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3 years ago

The mass of a giraffe on the moon would be

2 answers:

4 0

Mass is an independent quantity. It can neither be created nor be destroyed. So mass of giraffe will not change on moon.

nasty-shy [4]3 years ago

3 0

Mass is a constant.....so it can neither be change at any state......so it will remain constant

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"Ryan drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 7 hours. When Ryan drove

Flura [38]

Answer:

Distance between Ryan home and mountain will be 252 miles

Explanation:

Let the distance from home to mountains is d

It is given that it took 7 hours to go mountains from home

Let the speed in going to mountain is v

Then according to question speed in coming from mountain is v+27

And time taken in come from mountain to home is hour

As distance will be same in both case

Distance will be equal to $distance=speed\times time$

So $v\times 7=(v+27)\times 4$

$3v=108$

v = 36 miles per hour

So distance will be equal to $d=36\times 7=252miles$

8 0

4 years ago

Describe how earth distance from the sun changes throughout the year when is earth closest to the sun

yaroslaw [1]

The father earth is away from the sun the more cold.
The closer the sun gets the more hot.
When going threw these motions our seasons and days change as the earth moves around the sun while rotating. Hope this helped.

4 0

3 years ago

Two small spheres assumed to be identical conductors are placed at 30 cm from each other on a horizontal axis. the first S1 is l

charle [14.2K]

a) The electric force exerted by S1 on S2 is 21.58μN.

In this case we are talking about two different types of charges, a positive charge and a negative charge, therefore, they are sensing a force of attraction.

The magnitude of the force is determined by using the following formula:

$F_{e}=k_{e}\frac{|q_{1}||q_{2}|}{r^{2}}$

where:

= Electric force [N]

= Electric constant ()

= First charge [C]

= Second charge [C]

r = distance between the two charges

So, in this case, the force can be calculated like this:

$F_{e}=(8.99x10^{9}N\frac{m^{2}}{C^{2}})\frac{|12x10^{-9}C||18x10^{-9}C|}{(30x10^{-2}m)^{2}}$

So the force will be equal to:

$F=21.58x10^{-6}N$

which is the same as:

$F=21.58 \micro N$

b) The electric field created by S1 at the level of S2 is $1.20 \frac{kN}{C}$

The electric field tells us how many Newtons of force can be applied on a given point in space per unit of charge caused by an existing electric charge. From the concept, we can take the following formula for the electric field.

$E_{S1}=\frac{F_{e}}{q_{2}}$

where:

= electric field generated by the first sphere.

$E_{S1}=\frac{1.20 x10^{-6}N}{18x10^{-9}C}$

which yields:

$E_{S1}=1.20x10^{3} \frac{N}{C}$

$E_{S1}=1.20 \frac{kN}{C}$

When talking about electric fields, we know what their direction is if we suppose the electric field is always affecting a positive charge in the given point in space. In this case, since S1 is positive, we can asume the electric field is in a direction away from S1.

The electric potential created by S1 at the level or S2 is 360V

Electric potential is defined to be the amount of energy you will have at a given point per electric charge. This electric potential can be found by using the following formula:

V=Er

Where V is the electric potential and it is given in volts.

Volts are defined to be 1 Joule per Coulomb. Energy by electric charge.

So we can use the data found in the previous sections to find the electric potential:

$V=(1.20x10^{3} \frac{N}{C})(30x10^{-2}m)$

V=360V

d) The force exerted by S2 on S1 will be the same in magnitude as the force exerted by S1 on S2 but oposite in direction. This is because the force will depend on the two charges, and the distance between them, so:

The electric force exerted by S1 on S2 is 21.58μN.

The magnitude of the force is determined by using the following formula:

$F_{e}=k_{e}\frac{|q_{1}||q_{2}|}{r^{2}}$

$F_{e}=(8.99x10^{9}N\frac{m^{2}}{C^{2}})\frac{|12x10^{-9}C||18x10^{-9}C|}{(30x10^{-2}m)^{2}}$

So the force will be equal to:

$F=21.58x10^{-6}N$

which is the same as:

$F=21.58 \micro N$

e) The electric field generated by S1 in the middle of S1 and S2 is $4.79 \frac{kN}{C}$

In order to find the electric field generated by S1, we can make use of the following formula

$E=k_{e} \frac{q_{1}}{r_{1}^{2}}$

$E=(8.99x10^{9} N\frac{m^{2}}{C^{2}})(\frac{12x10^{-9}C}{(15x10^{-2}m)^{2}})$

which yields:

$E=4.79 \frac{kN}{C}$

f) The electric field in the middle of S1 and S2 is $11.99 \frac{kN}{C}$

In order to find the electric field generated by two different charges at a given point is found by using the following formula:

$E=k_{e} \sum \frac{q_{i}}{r_{i}^{2}}$

where:

$q_{i}$ = each of the charges in the system

$r_{i}$ = the distance between each of the charges and the point we are analyzing.

