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2 years ago

A sample of gold has a mass of 30.94 grams and density of 19.32g/cm^3. What volume of space will this sample of gold occupy?

2 answers:

6 0

Data:
mass, m = 30.94 g
density, d = 19.32 g/cm^3

Formula: d = m / v => v = m / d = 30.94 g / 19.32 g/cm^3 = 1.60 cm^3

Then, the answer is the option C.

Brums [2.3K]2 years ago

6 0

Answer:

1. 0.07

2. 1.15

Explanation:

plzz make brainliest

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With which field of science is Albert Einstein associated? biology chemistry medecine physics

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He was a theoretical physicist

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3 years ago

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Vilka [71]

Answer:

4.44m/s

Explanation:

Given parameters:

Number of laps = 4

Length of track = 400m

Time taken = 6min

Unknown:

Average speed = ?

Solution:

The average speed is the total distance covered divided by the time taken.

It is mathematically expressed as;

Average speed = $\frac{total distance}{time taken}$

Total distance = number of laps x length of track = 4 x 400 = 1600m

Now convert the time to seconds;

60s = 1 min

6 minutes will be 6 x 60 = 360s

So;

Average speed = $\frac{1600}{360}$ = 4.44m/s

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2 years ago

Need help with a physics question.

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3 years ago

An empty 2,500 kg train car is headed northbound at a velocity of 5 m/s. Ahead of the first car, an empty 1,500 kg car is headed

Tema [17]

Let the mass of 2500 kg car be $m_1$ and it's velocity be $v_1$ and the mass of 1500 kg car be $m_2$ and it's velocity be $v_2$ .

After the bumping the mass be M and it's velocity be V.

By law of conservation of momentum we have

$m_1v_1+m_2v_2 = MV$

2500 * 5 + 1500 * 1=4000 * V

V = 14000/4000 = 7/2 = 3.5 m/s

So the velocity of the two-car train = 3.5 m/s

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3 years ago

What is the minimum force require to move a 5kg wooden crate on a wooden floor?

kolbaska11 [484]

You need to know the coefficient of static friction between a wooden object and a wooden surface. I'll denote it with µ. If you're given a specific value you should obviously use that.

By Newton's second law, the horizontal and vertical net forces are

• net horizontal:

∑ F = p - f = 0

• net vertical:

∑ F = n - w = 0

where

p = magnitude of the pushing force

f = mag. of friction

n = mag. of the normal force

w = weight of the crate

The second equation gives

n = w = (5 kg) (9.8 m/s²) = 49 N

Friction is proportional to the normal force by a factor of µ, so

f = µ (49 N) = 49µ N

To overcome static friction, the push has to exceed this in magnitude, so that

p > 49µ N

For instance, if p = 0.25, then p would need to greater than 12.25 N. (This example isn't particularly helpful, though, since both possibly correct options are larger than 12.25 N...)

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3 years ago