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grandymaker [24]

3 years ago

11

What is the force exerted by a solid surface that opposes gravity?

2 answers:

jonny [76]3 years ago

6 0

Electromagnetic force between the molecules!

Ann [662]3 years ago

6 0

The force exerted upward that opposes gravity is the Normal Force.

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Identified in the consumer's service plan

fiasKO [112]

Pls only do one question

5 0

3 years ago

A girl has a weight of 450 N and her feet have a total area of 0.0300 m2. Calculate the pressure her feet put on the ground.

GenaCL600 [577]

Answer:

15000N/m²

Explanation:

Given parameters:

Weight = 450N

Total area = 0.03m²

Unknown:

Pressure under her feet = ?

Solution:

Pressure is the force per unit area on a body.

Pressure = $\frac{Force }{Area}$

Weight is a type of force;

Pressure = $\frac{450}{0.03}$ = 15000N/m²

4 0

3 years ago

10. A boy weighs 475 N. What is his mass? (acceleration due to gravity on Earth is 9.8m/s2 = g)

Darya [45]

Answer: mass = 48.47 kg.

Explanation:

Formula : Weight = mg , where m = mass of body , g= acceleration due to gravity .

Given: Weight = 475 N

$g= 9.8\ m/s^2$

Substitute all values in formula , we get

$475= m \times9.8\\\\\Rightarrow\ m = \dfrac{475}{9.8}\\\\\Rightarrow\ m \approx 48.47\ kg$

Hence, his mass = 48.47 kg.

7 0

3 years ago

The average distance of the planet mercury from the sun is 0.39 times the average distance of the earth from the sun. How long i

Stels [109]

Answer:

$T_1=0.24y$

Explanation:

Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:

$(\frac{T_1}{T_2})^2=(\frac{D_1}{D_2})^3$

Where $T_1$ and $T_2$ are the orbital periods of Mercury and Earth respectively. We have $D_1=0.39D_2$ and $T_2=1y$ . Replacing this and solving for

$T_1^2=T_2^2(\frac{D_1}{D_2})^3\\T_1^2=(1y)^2(\frac{0.39D_2}{D_2})^3\\T_1^2=1y^2(0.39)^3\\T_1^2=0.059319y^2\\T_1=0.24y$

5 0

3 years ago

Distance in scientific notation

n200080 [17]

Answer:

(See explanation for further information)

Explanation:

a) The distances of each planet with respect to the Sun is:

Mercury - $57.909 \times 10^{6}\,mi$

Venus - $67.240\times 10^{6}\,mi$

Earth - $92.960\times 10^{6}\,mi$

Mars - $141.600\times 10^{6}\,mi$

Jupiter - $483.800\times 10^{6}\,mi$

Saturn - $888.200\times 10^{6}\,mi$

Uranus - $1,787\times 10^{6}\,mi$

Neptune - $2,795\times 10^{6}\,mi$

b) The solutions are presented below:

A. The distance between Venus and Jupiter is:

$\Delta s = 483.800\times 10^{6}\,mi-67.240\times 10^{6}\,mi$

$\Delta s = 416.560\times 10^{6}\,mi$

B. The combined distance from the Sun is:

$\Delta s = 57.909\times 10^{6}\,mi + 67.240\times 10^{6}\,mi +92.960\times 10^{6}\,mi$

$\Delta s = 218.109\times 10^{6}\,mi$

Which is less than the distance from the Sun to Neptune ( $2,795\times 10^{6}\,mi$ ).

C. The new distance of Earth is $929.60 \times 10^{6}\,mi$ ( $929,600,000\,mi$ ). Saturn would be the closest planet to Earth, whose distance:

Scientific notation:

$\Delta s = 929.600\times 10^{6}\,mi-888.200\,\times 10^{6}\,mi$

$\Delta s = 41.4\times 10^{6}\,mi$

Standard notation:

$\Delta s = 929,600,000\,mi - 888,200,000\,mi$

$\Delta s = 41,400.000\,mi$

3 0

4 years ago