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2 years ago

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As a tractor pulls a plow across a field, it exerts a force on the plow in the forward direction. What is the "equal and opposit

e force" in this interaction, according to Newton's third law?

1 answer:

Karolina [17]2 years ago

7 0

Answer:

the force of the plow pulling backward on the tractor

Explanation:

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A block is attached to a horizontal spring. On top of this block rests another block. The two-block system slides back and forth

Minchanka [31]

Answer:

Angular frequency will increase

No change in the amplitude

Explanation:

At extreme end of the SHM the energy of the SHM is given by

$E = \frac{1}{2} (m_1 + m_2)\omega^2 A^2$

here we know that

$\omega^2 = \frac{k}{m_1 + m_2}$

now at the extreme end when one of the mass is removed from it

then in that case the angular frequency will change

$\omega'^2 = \frac{k}{m_1}$

So angular frequency will increase

but the position of extreme end will not change as it is given here that the top block is removed without disturbing the lower block

so here no change in the amplitude

6 0

2 years ago

What happens to the current supplied by the battery when you add an identical bulb in parallel to the original bulb?

Marysya12 [62]

Answer:

b. The current stays the same.

Explanation:

In the case given current is supplied by the battery to a bulb . Here, we should know that bulb also apply resistance to the flow of current .

Now, when an identical bulb is connected in parallel to the original bulb .

Therefore, both the resistance( bulb) are in parallel.

We know, when two resistance are in parallel , current through them is same and voltage is divided between them.

Therefore, in this case current stays same in the original bulb.

Hence, this is the required solution.

6 0

3 years ago

How much heat is needed to melt 16.5kg of silver that is initially at 20*c?

zubka84 [21]

The Specific Heat Capacity of silver is 230 J/kgK, melting point is 961.8 C so the difference is 941.8K. Now we simply do q=230J/kgK*16.5kg*941.8K and that is 3 574 131 J

8 0

3 years ago

When a car is parked in hot sunlight, the temperature of the car’s seats _____ and the thermal energy of the seats _____ .

defon

More info? I think the question is incomplete. Although, I believe the first 2 blanks are "rises"

4 0

3 years ago

Read 2 more answers

The 3.00 kg cube in fig. 15-47 has edge lengths d 6.00 cm and is mounted on an axle through its center. a spring (k 1200 n/m con

ira [324]

The Period of the resulting shm will be T=39.7

<u>Explanation:</u>

<u>Given data</u>

m=3kg

d=.06m

k=1200 N/m

Θ=3 °

T=?

we have the formulas,

I = (1/6)Md2

F = ma

F = -kx = -(mω2x)

k = mω2 τ = -d(FgsinΘ)

T=2 x 3.14/ √(m/k)

Solution for the given problem would be,

F=-Kx (where x= dsin Θ)

F=-k dsin Θ

F=-(1200)(.06)sin(3 °)

F=-10.16N

<u>By newton's second law.</u>

F = ma

a= F/m

a=(-10.16N)/3

a=3.38

<u>using the k=mω value</u>

k=mω

ω=k/m

ω=1200/3

ω=400

<u>Using F = -kx value</u>

x = F/-k

x=(-10.16)/1200

x=0.00847m

<u>Restoring the torque value </u>

τ = -dmgsinΘ where( τ = Iα so.).. Iα = -dmgsinΘ α = -(.06)(4)α =

α =(.06)(4)(9.81)sin(4°)

α=-1.781

<u>Rotational to linear form</u>

a = αr

r = .1131 m

a=-1.781 x .1131 m

a=-0.2015233664

<u>Time Period</u>

T=2 x 3.14/ √(m/k)

T=6.28/√(3/1200)

T=6.28/0.158

T=39.7

6 0

3 years ago