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2 years ago

5

Is a pair of sneakers a want or a need? Is the most expensive pair of sneakers a want or a need?

1 answer:

IgorLugansk [536]2 years ago

8 0

Explanation: A sneaker is a want because you don't actually need it to survive

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When a 58g tennis ball is served, it accelerates from rest to a constant speed of 36 m/s. The impact with the racket gives the b

inysia [295]

We first calculate the acceleration on the ball using:
2as = v² - u²; u = 0 because ball is initially at rest
a = (36)²/(2 x 0.35)
a = 1850 m/s²
F = ma
F = 0.058 x 1850
= 107.3 Newtons

5 0

3 years ago

The tendency of two masses alone in the universe to drift together is a result of

torisob [31]

It is as a result of gravity. (D)

This is as stated by Newton's law of universal gravitation. That two objects in the universe attract one another with a force that is proportional to the product of their masses and inversely proportional to the square of the distance apart.

The constant of proportionality is the Universal Gravitational Constant.

G = 6.673 × 10⁻¹¹ Nm²kg⁻²

4 0

3 years ago

Read 2 more answers

What is the heat flux (W/m^2), due to radiation heat transfer, from a black body if the surface temperature is 1000C? The convec

12345 [234]

Answer:

55000 W/m²

Explanation:

Parameters given:

Surface temperature, T = 1000°C

Hear transfer coefficient, h = 55 W/m²C

Convection heat transfer coefficient is given as:

h = Heat flux/Temperature

Hence, Heat Flux, q, is given as:

q = h * T

q = 55 * 1000 = 55000 W/m²C

7 0

3 years ago

Read 2 more answers

A parallel-plate capacitor has square plates that are 8.00cm on each side and 4.20mm apart. The space between the plates is comp

Sergeu [11.5K]

Answer:

$U=1.29\times 10^{-7}\ J$

Explanation:

Given that

a= 8 cm (square)

A= a ² = 64 cm²

d= 4.2 mm

d₁= 2.1 mm ,K₁= 4.7

d₂=2.1 mm , K₂=2.6

We know that capacitance given as

$C_1=\dfrac{K_1\varepsilon _oA}{d_1}$

$C_1=\dfrac{4.7\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}$

$C_1=1.26\times 10^{-10}\ F$

$C_2=\dfrac{K_2\varepsilon _oA}{d_2}$

$C_2=\dfrac{2.6\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}$

$C_2=0.701\times 10^{-10}\ F$

Net capacitance

$C=\dfrac{C_1C_2}{C_1+C_2}$

$C=\dfrac{1.26\times 10^{-10}\times 0.701\times 10^{-10}}{1.26\times 10^{-10}+0.701\times 10^{-10}}\ F$

$C=4.5\times 10^{-11}\ F$

We know that stored energy given as

$U=\dfrac{CV^2}{2}$

V= 76 V

$U=\dfrac{4.5\times 10^{-11}\times 76^2}{2}\ J$

$U=1.29\times 10^{-7}\ J$

3 0

2 years ago

Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o

IrinaVladis [17]

Answer:

The net gravitational force on the mass is $1.27\times 10^{-5}$

Explanation:

We have by Newton's law of gravity the force of attraction between masses $m_{1},m_{2}$

$F_{att}=G\frac{m_{1}m_{2}}{r^{2}}$

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

$F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N$

Force of attraction between 435 kg mass and 38 kg mass is

$F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N$

Thus the net force on mass 38.0 kg is $F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}$

8 0

3 years ago