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katrin2010 [14]

3 years ago

15

When observed from earth, how many oscillations does the pendulum on the spaceship undergo compared to the pendulum on earth in

a given time interval:__________

1 answer:

Roman55 [17]3 years ago

3 0

Answer: it’s Fewer oscillations.

Explanation:

You might be interested in

What minimum volume must the slab have for a 75.0 kgkg woman to be able to stand on it without getting her feet wet

Alisiya [41]

V = 0.9375 m3 is minimum volume .

<h3>What is density ?</h3>

We use the word "density" to indicate how much space (or "volume") an object or substance occupies in relation to the amount of matter contained therein (its mass).
Density can also be defined as the quantity of mass per unit of volume. A dense object is one that is both hefty and small.

Given,

1000 kg/m3 = density of water

920 kg/m3 = density of ice

BF = Woman Weight + Slab Weight

Pvg = 75 + Pvg

V = 75 kg / (1000 - 920)

V = 75 / 80

V = 0.9375 m3

Therefore, V = 0.9375 m3 is minimum volume .

Learn more about density

brainly.com/question/952755

#SPJ4

6 0

2 years ago

Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0

3 years ago

In a certain process, the energy of the system decreases by 250 kJ. The process involves 480 kJ of work done on the system. Find

gregori [183]

Answer:

Q = - 730 KJ

730 KJ is transferred out of the system

Explanation:

According to the first law of thermodynamics, energy can neither be created nor destroyed, but can be transformed from one form to another.

For a particular process/system, the first law is interpreted as

ΔU = Q + W (depending on convention, some textbooks give this relation as ΔU = Q - W)

ΔU is the change in internal energy of the system, in my convention, it is positive if the internal energy increases and negative otherwise.

Q = Heat transferred into or out of the system. Q is positive for heat transferred into the system and negative for heat transferred out of the system.

W = workdone by the system or work done on the system. W is positive for workdone on the system and W is negative for when work is done by the system.

ΔU decreases by 250 KJ, that is, ΔU = - 250 KJ

Q = ?

W = 480 KJ (Work is done on the system)

- 250 = Q + 480

Q = - 250 - 480 = - 730 KJ

The heat is transferred out of the system.

6 0

2 years ago

Read 2 more answers

A)iii) as the wheelbarrow was being pushed, the tyre went over a small stone. Describe and explain how the pressure changes.

slamgirl [31]

Answer:

it changed because the rock put the pressure all in one spot.

Explanation:

4 0

2 years ago

Khi vẽ đồ thị lưu ý nào sau đây không đúng

melomori [17]

Answer:

whatyyttyyyyy

Explanation:

pagtarong og ask

5 0

2 years ago