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3 years ago

When a driver touches a car’s steering wheel on a hot day, heat is transferred to the driver’s hands by ___________?

Physics

2 answers:

Alik [6]3 years ago

8 0

The correct answer to this question is : C) Conduction

EXPLANATION:

Before coming into any conclusion, first we have to understand conduction.

Conduction is the mode of transmission of heat in which heat will flow from a body at high temperature to a body at low temperature in contact or from hotter part of the body to its colder part, without any actual transport of particles.

As per the question, the heat is transferred to driver's hand when he touches the steering wheel.

As there is actual contact between hand and steering wheel, the heat transmission will be conduction.

Anna [14]3 years ago

5 0

The heat is transferred by conduction because the driver touches the car by contacting, the heat is transferred from hot to cold.

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A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of

Nastasia [14]

Answer:

Amplitude = 0.058m

Frequency = 6.25Hz

Explanation:

Given

Amplitude (A) = 8.26 x 10-2 m

Frequency (f) = 4.42Hz

Conversation of energy before split

½mv² = ½KA²

Make A the subject of formula

A = $v\sqrt{\frac{m}{k} }$

Conversation of energy after split

½(m/2)V'² = ½(m/2)V² = ½KA'²

½(m/2)V² = ½KA'²

Make A the subject of formula

First divide both sides by ½

(m/2)V² = KA'²

Divide both sides by K

$\frac{m}{2K}$ V² = A'²

$v\sqrt{\frac{m}{2k} }$ = A'

Substitute $v\sqrt{\frac{m}{k} }$ for A in the above equation

A' = A/√2

A' = 8.26 x 10^-2/√2

A' = 0.05840702012600882

Amplitude after split = 0.058 (Approximated)

Frequency (f') = f√2

f' = 4.42√2

f' = 6.25082394568908011

Frequency after split = 6.25Hz (approximated)

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2 years ago

During the deceleration of an ascending elevator, the normal force on the feet of a passenger is _____ her weight. During the de

gizmo_the_mogwai [7]

Answer: Smaller than ; larger than

Explanation:

When the elevator is moving in the upward direction, then the force acting on it is negative in nature because of

N= mg +ma, (g is gravity and a is acceleration)

here ma is negative so the N= mg-ma

Hence, it feels smaller than its original weight.

When the elevator is moving downward , then the force acting will be positive in nature

N= mg+ma,

here ma will be positive so it feels larger the original weight of passenger.

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B.

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ASHA 777 [7]

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