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Charra [1.4K]
3 years ago
12

A 10.1 g bullet leaves the muzzle of a rifle with a speed of 425 m/s. What constant force is exerted on the bullet while it is t

raveling down the 0.6 m length of the barrel of the rifle?
Physics
1 answer:
PIT_PIT [208]3 years ago
5 0

Answer

Force will be 1520.2604 N

Explanation:

We have given mass of the bullet m = 10.1 gram = 0.0101 kg

Distance S= 0.6 m

Final velocity v = 425 m/sec

Initial velocity u will be 0 m/sec

From third equation of motion v^2=u^2+2as

425^2=0^2+2\times a\times 0.6

a=150520.833m/sec^2

We know that force is given by F = ma

So force = 0.0101×150520.833 = 1520.2604 N

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d. fishing

The other options all directly affect algal bloom production because they affect the nutrients in the water, and an overabundance of certain nutrients in the water is what causes algal bloom.

6 0
3 years ago
Please awnser and show the ways​
topjm [15]

Answer:

Answers in solutions.

Explanation:

<u>Question 6:</u>

The density of gold is 19.3 g/cm³

The density of silver is 10.5 g/cm³

  • The density of the substance in Crown A;

Density = mass ÷ volume = \frac{1930}{100} = 19.3 g/cm³

Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.

  • The density of the substance in Crown B;

Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043  g/cm³ ≈ 10.5 g/cm³  (answer rounded up to one decimal place)

Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.

  • The density of substance in Crown C;

Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)

<h3><u>The density of the mixture:</u></h3><h3 />

For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g

∴ for 1 cm³ of the mixture, its mass equal to \frac{29.8}{2} = 14.9 g

Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.

* Crown C is not made up of a mixture of gold and silver.

<u>Question 7:</u>

<u />

  • An empty masuring cylinder has a mass of 500 g.
  • Water is poured into measuring cylinder until the liquid level is at the 100 cm³ mark.
  • The total mass is now 850 g

The mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g

Density of the liquid (water) poured into the container = mass ÷  volume = 350 g ÷ 100 cm³ = 3.5g/cm³

<u>Question 8:</u>

<u />

A tank filled with water has a volume of 0.02 m³

(a) 1 liter = 0.001 m³

How many liters? = 0.02 m³ ?

Cross multiplying gives:

\frac{0.02 * 1}{0.001} =  20 liters

(b) 1 m³ = 1,000,000 cm³

0.02 m³ = how many cm³ ?

Cross-multiplying gives;

\frac{0.02 * 1,000,000}{1} = 20,000 cm³

(c) 1 cm³ = 1 ml

∴ 0.02 m³ of the water = 20,000 cm³ = 20,000 ml

<u>Question 9:</u>

<u />

Caliper (a) measurement = 3.2 cm

Caliper (b) measurement = 3 cm

<u>Question 10:</u>

<u />

  • A stone is gently and completely immersed in a liquid of density 1.0 g/cm³
  • in a displacement can
  • The mass of liquid which overflow is 20 g

The mass of the liquid which overflow = mass of the stone = 20 g

1 gram of the liquid occupies 1 cm³ of space.

20 g of the liquid will occupy; \frac{20 * 1}{1} = 20 cm³

(a) Since the volume of the water displaced is equal to the volume of the stone.

∴ The volume of the stone = 20 cm³

(b) Mass = density ×  volume

Density of the stone = 5.0 g/cm³

Volume of the stone = 20 cm³

Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g

7 0
3 years ago
The MAC is 58 inches, The CG limits are from 26% to 43% MAC. If the CG is found to be
eimsori [14]

By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.

<h3>What are the limits?</h3>

First, we need to find the limits.

We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.

So if 58 in is the 100%, the 26% and 43% of that are:

  • 26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
  • 43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.

But we know that the CG is found to be 45.5% MAC, then it measures:

(45.5%/100%)*58in = 0.455*58in = 26.39 in

We need to compare it with the largest limit, so we get:

26.39 in - 24.94 in = 1.45 in

This means that the CG is 1.45 inches out of limits.

If you want to learn more about percentages, you can read:

brainly.com/question/14345924

6 0
2 years ago
Is Heredity nature or nurture?
Luden [163]
The answer is nature
8 0
3 years ago
Read 2 more answers
A spring of force constant 1500 Nm-l is acted
olchik [2.2K]

Answer:

1.876 J

Explanation:

First, let’s calculate the compression of the spring from the Hooke’s law:

F=kx,

here, F=75 N is the force acted on the spring, k=1500 N⁄m is the force constant of the spring, x is the compression of the spring.

Then, we get:

x=F/k=(75 N)/(1500 N/m)=0.05 m.

Finally, we can find the potential energy stored in the spring:

PE=1/2 kx^2=1/2∙1500 N/m∙(0.05 m)^2=1.875 J.

correct my answer if it's wrong ^^

7 0
3 years ago
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