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Tamiku [17]
1 year ago
7

What two vertical forces act on a falling leaf?(1 point)

Physics
1 answer:
Sloan [31]1 year ago
5 0

Option C. Weight, Friction are the vertical forces that act on the leaf falling.

<h3>What are vertical forces?</h3>

These are forces that act in the vertical plane, which is perpendicular to the ground. There are at least two vertical forces acting. Gravitational force is provided by the earth and always acts downwards.

The weight, W (force) acting on the leaf is as a result of the product of the mass(m) of the leaf and the acceleration due to gravity (g).

Mathematically; W = mg

Friction in the form of air resistance also try to stop or reduce this acceleration due to gravity (g) on the falling leaf and it acts upwards.

In conclusion, Weight and Friction (air resistance) are the two vertical forces acting on a falling leaf.

Learn more about forces: brainly.com/question/12970081

#SPJ1

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astraxan [27]

Answer:

Explanation:

3. The word circuit means "go around", therefore a circuit is a pathway or closed path around which electricity (or water) flows.

4. Electrons flowing through a wire can be compared to water flowing through a hose. Once the flow of electrons or water is going, work, is performed.

5. You would get shocked in a bumper car by touching the floor and the ceiling at the same time. This means you are completing the circuit allowing electricity to flow.

6. Electricity from a wall outlet has enough energy to stop your heart.

7. Electricity is the flow of electrons, because electrons move or jump from atom to atom.

8. Materials that allow electrons to move easily from atom to atom are called conductors.

9. Materials that do not allow electrons to flow easily are called insulators.

10. Semi- conductors are materials that are somewhere in between .

11. Voltage is the force or pressure of electricity and is compared to the amount of water pressure in a hose.

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4 0
2 years ago
Point P is on the rim of a wheel of radius 2.0 m. At time t = 0, the wheel is at rest, and P is on the x-axis. The wheel undergo
oksian1 [2.3K]

Answer:

e). a = 0.066 m/s^2

Explanation:

As we know that wheel is turned by 90 degree angle

so the angular speed of the wheel is given as

\omega_f^2 - \omega_i^2 = 2\alpha \theta

now we have

\omega_f^2 - 0 = 2(0.01)(\frac{\pi}{2})

\omega = 0.177 rad/s

now the centripetal acceleration of the point P is given as

a_c = \omega^2 R

a_c = (0.177)^2(2)

a_c = 0.063 m/s^2

tangential acceleration is given as

a_t = R\alpha

a_t = 2(0.01)

a_t = 0.02 m/s^2

now net acceleration is given as

a = \sqrt{a_t^2 + a_c^2}

a = \sqrt{0.02^2 + 0.063^2}

a = 0.066 m/s^2

8 0
3 years ago
What is the relationship between lightning and thunder
PtichkaEL [24]
Whenever lightning strikes it separates the air where it goes. This air then rushes back together making a loud noise when it connects, creating thunder.
6 0
3 years ago
A pilot heads his jet due east. The jet has a speed of 475 mi/h relative to the air. The wind is blowing due north with a speed
oksian1 [2.3K]

Explanation:

a. The velocity of the wind as a vector in component form will be represented as v vector:

    v=30j

b.The velocity of the jet relative to the air as a vector in component form will be represented as u vector

    u=475i

c. The true velocity of the jet as a vector will be represented as w:

  w=u+v

  w=475i+30j

d.  The true speed of the jet will be calculated as:

    IwI=\sqrt{(475)^2+(30)^2}

    IwI=\sqrt{225625+900}

    IwI=\sqrt{226525}

    IwI=476 mi/h

e. The direction of the jet will be:

tita=tan^{-1}\frac{30}{475}

tita=tan^{-1}(0.0632)

tita=3.62degrees,or,N86.38degreesS

7 0
3 years ago
A veritical brass rod of circular section is loaded by placing a 10 kg wt on top of it .if it's length is 1 m. it's radius of cr
Inga [223]

Answer:

4.37 * 10^-4 J

Explanation:

Energy stored :

mgΔl / 2

m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m

Δl = mgl / πr²Y

Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m

Δl = (10 * 9.8 * 1) / π * (1 * 10^-2)²* 3.5 ×10^10

Δl = 98 / 3.5 * π * 10^6

Δl = 0.00000891267

Energy stored :

mgΔl / 2

(10 * 9.8 * 0.00000891267) / 2

= 0.00043672083 J

4.37 * 10^-4 J

3 0
2 years ago
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