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spin [16.1K]
3 years ago
5

A test car starts from rest on a horizontal circular track of 85-m radius and increases its speed at a uniform rate to reach 115

km/h in 13 seconds. Determine the magnitude a of the total acceleration of the car 11 seconds after the start.
Physics
1 answer:
Mashutka [201]3 years ago
4 0

Answer:

27.03 m/s.

Explanation:

vo = 0 m/s

vi = 115 km/h

Converting to m/s,

115 km/h * 1000 m/1 km * 1 h/3600 s

= 31.94 m/s.

Using equation of motion,

vi = vo + at

a = 31.94/13

= 2.46 m/s^2

After t = 11 s,

vi = 2.46 * 11

= 27.03 m/s.

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The medium through which a mechanical wave passes can be a solid, liquid, or gas. Properties of a wave change when it moves from
Tpy6a [65]

Answer:

Option A

Explanation:

Mechanical waves requires some medium to travel through. They travel faster in the dense medium as compared to a free medium.

The speed of a mechanical wave is fastest in the solid medium and the slowest in the gaseous medium. Hence, as the wave traverses from gaseous medium to the solid medium, its speed increases.

Thus, option A is correct

3 0
3 years ago
A paper-filled capacitor is charged to a potential difference of V0=2.5 V. The dielectric constant of paper is k=3.7 . The capac
Pavlova-9 [17]

Answer:

k =  6.72

Explanation:

K of paper = 3.7

k of air = 1

Given that charge Q on the capacitor is constant because cell is disconnected from the circuit. So

V = Q / C = 2.5

Capacity becomes C / 3.7 in air .

capacity becomes C/3.7 when paper is replaced by air .

V₁ = Q / (C/3.7)

= 3.7 Q/C

3.7 x 2.5

= 9.25 V

In the second case ,

capacitance  due to new unknown dielectric k

= C/3.7 x k

= kC / 3.7 ( Capacitance in air is C/3.7 )

V ( new ) = Q / ( kC/3.7 )

= 3.7 Q/kC

.55 x 2.5 = 3.7 x( 2.5 / k )

k = 3.7 / .55

= 6.72

7 0
3 years ago
Which stage occurs just before ignition in an internal combustion engine?
ddd [48]
I think it is this because compression stroke it needs to be compressed then open up when started.
5 0
4 years ago
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Though some self-report error may exist, surveys/interviews are useful in that they allow researchers to collect large amounts o
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4 0
3 years ago
A particular car engine operates between temperatures of 440°C (inside the cylinders of the engine) and 20°C (the temperature of
Step2247 [10]

One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.

From the temperature the value is given as

\eta = 1-\frac{T_L}{T_H}

Where,

T_L = Cold focus temperature

T_H = Hot spot temperature

Our values are given as,

T_L = 20\° C = (20+273) K = 293 K

T_H = 440\° C = (440+273) K = 713 K

Replacing we have,

\eta = 1-\frac{T_L}{T_H}

\eta = 1-\frac{293}{713}

\eta = 0.589

Therefore the maximum possible efficiency the car can have is 58.9%

4 0
3 years ago
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