Since the electric field is a vector, we need to take into account the individual electric fields' directions. In this case we suppose we have a positive test charge between the two charges. We can see that the positive test charge will sense a force in the same direction independently on if the force is excerted by the positive charge or the negative charge. Therefore both electric fields will have the same direction. We'll suppose the electric fields will be positive then, so:

$E=(8.99x10^{9} N\frac{m^{2}}{C^{2}})[\frac{12x10^{-9}C}{(15x10^{-2}m)^{2}}+\frac{18x10^{-9}C}{(15x10^{-2}m)^{2}}]$

which yields:

$E=11.99 \frac{kN}{C}$

g) The electric potential in the middle of S1 and S2 is 1.80 kV

Since we know what the electric field is from the previous question, we can make use of the same formula we used before to find the electric potential in the middle of S1 and S2

So let's take the formula:

V=Er

So we can use the data found in the previous sections to find the electric potential:

$V=(11.99x10^{3} \frac{N}{C})(15x10^{-2}m)$

V=1.80kV

The electric potential generated by S2 on the position of S1 is 539.4V and can be found by using the following formula:

$V=k_{e}\frac{q_{2}}{r}$

So we can use the data provided by the problem to find the electric potential.

$V=(8.99x10^{9} N\frac{m^{2}}{C^{2}})(\frac{18x10^{-9}C}{30x10^{-2}m})$

V=539.4V

8 0

3 years ago

A light spring of constant 163 N/m rests vertically on the bottom of a large beaker of water.

LUCKY_DIMON [66]

Answer:

24.71cm

Explanation:

We approach this problem base don Hooke's law which states the elongation produced in an elastic material is proportional to the applied load or force provided that its elastic limit is not exceeded. This is expressed mathematically as follows;

$F=ke................(1)$

where F is the applied force, k is the force constant and e is the elongation or extension of the material.

In this problem, the applied force F is the weight of the wood which is calculated as follows,

$F=mg.............(2)$

m = 4.11kg

$g=9.8m/s^2$

Hence,

$F=4.11*9.8\\F=40.278N$

Given that k = 163N/m, we make appropriate substitutions into equation (1) to obtain the following;

$40.278=163*e\\e=\frac{40.78}{163}\\e=0.2471m$

Since it is required in cm, we perform the conversion as follows, knowing that 100cm = 1m

$0.2471m=0.2471*100cm=24.71cm$

NB: We do not necessarily need the the density of the wood to perform our calculations since other parameters were given from which we were able to obtain its weight.

7 0

4 years ago

IMPORTANT 3 QUESTIONS!

Brums [2.3K]

Answer:

7. 20,000,000 mL.

8. 8,000 m.

9. 120,000 secs.

10. 4

Explanation:

7. Determination of the volume in millilitres (mL)

Volume in litre (L) = 20,000 L

Volume in millilitres (mL) =..?

The volume in mL can be obtained as follow:

1 L = 1,000 mL

Therefore,

20,000 L = 20,000 x 1,000 = 20,000,000 mL.

Therefore, 20,000 L is equivalent to 20,000,000 mL.

8. Determination of the distance in metre (m)

Distance in mile = 5 mile

Distance in metre =?

First, we shall convert from mile to kilometre.

This can be done as follow:

1 mile = 1.6 km

Therefore,

5 mile = 5 x 1.6 = 8 km

Finally, we shall convert 8 km to metre (m).

This is illustrated below:

1 km = 1,000 m

Therefore,

8 km = 8 x 1,000 = 8,000 m

Therefore, 5 miles is equivalent to 8,000 m.

9. Determination of the time in seconds.

Time = 400 minutes for 5 days.

First, we shall convert 400 mins to hour.

This is illustrated below:

60 minutes = 1 hour

Therefore,

400 mins = 400/60 = 20/3 hours

The time (hours) is 20/3 hours in 1 day.

Therefore, the time (hours) in 5 days will be = 20/3 x 5 = 100/3 hours.

Next, we shall convert 100/3 hours to minutes.

This is illustrated below:

1 hour = 60 minutes

Therefore,

100/3 hours = 100/3 x 60 = 2000 mins

Finally, we shall convert 2000 mins to seconds.

This is illustrated below:

1 mins = 60 secs

2000 mins = 2000 x 60 = 120,000 secs.

Therefore, the time is 120,000 secs.

10. Determination of the number of significant figures.

To obtain the significant figures of a number, we simply count all the numbers available.

Therefore, the number of significant figures for 9876 is 4.

6 0

4 years ago