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Vadim26 [7]
3 years ago
13

Suppose the force of the air drag on an object is proportional to the speed of the object and in the direction opposite the obje

ct's velocity. if you throw an object upward, the magnitude of its acceleration is greatest
Physics
2 answers:
weeeeeb [17]3 years ago
8 0
When the object leaves your hand.

Bingel [31]3 years ago
8 0

Answer:

The acceleration of the object is at is greatest in two moments:

Just when you throw the object (you are appliying force to the objet that accelerates it)

If we only consider the object when is moving, then:

Right after you throw it up, it is traveling upwards, so the force of the air is pusing down, and also the force of gravity, so here the two forces are pulling in the same direction, so in this moment is when the magnitude of the acceleration is greater.

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xxTIMURxx [149]
I think its C not sure tho

8 0
3 years ago
A converging lens of focal length 20 cm is placed in contact with a diverging lens of focal length 30 cm. The focal length of th
kramer

Answer:

The focal lenth (F) =+10.0cm

Explanation:

The formular for combined focal length (F) is given as;

1/f = 1/f_{1} + 1/f_{2}

In this  question,

F1 = 20cm

F2  = -30cm

Plugging the values into the formuar above,

1/f = 1/20 + 1/(-30)

1/f = 0.05 - 0.033[tex]1/f = -0.017f = [tex]1/ -0.017

f = 58.82cm

i.e. the combination behaves as a converging lens (because of the postive sign) of focal length 58.82cm .

7 0
3 years ago
Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum
cestrela7 [59]

Answer:

T_{2}=278.80 K

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}.

Now, let's use the ideal gas equation to the initial and the final state:

\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}

Let's recall that the term nR is a constant. That is why we can match these equations.  

We can find a relation between the volumes of the initial and the final state.

\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}

Combining this equation with the first equation we have:

(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}

(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}

Now, we just need to solve this equation for T₂.

T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40

Finally, T2 will be:

T_{2}=278.80 K

6 0
3 years ago
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Answer:

a

Explanation:

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What is a newton equal to in terms of units of mass and acceleration
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1 newton is the force needed to accelerate 1 kilogram of mass
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It's a force equal to roughly  3.6 ounces.

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3 years ago
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