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ICE Princess25 [194]
1 year ago
13

A miner finds a small mineral fragment with a volume of 5.74 cm^3 and a mass of 28.7 g. What is the density of that mineral frag

ment?
Physics
2 answers:
statuscvo [17]1 year ago
8 0
P=M÷V
p=28.7÷5.74
p=5

The density is 5.
Shkiper50 [21]1 year ago
7 0

Explanation:

Density is defined as mass per unit volume.

Mathematically,       d = \frac{m}{V}

where,             d = density

                       m = mass

                       V = volume

It is given that volume is 5.74 cm^{3} and mass is 28.7 g. Then calculate the density as follows.

                d = \frac{m}{V}

                          = \frac{28.7 g}{5.74 cm^{3}}

                          = 5 g/cm^{3}

Therefore, we can conclude that the density of that mineral fragment is 5 g/cm^{3}.

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A 0.45 Caliber bullet (m = 0.162 kg) leaving the muzzle of a gun at 860
Charra [1.4K]

Answer:139.32kgm/s

Explanation:

Mass(m)=0.162kg

Velocity(v)=860m/s

Momentum=mass x velocity

Momentum=0.162 x 860

Momentum=139.32kgm/s

3 0
1 year ago
Caelyn wanted to find out what shampoo made her hair the shiniest . Everyday she washed her hair with different shampoos and the
Arte-miy333 [17]

Answer:

IV: type of shampoo used

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control: water

Constant: rated from scale from 1-10

Explanation:

8 0
2 years ago
Official (Closed) - Non Sensitive
Pavlova-9 [17]

Answer:

The minimum running time is 319.47 s.

Explanation:

First we find the distance covered and time taken by the train to reach its maximum speed:

We have:

Initial Speed = Vi = 0 m/s    (Since, train is initially at rest)

Final Speed = Vf = 29.17 m/s

Acceleration = a = 0.25 m/s²

Distance Covered to reach maximum speed = s₁

Time taken to reach maximum speed = t₁

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

t₁ = (29.17 m/s - 0 m/s)/(0.25 m/s²)

t₁ = 116.68 s

Using 2nd equation of motion:

s₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

s₁ = (0 m/s)(116.68 s) + (0.5)(0.25 m/s²)(116.68 s)²

s₁ = 1701.78 m = 1.7 km

Now, we shall calculate the end time and distance covered by train, when it comes to rest on next station.

We have:

Final Speed = Vf = 0 m/s    (Since, train is finally stops)

Initial Speed = Vi = 29.17 m/s     (The train must maintain max. speed for min time)

Deceleration = a = - 0.7 m/s²

Distance Covered to stop = s₂

Time taken to stop = t₂

Using 1st equation of motion:

Vf = Vi + at₂

t₂ = (Vf - Vi)/a

t₂ = (0 m/s - 29.17 m/s)/(- 0.7 m/s²)

t₂ = 41.67 s

Using 2nd equation of motion:

s₂ = (Vi)(t₂) + (0.5)(a)(t₂)²

s₂ = (29.17 m/s)(41.67 s) + (0.5)(- 0.7 m/s²)(41.67 s)²

s₂ = 607.78 m = 0.6 km

Since, we know that the rest of 7 km, the train must maintain the maximum speed to get to the next station in minimum time.

The remaining distance is:

s₃ = 7 km - s₂ - s₁

s₃ = 7 km - 0.6 km - 1.7 km

s₃ = 4.7 km

Now, for uniform speed we use the relation:

s₃ = vt₃

t₃ = s₃/v

t₃ = (4700 m)/(29.17 m/s)

t₃ = 161.12 s

So, the minimum running time will be:

t = t₁ + t₂ + t₃

t = 116.68 s + 41.67 s + 161.12 s

<u>t = 319.47 s</u>

5 0
2 years ago
If Pete ( mass=90.0kg) weights himself and finds that he weighs 30.0 pounds, how far away from the surface of the earth is he
shutvik [7]

Answer: 9938.8 km

Explanation:

1 pound-force = 4.48 N

30.0 pounds-force = 134.4 N

The force of gravitation between Earth and object on the surface of is given by:

F = \frac{GMm}{R^2} = mg

Where M is the mass of the Earth, m is the mass of the object, R (6371 km) is the radius of the Earth.

At height, h above the surface of the Earth, the weight of the object:

(mg)'= \frac{GMm}{(R+h)^2}

we need to find "h"

taking the ratio of two:

\frac{mg}{(mg)'}=\frac{(R+h)^2}{R^2}\\ \Rightarrow \frac{90kg \times 9.8 m/s^2}{134.4 N}=\frac{(R+h)^2}{R^2}\\ \Rightarrow 6.56 R^2= (R+h)^2 \Rightarrow h= (2.56-1)R\\ \Rightarrow h = 1.56 R = 1.56 \times 6371 km = 9938. 8 km

Hence, Pete would weigh 30 pounds at 9938.8 km above the surface of the Earth.

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2 years ago
There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the differe
Usimov [2.4K]

An example that illustrates the difference is the circular motion

Explanation:

Let's start by reminding the definition of the two quantities:

- Speed is a scalar quantity that tells "how fast" an object is moving, regardless of its direction of motion.

Speed can be  calculate as:

speed = \frac{d}{t}

where:

d is the distance travelled

t is the time taken

- Velocity is instead a vector quantity, given by:

velocity = \frac{d}{t}

where;

d is the displacement of the object (displacement is a a vector connecting the initial position to the final position of motion)

t is the time taken

Since it is a vector, velocity has both a magnitude and a direction, therefore it also takes into account the direction of motion of the object.

For an object in motion in a straight line, speed and velocity are the same. However, this is not always the case.

In fact, an example of motion in which the two quantities are different is the circular motion. Consider for example the object making one complete revolution along the circle. Therefore, its average speed is the ratio between the length of the perimeter (the distance) divided by the time taken:

speed = \frac{2\pi r}{t}

where r is the radius of the circle.

However, the displacement of the object is zero (because the object returns to the starting point), and so the average velocity is also zero:

velocity = \frac{0}{t}=0

Learn more about speed and velocity:

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#LearnwithBrainly

5 0
2 years ago
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