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vova2212 [387]
3 years ago
14

A child kicks a ball horizontally with a speed of 2.8 m/s from the end of a deck that is 8.5 m high.

Physics
1 answer:
Bond [772]3 years ago
7 0

Answer:

Explanation:

time =d/s

therefore, t =8.5/2.8

t=3.02s

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A bicyclist in the Tour de France crests a mountain pass as he moves at 18 km/h. At the bottom, 4.0 km farther, his speed is at
Allisa [31]

We are given:

v0 = initial velocity = 18 km/h

d = distance = 4 km

v = final velocity = 75 km/h

a =?

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We can solve this problem by using the formula:</span>

v^2 = v0^2 + 2 a d

 

75^2 = 18^2 + 2 (a) * 4

5625 = 324 + 8a

<span>a = 662.625 km/h^2</span>

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Is vapor pressure independent of atmospheric pressure? - This is what my instructor told me, but I don't understand why. ...?
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Vapor pressure<span> or equilibrium </span>vapor pressure<span> is defined as the </span>pressure<span> exerted by a </span>vapor<span> in thermodynamic equilibrium with its condensed phases at a certain temperature. It is independent with atmospheric pressure since it does not change by changing the atmospheric pressure only. </span>
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3 years ago
QUESTION 1
mario62 [17]
It's D. If it absorbed it would be turning to steam. I am taking honors chem in high school we are learning this.
7 0
3 years ago
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How does mass affect density? How does volume affect density?
Musya8 [376]

Answer:

Here's the Density Formula:  D = M/V

Q: How does mass affect density?

A: <em>Mass is a factor in density, the density is proportional to the mass. So as the mass increases, so does the density, provided the volume remains constant.</em>

Q: How does volume affect density?

A:<em> If an object has a larger mass than its volume it has a high density, if an object has a smaller mass than its volume it has a lower density.</em>

Explanation:

<em><u>I really Hope this Helps!!</u></em>

7 0
2 years ago
What is the total entropy change if a 0.280 efficiency engine takes 3.78E3 J of heat from a 3.50E2 degC reservoir and exhausts i
babymother [125]

Answer:

\Delta S=1.69J/K

Explanation:

We know,

\eta=1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}      ..............(1)

where,

η = Efficiency of the engine

T₁ = Initial Temperature

T₂ = Final Temperature

Q₁ = Heat available initially

Q₂ = Heat after reaching the temperature T₂

Given:

η =0.280

T₁ = 3.50×10² °C = 350°C = 350+273 = 623K

Q₁ = 3.78 × 10³ J

Substituting the values in the equation (1) we get

0.28=1-\frac{Q_2}{3.78\times 10^{3}}

or

\frac{Q_2}{3.78\times 10^3}=0.72

or

Q_2=3.78\times 10^3\times0.72

⇒ Q_2 =2.721\times 10^3 J

Now,

The entropy change (\Delta S) is given as:

\Delta S=\frac{\Delta Q}{T_1}

or

\Delta S=\frac{Q_1-Q_2}{T_1}

substituting the values in the above equation we get

\Delta S=\frac{3.78\times 10^{3}-2.721\times 10^3 J}{623K}

\Delta S=1.69J/K

7 0
3 years ago
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