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Scilla [17]
3 years ago
8

Suppose a candy bar is 8 cm long, 1 cm high, and 5 cm wide. How many whole candy bars will you be able to fit in a box with a vo

lume of 300 cm3?
A.8 candy bars
B.7.5 candy bars
C.7 candy bars
Physics
1 answer:
Illusion [34]3 years ago
7 0
C: 7 candy bars is the answer i am pretty sure
You might be interested in
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
Velocity has two pieces of information.what are they
seraphim [82]
They are speed and direction.
3 0
3 years ago
20. A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kPa. When the gas is allowed to
aliina [53]
Given:\\V_1=4m^3\\T_1=290K\\p_1=475kPa\\\\V_2=6.5m^3\\T_2=277K\\\\Find:\\p_2=?\\\\Solution:\\\\ \frac{pV}{T} =const.\\\\ \frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \\\\\frac{p_1V_1T_2}{T_1}=p_2V_2\\\\\frac{p_1V_1T_2}{T_1V_2}=p_2\\\\p_2=p_1 \frac{V_1}{V_2}  \frac{T_2}{T_1} \\\\\\p_2=475kPa\cdot  \frac{4m^3}{6.5m^3} \cdot  \frac{277K}{290K} \approx 279.2kPa\\\\Correct\;is\;answer\;\;C
7 0
2 years ago
6.
yaroslaw [1]

Answer:

12 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 7.6 kg

Distance (d) = 6 m

Velocity (v) = 5 m/s

Force (F) = 2 N

Workdone (Wd) =.?

Workdone can be defined as the product of force and distance moved in the direction of the force. Mathematically, it is expressed as:

Workdone = Force × distance

Wd = F × d

With the above formula, we can obtain the workdone as follow:

Distance (d) = 6 m

Force (F) = 2 N

Workdone (Wd) =.?

Wd = F × d

Wd = 2 × 6

Wd = 12 J

Thus, the workdone is 12 J

6 0
2 years ago
a man is trying to move a piano by himself, but it will not move, what force must they overcome to start the piano moving?
m_a_m_a [10]

Answer:

The friction of the piano and the weight

Explanation:

7 0
2 years ago
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