

Mass = 52kg

Therefore

1800 = 52 x a

Divide both sides by 52

1800/52 = 52/52 x a

34.62 = a

a = 34.62m/s^2

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3 years ago
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In a longitudinal wave, amplitude can be measured
nignag [31]

Answer:

Compressions

Explanation:

A longitudinal wave is a wave in which the direction of propagation of the wave and the direction of the displacement of the particles of the wave are parallel.

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4 0
3 years ago
A 100 kg mass is pulled along a frictionless surface by a horizontal force F such that its acceleration is 10.0 m/s2. A 20 kg ma
DIA [1.3K]

Answer

given,

mass = 100 kg

acceleration = 10 m/s²

A mass 20 kg slides over 100 kg block

acceleration = 3 m/s²

horizontal friction exerted by the 100 kg block on 20 kg

using newton's second law

F - f = 0

F = f

f = ma

f = 20 × 3

f = 60 N

now net force acting on the 100 kg block

F_net = m a

F_net = 100 x 10

F_net = 1000 N

after 20 kg block falls the acceleration of the bock

F = 1000 +60

F = 1060 N

acceleartion on the block

a = \dfrac{F}{m}

a = \dfrac{1060}{100}

a = 10.60 m/s²

3 0
3 years ago
If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the
PolarNik [594]

Answer:

a. v₁ = 16.2 m/s

b. μ = 0.251

Explanation:

Given:

θ = 15 ° , r = 100 m , v₂ = 15.0 km / h

a.

To determine v₁ to take a 100 m radius curve banked at 15 °

tan θ  = v₁² / r * g

v₁ = √ r * g * tan θ

v₁ = √ 100 m * 9.8 m/s² * tan 15° = 16.2 m/s

b.

To determine μ friction needed for a frightened

v₂ = 15.0 km / h * 1000 m / 1 km * 1h / 60 minute * 1 minute / 60 seg

v₂ = 4.2 m/s

fk = μ * m * g

a₁ = v₁² / r = 16.2 ² / 100 m = 2.63 m/s²

a₂ = v₂² / r = 4.2 ² / 100 m = 0.18 m/s²

F₁ = m * a₁  ,  F₂ = m * a₂

fk = F₁ - F₂   ⇒  μ * m * g = m * ( a₁ - a₂)

μ * g = a₁ - a₂   ⇒  μ = a₁ - a₂ / g

μ = [ 2.63 m/s² - 0.18 m/s² ] / (9.8 m/s²)

μ = 0.251

3 0
3 years ago
An object in a certain direction with an acceleration in the perpendicular direction
geniusboy [140]

Answer:

An object moving in certain direction with an acceleration in the perpendicular direction. The above condition is possible . Example of such situation in life would be when stone tied to a string whirling in a circular path

Hope this helps and pls mark as BRAINLIEST :)

3 0
3 years ago
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