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strojnjashka [21]

3 years ago

15

What is the weight of an object (mass = 60 kilograms) on Mars, where the acceleration due to gravity is 3.75 meters/second2?. Se

lect one of the options below as your answer:. A. 1.6 newtons. B. 16 newtons. C. 22.5 newtons. D. 225 newtons. E. 2250 newtons.

1 answer:

77julia77 [94]3 years ago

7 0

Weight = mass * gravity = 60 kg * 3.75 m/s² = 225 N

<span>Option D.</span>

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The use of non-renewable energy resources in the UK has changed in the last 30 years.

soldi70 [24.7K]

Answer: Varies

Explanation: It became more relied upon, technology is responsible for this.

3 0

3 years ago

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Communications satellites are placed in orbits so that they always remain above the same point of the earth's surface.A Part com

masya89 [10]

Answer:

24 hours

$7\times 10^{-5}\ rad/s$

Explanation:

If a satellite is in sync with Earth then the period of each satellite is 24 hours.

$T=24\times 60\times 60\ s$

Angular velocity is given by

$\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{24\times 60\times 60}\\\Rightarrow \omega=7\times 10^{-5}\ rad/s$

The angular velocity of the satellite is $7\times 10^{-5}\ rad/s$

3 0

3 years ago

Find the distance from a point charge q=100nC where the field intensity is equal to E=6kN/C. please include description of the r

madam [21]

Answer: 0.3872m

Explanation:

q= 100nC --> 100x10^-9 C

k= 9x10^9 Nm^2/C^2

E= 6kN/C --> 600 N/C

r=?

$E= K\frac{q}{r^{2} }$ --> $r=\sqrt{\frac{kq}{E} }$ Despejas "r"

Resuelves

<h3> $r=\sqrt{\frac{(9x10^9 Nm^2/C^2)(100x10x^{-9} C)}{6000N/C} }$ (la x es por, no es una variable)</h3><h3>r= 0.3872983346m</h3>

4 0

3 years ago

A uniformly charged sphere has a potential on its surface of 450 V. At a radial distance of 7.2 m from this surface, the potenti

Yakvenalex [24]

Answer:

The radius of the sphere is 3.6 m.

Explanation:

Given that,

Potential of first sphere = 450 V

Radial distance = 7.2 m

If the potential of sphere =150 V

We need to calculate the radius

Using formula for potential

For 450 V

$V=\dfrac{kQ}{r}$

$450=\dfrac{kQ}{r}$ ....(I)

For 150 V

$150=\dfrac{kQ}{r+7.2}$ ....(II)

Divided equation (I) by equation (II)

$\dfrac{450}{150}=\dfrac{\dfrac{kQ}{r}}{\dfrac{kQ}{r+7.2}}$

$3=\dfrac{(r+7.2)}{r}$

$3r=r+7.2$

$r=\dfrac{7.2}{2}$

$r=3.6\ m$

Hence, The radius of the sphere is 3.6 m.

3 0

3 years ago

15) On a cold day, you take in 4.2 L (i.e., 4.2 x 10-3 m3) of air into your lungs at a temperature of 0°C. If you hold your brea

Rudiy27

Answer:

4.8L ( i.e 4.8 x 10^-3 m3)

Explanation:

Step 1:

Data obtained from the question.

Initial volume (V1) = 4.2L

Initial temperature (T1) = 0°C

Final temperature (T2) = 37°C

Final volume (V2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below

K = °C + 273

T1 = 0°C = 0°C + 273 = 273K

T2 = 37°C = 37°C + 273 = 310K

Step 3:

Determination of the final volume.

Since the pressure is constant,

Charles' Law equation will be applied as shown below:

V1 /T1 = V2/T2

4.2/273 = V2 /310

Cross multiply to express in linear form

273 x V2 = 4.2 x 310

Divide both side by 273

V2 = (4.2 x 310)/273

V2 = 4.8L ( i.e 4.8 x 10^-3 m3)

Therefore, the volume of the air in the lungs at that point is 4.8L ( i.e 4.8 x 10^-3 m3)

3 0

4 years ago