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Mars2501 [29]
3 years ago
7

What kind of image is formed by an object that is placed beyond the center of curvature on the principal axis of a concave mirro

r?
A.
real, inverted, the same size as the object, and at the same distance as the object
B.
real, inverted, larger than the object, and farther from the mirror than the object
C.
virtual, upright, left-right reversal, the same size as the object, and at the same distance as the object
D.
real, inverted, smaller than the object, and closer to the mirror than the object
E.
virtual, upright, larger than the object, and farther from the mirror than the object
Physics
1 answer:
aev [14]3 years ago
7 0

Answer:D.

Real, inverted, smaller than the object, and closer to the mirror than the object

Explanation: Real images will be inverted. As it is beyond the centre of curvature it is also beyond 2F which means that the image is inside the centre of curvature ( between F and 2F from the mirror ) As the image is closer to the mirror than the object it must be diminished in size

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If an airplane flies at 364 km/h, how long will it take to get from New York to Washington DC (288 km)? (If needed, round to the
Kazeer [188]

Answer:

0.791 h

Explanation:

Distance = rate × time

288 km = 364 km/h × t

t = 0.791 h

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3 years ago
At high pressures, what two factors will cause deviations during ideal gas law calculations?
Tju [1.3M]

At high pressures, the two factors that cause deviation during ideal gas law calculation are the size of molecular and intermolecular force.

The high pressure causes the molecules to approach each other at a very close distance. In that case, if the intermolecular force of attraction is high, the molecules may undergo a state transition, which will result in a completely different outcome as predicted by Ideal gas law.

If the size of the molecule is more, that is for heavy gases like refrigerants, the ideal gas law deviates due to the fact that, with increase in pressure, the volume of gas can no longer be considered as negligible.

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4 0
2 years ago
A constant force of 560 n was applied to a rock. If 2700j of energy was used to move the rock, how far did the rock move
larisa [96]

Answer:

Work done = Force X Distance

Work done/Force = Distance

2700/560 = Distance

Distance = 4.8 meters

4 0
4 years ago
When does a chemical indicator change?
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Chemical indicator, any substance that gives a visible sign, usually by a colour change, of the presence or absence of a threshold concentration of a chemical species, such as an acid or an alkali in a solution. An example is the substance called methyl yellow, which imparts a yellow colour to an alkaline solution.
8 0
3 years ago
a 3.00 kg block moving 2.09 m/s right hits a 2.22 kg block moving 3.92 m/s left. afterward, the 3.00 kg block moves 1.11 m/s lef
LuckyWell [14K]

The momentum of the 2.22 kg afterwards is 0.91 kg m/s to the right

Explanation:

We can solve this problem by applying the law of conservation of momentum.

In fact, the total momentum before and after the collision must be conserved. So, we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where:

m_1 = 3.00 kg is the mass of the first block

u_1 = 2.09 m/s is the initial velocity of the first block (we take the right as positive direction)

v_1 = -1.11 m/s is the final velocity of the first block (to the left)

m_2 = 2.22 kg is the mass of the second block

u_2 = -3.92 is the initial velocity of the second block

v_2 is the final velocity of the second block

Re-arranging the equation and substituting the values, we find: the final velocity of the second block:

v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(3.00)(2.09)+(2.22)(-3.92)-(3.00)(-1.11)}{2.22}=0.41 m/s

(to the right, since it is positive)

And so, the momentum of the 2.22 kg block afterwards is:

p_2 = m_2 v_2 = (2.22)(0.41)=0.91 kg m/s (to the right)

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8 0
3 years ago
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