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vampirchik [111]
2 years ago
8

Please help!!

Physics
1 answer:
sukhopar [10]2 years ago
3 0

Answer:

<u>: WHY DIDN'T THE POD DOCK LIKE IT WAS SUPPOSED TO DO?</u><u> </u>

<u>ANSWER</u><u>;</u>

The force exerted by the thrusters caused the pod to change direction.

WHAT NEW THEORIES DO YOU HAVE?

ANSWER;

This pod moved differently because it was more massive.

<em><u>C</u></em><em><u>A</u></em><em><u>R</u></em><em><u>R</u></em><em><u>Y</u></em><em><u>O</u></em><em><u>N</u></em><em><u>L</u></em><em><u>E</u></em><em><u>A</u></em><em><u>R</u></em><em><u>N</u></em><em><u>I</u></em><em><u>N</u></em><em><u>G</u></em><em><u>:</u></em><em><u>)</u></em>

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these are the types of attractions found between molecules. their strength determines the state of the substance at room tempera
Elina [12.6K]
<h2>Answer</h2>

The physical state of the elements depends upon the <u>attraction forces </u>and their <u>kinetic energy</u>.

<h2>Explanation</h2>

The elements or substances are fixed with each other with the help of different chemical forces including ionic bonding, covalent bonding, H- bonding etc. The strength of these forces is also one of the factors that affect their physical natures. For example, covalent or ionic bonds are the strongest bonds than all other bonds and metals that contain these forces are mostly in solid form. The kinetic motion of electrons in the element also affects the physical state of the element and potential of bonding.

7 0
3 years ago
Read 2 more answers
During each cycle, a refrigerator ejects 610 kJ of energy to a high-temperature reservoir, and takes in 505 kJ of energy from a
jenyasd209 [6]

Answer

A. the work done on the refrigerant in each cycle is 105kJ

B the coefficient of performance of the refrigerator is 4.8

Explanation

Given data

Work done at high temperature T2 Qh=610kJ

Work done at low temperature T1 Ql=505kJ

We know that the net work done by the refrigerator is expressed as

Wnet= Qh-Ql

=610-505

=105kJ

Also we know that the coefficient of performance is expressed as

COP= Ql/Wnet

COP= 505/105

= 4.8

8 0
3 years ago
Which of these statements best describes a similarity between the two galaxies?
Jlenok [28]
Elliptical and Spiral have some similarities, they both are huge and contain lost of dust and also they are held by gravitational forces. 
8 0
3 years ago
Read 2 more answers
A simple pendulum has a period of 3.45 second, when the length of the pendulum is shortened by 1.0m, the period is 2.81 second c
den301095 [7]

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, L_1=L-1

New time period of the pendulum is, T_1=2.81\ s

We know that, the time period of a simple pendulum of length 'L' is given as:

T=2\pi\sqrt{\frac{L}{g}}-------------- (1)

So, for the new length, the time period is given as:

T_1=2\pi\sqrt{\frac{L_1}{g}}------------ (2)

Squaring both the equations and then dividing them, we get:

\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1

Now, plug in the given values and calculate 'L'. This gives,

L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m

Therefore, the original length of the simple pendulum is 2.97 m

4 0
3 years ago
A 2.5 m -long wire carries a current of 8.0 A and is immersed within a uniform magnetic field B⃗ . When this wire lies along the
leva [86]

Answer:

Explanation:

Let the magnetic field be B = B₁i + B₂j + B₃k

Force = I ( L x B )  , I is current , L is length and B is magnetic field .

In the first case

force = - 2.3 j N

L = 2.5 i

puting the values in the equation above

- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]

= - 20 B₃ j + 20 B₂ k

comparing LHS and RHS ,

20B₃ = 2.3

B₃ = .115

B₂ = 0

In the second case

L = 2.5 j

Force = I ( L x B )

2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )

=  - 20 B₁ k + 20B₃ i

2.3i−5.6k = - 20 B₁ k + 20B₃ i

B₃ = .115

B₁ = .28

So magnetic field B = .28 i + .115 B₃

Part A

x component of B = .28 T

Part B

y component of B = 0

Part C

z component of B = .115 T .

8 0
3 years ago
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