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mixas84 [53]
3 years ago
10

Question 1-1: In each case, lifting or pushing, why must you exert a force to keep the object moving at a constant velocity?

Physics
1 answer:
VMariaS [17]3 years ago
5 0

Answer:

We must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).

Explanation:

LIFTING:

When an object is lifted, we first need to overcome the force exerted on it by the field of gravity. Due to this force, which is also called the weight of object, we must apply a force on the object to keep it moving at constant speed, otherwise the gravity force will cause the object to slow down and eventually fall back on ground.

PUSHING:

When pushing an object the person must apply the force to first overcome the frictional force. The frictional force acts in opposite direction of motion. Thus, to move the object at constant speed we must apply force to it.

<u>Hence, we must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).</u>

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A 10248 kg car is pulled by a low tow truck that has an acceleration of 2.0m/s what is the net force on the car
Ksenya-84 [330]

Answer: 20496N

Explanation:

The formula to calculate the net force will be given as:

Net force = Acceleration × Mass

Since 10248 kg car is pulled by a low tow truck that has an acceleration of 2.0m/s, the net force would be:

= 10248 × 2

= 20496N

8 0
3 years ago
A photon detector captures a photon with an energy of 4.29 ✕ 10−19 J. What is the wavelength, in nanometers, of the photon?
serious [3.7K]

Answer :  The wavelength of photon is, 4.63\times 10^{2}nm

Explanation : Given,

Energy of photon = 4.29\times 10^{-19}J

Formula used :

E=h\times \nu

As, \nu=\frac{c}{\lambda}

So, E=h\times \frac{c}{\lambda}

where,

\nu = frequency of photon

h = Planck's constant = 6.626\times 10^{-34}Js

\lambda = wavelength of photon  = ?

c = speed of light = 3\times 10^8m/s

Now put all the given values in the above formula, we get:

4.29\times 10^{-19}J=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{\lambda}

\lambda=4.63\times 10^{-7}m=4.63\times 10^{-7}\times 10^9nm=4.63\times 10^{2}nm

Conversion used : 1nm=10^{-9}m

Therefore, the wavelength of photon is, 4.63\times 10^{2}nm

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3 years ago
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Answer:

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Explanation:

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3 years ago
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When light reaches the end of a barrier on the edge of an opening it diffracts, which means...
Mama L [17]
Hey there Evan!

Let's remember, these are (light waves). So, we you learn about waves, you would remember that when light does reach an end, they will most likely just (bounce back). That's all it would actually do.

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3 0
3 years ago
A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th
irina [24]

The angular momentum is defined as,

L=I\omega

Acording to this text we know for conservation of angular momentum that

L_i=L_f

Where L_iis initial momentum

L_f is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

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At the point 0 we have that,

L_i=mvl

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

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L_f=(I_b+I_s)\omega

Substituting

L_f=(ml^2+\frac{10ml^2}{3})\omega

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m(0.65)l=\frac{13}{10}ml^2 \omega

\omega=\frac{1}{2l}

Applying conservative energy equation we have

\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'

\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)

Replacing the values and solving

l=\frac{13}{0.54g}

Substituting

l=\frac{13}{0.54(9.8)}

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7 0
4 years ago
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