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Otrada [13]
3 years ago
9

A grapefruit falls from a tree and hits the ground 0.76 s later.

Physics
1 answer:
Tomtit [17]3 years ago
7 0

Answer:

2.8 m 7.4 m/s

Explanation:

write all the values then use the equations of motion to find the distance and speed. please see attached photo

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According to Coulomb's law, what happens to the attraction of two oppositely charged objects as their distance of separation inc
NARA [144]

Answer:

Option B. Decreases

Explanation:

Coulomb's law states that:

F = Kq₁q₂ / r²

Where:

F => is the force of attraction between two charges

K => is the electrical constant.

q₁ and q₂ => are the two charges

r => is the distance apart.

From the formula:

F = Kq₁q₂ / r²

The force of attraction (F) is inversely proportional to the square of their separating distance (r).

This implies that as the distance between them increase, the force of attraction between the two charges will decrease and as the distance between two charges decrease, the force of attraction between them will increase.

Considering the question given above and the illustration given above, the force of attraction will decrease as their distance of separation increases.

Option B gives the right answer to the question.

7 0
3 years ago
In the formula Work=Force X Distance, what happens to the amount of work done if either Force or Distance is increased?
AfilCa [17]

Explanation:

If force or distance is increased, then amount of workdone will also increase.

6 0
3 years ago
I will mark brainliest if you answer please im givingalot of points also
Tamiku [17]

A. two forces

B. P-f

Hope this helps!

6 0
3 years ago
A force, F1, of magnitude 2.0 N and directed due east is exerted on an object. A second force exerted on the object is F2 = 2.0
IgorLugansk [536]

Answer:

Magnitude of the force is

F_3 = 2.83

direction of the force is given as

\theta = 45 degree West of South

Explanation:

As we know that force is a vector quantity and in order to find the resultant of two or more forces we need to add them vectorialy

So here we have

\vec F_1 + \vec F_2 + \vec F_3 = 0

here we know that first force is of magnitude 2 N towards east

\vec F_1 = 2 \hat i N

second force is also of 2.0 N due North

\vec F_2 = 2 \hat j

now from above equation

2\hat i + 2\hat j + \vec F_3 = 0

\vec F_3 = -2\hat i - 2\hat j

so magnitude of the force is given as

F_3 = \sqrt{2^2 + 2^2}

F_3 = 2.83

direction of the force is given as

\theta = tan^{-1}\frac{F_y}{F_x}

\theta = tan^{-1}\frac{-2}{-2}

\theta = 45 degree West of South

3 0
3 years ago
Read 2 more answers
When UV light of wavelength 248 nm is shone on aluminum metal, electrons are ejected withmaximum kinetic energy 0.92 eV. What ma
Lina20 [59]

Answer:

The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm

Explanation:

Given;

wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m

maximum kinetic energy of the ejected electron, K.E = 0.92 eV

let the work function of the aluminum metal = Ф

Apply photoelectric equation:

E = K.E + Ф

Where;

Ф is the minimum energy needed to eject electron the aluminum metal

E is the energy of the incident light

The energy of the incident light is calculated as follows;

E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J

The work function of the aluminum metal is calculated as;

Ф = E - K.E

Ф = 8.02 x 10⁻¹⁹  -  (0.92 x 1.602 x 10⁻¹⁹)

Ф =  8.02 x 10⁻¹⁹ J   -  1.474 x 10⁻¹⁹ J

Ф = 6.546 x 10⁻¹⁹ J

The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;

\phi = hf = \frac{hc}{\lambda_{max}} \\\\\lambda_{max} = \frac{hc}{\phi} \\\\\lambda_{max} = \frac{(6.626\times 10^{-34}) \times (3 \times 10^8) }{6.546 \times 10^{-19}} \\\\\lambda_{max} = 3.037 \times 10^{-7} m\\\\\lambda_{max} = 303.7 \ nm

3 0
3 years ago
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