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3 years ago

If a car travels 400m in 20 seconds how fast is it going? 20 m/s

Physics

1 answer:

scoundrel [369]3 years ago

5 0

Answer:

20m/s

Explanation:

Speed = distance / time

Speed = 400/20

Speed = 20

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A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

3 years ago

Consider an oil droplet of mass m and charge q. We want to determine the charge on the droplet in a Millikan-type experiment. We

Katen [24]

Answer:

$q=\frac{mg}{E_o}$

Explanation:

Given:

Charge = <em>q</em>

Electric field strength = $E_o$

weight of the droplet = <em>mg</em>

The charge is suspended motionless. This is because the electric force on the charge is balanced by the weight of the droplet.

electric force on charged droplet, $F=qE_o$

This is balanced by the weight, $mg$

Equating the two:

$qE_o=mg\\\Rightarrow q=\frac{mg}{E_o}$

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3 years ago

True or False? Materials that are good conductors of heat are usually poor conductors of electricity.

alexandr402 [8]

False

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5 0

3 years ago

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The stiffness of a particular spring is 42 N/m. One end of the spring is attached to a wall. A force of 2 N is required to hold

Liula [17]

Answer:

$L_{o}=0.1224m$

Explanation:

Given data

Force F=2 N

Length L=17 cm = 0.17 m

Spring Constant k=42 N/m

To find

Relaxed length of the spring

Solution

From Hooke's Law we know that

$F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m$

6 0

4 years ago

How much physical activity should an adult have each week?

LuckyWell [14K]

Answer:d

Explanation:

7 0

3 years ago

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