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3 years ago

What's needed to know

2 answers:

8 0

You're a little late. But if you want some short, quick rules, then these are
a couple that I would take in with me (stored only in my brain, of course):

-- If something is not accelerating or moving at all, then all the forces on it
must add up to zero. That could even mean a hanging rope.

-- In a vertical rope, the tension in it is the same everywhere in the rope.
The tension is the weight of whatever is hanging from the bottom.

That's really all I'm sure of, based on your hazy, fuzzy description of
what you've been doing in class. I don't want to get into things that
you might not have learned yet, and confuse you.

Orlov [11]3 years ago

5 0

You're absolute best bet is to read and re-read the entire chapter one of the book. This is for an absolute best strategy. On the other hand, there are usually key words and terms that are highlighted that most teachers use as a frame for test taking. Studying those highlighted words would be the next best bet. If there are no highlighted words, tough luck. You need to figure out what is important. Which is part of the learning process, and also why I won't directly answer your question. I care more about you learning than you getting a good grade.

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When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is referred to as

Dafna11 [192]

When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is called "load triangle". Because of a motorcycle's size and weight and the fact that it has only two wheels, how to carry extra load is very important. One has to make sure that they are keeping the weight low and close to the middle of the motorcycle and keep the load evenly from side to side. Heavier items should be in the "load triangle".

3 0

3 years ago

Please post detailed answers to the following questions. Please use complete sentences. What are some of the challenges that you

Degger [83]

As what Douglas Adams says, "Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is. I mean, you may think it's a long way down the road to the chemist, but that's just peanuts to space."

The fact that the stars are very far from Earth, one can't fully understand the stars as it would require hundred of years to closely travel from earth to the stars because it is located million miles away from the Earth. Aside from its location, the stars are also hard to closely study because it emits lights that are harm for people when exposed longer.

8 0

3 years ago

A cube of metal has a mass of 11 grams and a volume of 1 cm . When fully submerged in water this metal cube hanging from an accu

Anarel [89]

Answer:

0.098 N

Explanation:

From the question,

Spring scale reading = W-U............... Equation 1

Where W = weight of the cube, U = upthrust.

W = mg

Where m = mass of the cube, g = acceleration due to gravity.

Given: m = 11 g = 0.011 kg, g = 9.8 m/s².

W = 0.011(9.8)

W = 0.1078 N.

From Archimedes principle,

Upthrust = weight of water displaced.

U = (Density of water×volume of metal cube)×acceleration due to gravity.

U = (D×V)g

Given: D = 1000 kg/m², V = 1 cm³ = (1/1000000) = 1×10⁻⁶ m³, g - 9.8 m/s²

U = 1000(9.8)(10⁻⁶)

U = 0.0098 N.

Substitute the value of W and U into equation 1

Reading of the spring scale = 0.1078-0.0098

Reading of the spring scale = 0.098 N

7 0

3 years ago

Max ran 100 feet in 10 seconds.

Ivanshal [37]

Answer:

for max :

100 feet in 10 secs

for molly :

60 feet in 5 secs = 120 feet in 10 secs

so, molly ran farther in the same time interval i.e. covered 120 feet where as Max covered 100 feet

Explanation:

brainliest plz

8 0

3 years ago

Read 2 more answers

A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If

Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

$\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s$

The induced emf in the shorter coil is calculated as;

$E = NA\frac{\delta B}{\delta t}$

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

$E = NA\frac{\delta B}{\delta t}$

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

8 0

3 years ago