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Jlenok [28]
4 years ago
6

If an irregularly shaped object (such as a wrench) is dropped from rest in a classroom and feels no air resistance, it will:

Physics
1 answer:
Maslowich4 years ago
7 0

Answer:

D) accelerate but will not spin.

Explanation:

On the off chance that there is no air resistance the object will accelerate yet won't turn, this is on the grounds that without air resistance same force is applied on each bit of the object. Force on each segment is coordinated descending i.e parallel force. So there is no force to deliver spin movement in it.  

It will accelerate because of gravity

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What is the potential energy of two electrons that are separated by a distance of 3.5 x 10^-11m ?
Simora [160]

Answer:

6.58×10⁻¹⁸ J

Explanation:

Applying

E = kq²/r.................. Equation 1

Where E = potential energy, q = charge on each electron, r = distance between the electron, k = coulomb's constant.

From the question,

Given: r = 3.5×10⁻¹¹ m,

Constant: q = 1.6×10⁻¹⁹ C, k = 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

E = (1.6×10⁻¹⁹)²(8.99×10⁹)/(3.5×10⁻¹¹)

E = 6.58×10⁻¹⁸ J

4 0
3 years ago
Which two things does force depend on?​
saveliy_v [14]

Answer:

Mass and Gravity

Explanation:

Objects with mass exert forces on each other via the force of gravity. This force is proportional to the mass of the two interacting objects and is inversely proportional to the square of the distance between them.

3 0
3 years ago
how do scientists know that black holes exist A. by running experiments on the sun B. by observing objects and light around blac
NNADVOKAT [17]

Answer:

E. none of the above

Explanation:

I hope thats right

8 0
3 years ago
Read 2 more answers
It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet
TEA [102]

Answer:

v=346.05\ m.s^{-1}

Explanation:

Given:

initial temperature of the lead bullet, T_i=43^{\circ}C

latent heat of fusion of lead, L_f=2.32\times 10^4\ J.kg^{-1}

melting point of lead, T_m=327.3^{\circ}C

We have:

specific heat capacity of lead, c=129\ J.kg^{-1}.K^{-1}

<em>According to question the whole kinetic energy gets converted into heat which establishes the relation:</em>

\rm KE=(heat\ of\ rising\ the\ temperature\ from\ 43\ to\ 327.3\ degree\ C)+(heat\ of\ melting)

\frac{1}{2} m.v^2=m.c.\Delta T+m.L_f

\frac{1}{2} m.v^2=m(c.\Delta T+L_f)

\frac{v^2}{2} =129\times(327.3-43)+23200

v=346.05\ m.s^{-1}

3 0
3 years ago
Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the
ale4655 [162]

I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)

Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.

The 2.0-kg block feels

• the downward pull of its own weight, (2.0 kg) <em>g</em>

• the upward normal force of the surface, magnitude <em>n₁</em>

• kinetic friction, mag. <em>f₁</em> = 0.30<em>n₁</em>, pointing in the negative horizontal direction

• the contact force of the larger block, mag. <em>c₁</em>, also pointing in the negative horizontal direction

• the applied force, mag. <em>F</em>, pointing in the positive horizontal direction

Meanwhile the 3.0-kg block feels

• its own weight, (3.0 kg) <em>g</em>, pointing downward

• normal force, mag. <em>n₂</em>, pointing upward

• kinetic friction, mag. <em>f₂</em> = 0.30<em>n₂</em>, pointing in the negative horizontal direction

• contact force from the smaller block, mag. <em>c₂</em>, pointing in the <u>positive</u> horizontal direction (this is the force that is causing the larger block to move)

Notice the contact forces form an action-reaction pair, so that <em>c₁</em> = <em>c₂</em>, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:

• net vertical force:

<em>n₂</em> - (3.0 kg) <em>g</em> = 0   ==>   <em>n₂</em> = (3.0 kg) <em>g</em>   ==>   <em>f₂</em> = 0.30 (3.0 kg) <em>g</em>

• net horizontal force:

<em>c₂</em> - <em>f₂</em> = 0   ==>   <em>c₂</em> = 0.30 (3.0 kg) <em>g</em> ≈ 8.8 N

4 0
3 years ago
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