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Shalnov [3]
3 years ago
8

An electric drag racer is much like its piston engine counterpart, but instead it is powered by an electric motor running off of

onboard batteries. These vehicles are capable of covering a 1/4 mile straight-line track in 12 s .
A) Determine the acceleration of the drag racer in units of m/s2. (Assume that the acceleration is constant throughout the race.)
B) How does the value you get compare with the acceleration of gravity?
C) Calculate the final speed of the drag racer in mi/h.
Physics
1 answer:
Lapatulllka [165]3 years ago
6 0

Answer:

5.5879 m/s²

0.56961g

149.9976 mph

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement = 0.25 miles

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 0.25\times 1609.34=0\times t+\frac{1}{2}\times a\times 12^2\\\Rightarrow a=\dfrac{0.25\times 1609.34\times 2}{12^2}\\\Rightarrow a=5.5879\ m/s^2

The acceleration is 5.5879 m/s²

Dividing by g

\dfrac{a}{g}=\dfrac{5.5879}{9.81}\\\Rightarrow a=0.56961g

The acceleration of these cars is 0.56961 times g

v=u+at\\\Rightarrow v=0+5.5879\times 12\\\Rightarrow v=67.0548\ m/s

Converting mph

67.0548\times \dfrac{3600}{1609.34}=149.9976\ mph

The speed is 149.9976 mph

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Answer:

f = 12 cm

Explanation:

<u>Center of Curvature</u>:

The center of that hollow sphere, whose part is the spherical mirror, is known as the ‘Center of Curvature’ of  mirror.

<u>The Radius of Curvature</u>:

The radius of that hollow sphere, whose part is the spherical mirror, is known as the ‘Radius of Curvature’ of  mirror. It is the distance from pole to the center of curvature.

<u>Focal Length</u>:

The distance between principal focus and pole is called ‘Focal Length’. It is denoted by ‘F’.

The focal length of the spherical (concave) mirror is approximately equal to half of the radius of curvature:

f = \frac{R}{2}

where,

f = focal length = ?

R = Radius of curvature = 24 cm

Therefore,

f = \frac{24\ cm}{2}

<u>f = 12 cm</u>

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2 years ago
Of the planets with atmospheres, which is the warmest?<br> a. Venusb. Earthc. Marsd. Jupiter
tangare [24]

The Atmosphere in Jupiter is full of gases that move at high speeds in giant eddies. Its atmosphere consists mostly of gases such as hydrogen that generate a temperature fluctuation of around 128K.

On Earth, due to the protection of the Ozone Layer and the presence of Nitrogen and Oxygen, the temperature fluctuates by an average of 300K.

In the case of Mars, its atmosphere is thin, mostly composed of Carbon Dioxide and Diatomic Nitrogen, which allow a temperature oscillation of 210K.

In contrast, the atmosphere of Venus is thick and is composed of carbon dioxide that does not allow the sun's rays to escape, generating an extreme 'greenhouse effect' with temperatures ranging from 737K,

Correct Answer is A.

7 0
3 years ago
If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu
Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}.

Volume of water:

V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}.

Mass of water:

m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}.

Amount of heat that the 15 L water absorbed:

Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}.

What's the mass of the hot steel tool?

The specific heat of carbon steel is 0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}.

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}.

\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}.

m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}.

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3 years ago
The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m
Sonbull [250]

Answer:

density is 10^{6} Mg/µL

Explanation:

given data

density of nuclear = 10^{18} kg/m³

1 ml = 1 cm³

to find out

density of nuclear matter in Mg/µL

solution

we know here

1 Mg = 1000 kg

so

1 m³ is equal to 10^{6} cm³

and here 1 cm³ is equal to  1 mL

so we can say 1 mL is equal to 10³ µL

so by these we can convert density

density = 10^{18} kg/m³

density = 10^{18} kg/m³ × \frac{10^{-3} }{10^{6} }  Mg/µL

density =  10^{6} Mg/µL

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3 years ago
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laiz [17]

Answer:

The 39.

Explanation:

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2 years ago
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