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bulgar [2K]
2 years ago
5

A car starts from rest and accelerates for 5.2 s with an acceleration of 2.8 m/s 2 . How far does it travel? Answer in units of

m.
Physics
1 answer:
Mumz [18]2 years ago
5 0
The final velocity is 5.2^2. The initial velocity is 0 . You are using Kinematic equation 3. (V^2=Vo^2+2a(x)) . (5.2^2=2(2.8)x)
27.04=5.6x
27.04/5.6=5.6/5.6x
4.83=x
The answer is 4.83m:)
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When you cook a marshmallow on a metal poker tool over an open flame, energy is transferred. Identify the three different ways t
sergejj [24]

Answer:

The three ways thermal energy is transferred are;

1) Conduction

2) Convection

3) Radiation

Explanation:

1) The conduction of the heat from the open flame to the marshmallow is through the direct contact of the flame with the marshmallow, such that the flame the region of the combustion reaction, that produces light and heat touches the marshmallow

2) The convection process is the transfer of heat from the rising heated combustion products, as well as the heated air that rises from the flame

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3 years ago
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3 0
2 years ago
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barxatty [35]

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8 0
2 years ago
A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit
vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

\omega =\sqrt{\dfrac{K}{m}}

The maximum speed v given as

v=\omega A

A=Amplitude

v=\sqrt{\dfrac{K}{m}}\times A

0.32=\sqrt{\dfrac{13.5}{0.75}}\times A

A=0.075 m

A= 0.75 cm

The speed at distance x

v=\omega \sqrt{A^2-x^2}

v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}

v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}

v= 0.21 m/s

5 0
2 years ago
Which factors affect the strength of the electric force between two objects
Margaret [11]
<span>-- the product of the net charges on the objects;. -- the distance between the centers of their net charges. (Pretty much identical to the formula for gravitational force)</span>
8 0
3 years ago
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