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bixtya [17]
3 years ago
8

Which chemicals are major contributors to ozone layer destruction? Select two answer choices.

Physics
2 answers:
yarga [219]3 years ago
6 0

Answer:

c and d

Explanation:

Nastasia [14]3 years ago
3 0

Answer:

what are the answers to select? or do you just say them

Explanation:

???

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Martin is conducting an experiment. His first test gives him a yield of 5.2 grams. His second test gives him a yield of 1.3 gram
Bogdan [553]

Complete Question

Martin is conducting an experiment. His first test gives him a yield of 5.2 grams. His second test gives him a yield of 1.3 grams. His third test gives him a yield of 8.5 grams. On average, his yield is 5.0 grams, which is close to the known yield of 5.1 grams of substance. Which of the following are true?

A His results are accurate but not precise.

B His results are neither accurate nor precise.

C His results are both accurate and precise

D His results are precise but not accurate.

Answer:

Correct option is A

Explanation:

From the question we are told that  

   The  yield of the first test k  =  5.2 \  g

   The  yield of the second  is  u =  1.3 \  g

   The  third yield is  p =  8.5 \  g

   The  average yield  A = 5.0 \ g

    The  know yield is  A_S =  5.1 \  g

From the data given we see that

        A_S \ne A

Since his average yield is closer to the known yield then the answer is accurate

But since the yield for each test are not repeated the answer is not precise

So the answer is accurate but not precise  

4 0
3 years ago
Metal surfaces on spacecraft in bright sunlight develop a net electric charge.
koban [17]

Answer:

Positive.

Explanation:

As a consequence of the photoelectric effect, electrons that will get hit by sufficiently energetic photons will abandon the metal surfaces exposed to bright sunlight. This decreases the negative charge of the surface, thus causing it to develop a positive net charge.

5 0
3 years ago
The cannon on a battleship can fire a shell a maximum distance of 36.0 km.
Paraphin [41]

(a)The initial velocity of the shell will be 594.27 m/sec

(b)The maximum height it reaches will be 9000 m.

c)101.249 m meters lower will its surface be 36.0 km from the ship along a horizontal line parallel to the surface of the ship.

d)The error could be significant compared to the size of a target. Option C is correct.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

m is the mass of the block = Kg.

u is the initial velocity of fall = m/sec

h is the distance of fall =  m

g is the acceleration of free fall = m/sec²

v is the hitting velocity of =?

a)

The range of the projectile is;

\rm R = \frac{u^2 sin 2 \theta }{g} \\\\ 36 \times 10^ 3 = \frac{U^2 sin 45^0}{9.81} \\\\ U= 594.27  \ m/sec

b)

The maximum height of the projectile is;

\rm H = \frac{u^2 sin 2 \theta }{2g} \\\\ H = \frac{(594.27)^2\times (sin 45)^2}{2 \times 9.81 } \\\\ H = 9000 \ m

c)

The distance between its surface and the ship, measured in a horizontal arc parallel to the surface, will be 36.0 kilometers. The distance from the lower surface is found as;

\rm( R_e + h)^2 = R_e^2+(36)^2 \\\\ (R_e)^2 = h^2+2R_e h= R_e^2 + 12196 \\\\ h^2 + 12800 h - 1296 = 0 \\\\ h = 101.249 \ m

d)

An error is a mistaken or erroneous action. In some contexts, an error is interchangeable with a mistake.

The difference between the calculated value and the original value is known as the error. The inaccuracy may be large in comparison to the target's size. Option C is correct.

Hence the initial velocity of the shell, maximum height, and the distance from the lower surface will be 594.27 m/sec,9000 m, and 101.249 m and option c for question d are correct respectively.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

4 0
2 years ago
A 10-g bullet moving horizontally with a speed of 2.0 km/s strikes and passes through a 4.0-kg block moving with a speed of 4.2
SVEN [57.7K]

Answer:

K=512J

Explanation:

Since the surface is frictionless, momentum will be conserved. If the bullet of mass m_1 has an initial velocity v_{1i} and a final velocity v_{1f} and the block of mass m_2 has an initial velocity v_{2i} and a final velocity v_{2f} then the initial and final momentum of the system will be:

p_i=m_1v_{1i}+m_2v_{2i}

p_f=m_1v_{1f}+m_2v_{2f}

Since momentum is conserved, p_i=p_f, which means:

m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}

We know that the block is brought to rest by the collision, which means v_{2f}=0m/s and leaves us with:

m_1v_{1i}+m_2v_{2i}=m_1v_{1f}

which is the same as:

v_{1f}=\frac{m_1v_{1i}+m_2v_{2i}}{m_1}

Considering the direction the bullet moves initially as the positive one, and writing in S.I., this gives us:

v_{1f}=\frac{(0.01kg)(2000m/s)+(4kg)(-4.2m/s)}{0.01kg}=320m/s

So kinetic energy of the bullet as it emerges from the block will be:

K=\frac{mv^2}{2}=\frac{(0.01kg)(320m/s)^2}{2}=512J

6 0
4 years ago
The orbital motion of a deep water wave extends to a depth equal to.
Amanda [17]
1/2 the wavelength.......
8 0
2 years ago
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