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Jlenok [28]
3 years ago
6

Sam was investigating a container of water. Water can be a solid, liquid, or gas. At first, Sam said the water molecules were mo

ving around each other. Later, the water molecules were moving in place. What change did they observe to the water?
At first, the water was a gas. Later, it was a liquid.
At first, the water was a liquid. Later, it was a solid.
At first, the water was a solid. Later, it was a liquid.
At first, the water was a liquid. Later, it was a gas.
Physics
2 answers:
Andreyy893 years ago
8 0

Answer:

At first, the water was a gas. Later, it was a liquid.

Explanation:

aleksandr82 [10.1K]3 years ago
4 0

Answer:

A) At first, the water was a gas. Later, it was a liquid.

Explanation:

The molecules of a solid cant move at all, so therefore we can knock out B and C. Moleculaes move more freely in a gas form than in a water form, so therefore the firt form has to be a gas, so knock out the answer choice D, which keaves you with A. I hope this helps :)

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Two technicians are discussing torque wrenches. Technician A says that a torque wrench is capable of tightening a fastener with
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3 years ago
A man has a mass of 110 kg. What is his weight?<br> A. 110N<br> B. 1325 N<br> C. 559 N<br> D. 1078 N
Slav-nsk [51]

Answer:

Option D

Explanation:

<u><em>Given:</em></u>

Mass = m = 110 kg

Acceleration due to gravity = g = 9.8 m/s

<u><em>Required:</em></u>

Weight = W = ?

<u><em>Formula</em></u>

W = mg

<u><em>Solution:</em></u>

W = (110)(9.8)

W = 1078 N

7 0
3 years ago
Help ME QUICK PLEASE QUESTION 2 AND 4 PLEASE 58 POINTS
monitta

Answer:

For number 4: A vector pointing to the right with a magnitude of 2.0

Explanation:

Very simple- just subtract 6-2

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6 0
3 years ago
A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the
andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and  how far away from the overpass is the roadrunner when the coyote drops the net.

d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0
3 years ago
A 132 cm wire carries a current of 2.2 A. The wire is formed into a circular coil and placed in a B Field of intensity 1 T. a) F
EastWind [94]

Given Information:

Length of wire = 132 cm = 1.32 m

Magnetic field = B =  1 T

Current = 2.2 A

Required Information:

(a) Torque = τ = ?

(b) Number of turns = N = ?

Answer:

(a) Torque = 0.305 N.m

(b) Number of turns = 1

Explanation:

(a) The current carrying circular loop of wire will experience a torque given by

τ = NIABsin(θ)   eq. 1

Where N is the number of turns, I is the current in circular loop, A is the area of circular loop, B is the magnetic field and θ is angle between B and circular loop.

We know that area of circular loop is given by

A = πr²

where radius can be written as

r = L/2πN

So the area becomes

A = π(L/2πN)²

A = πL²/4π²N²

A = L²/4πN²

Substitute A into eq. 1

τ = NI(L²/4πN²)Bsin(θ)

τ = IL²Bsin(θ)/4πN

The maximum toque occurs when θ is 90°

τ = IL²Bsin(90)/4πN

τ = IL²B/4πN

torque will be maximum for N = 1

τ = (2.2*1.32²*1)/4π*1

τ = 0.305 N.m

(b) The required number of turns for maximum torque is

N = IL²B/4πτ

N = 2.2*1.32²*1)/4π*0.305

N = 1 turn

8 0
3 years ago
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