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serious [3.7K]
3 years ago
13

Joey throws a football 30 meters with a force of 10 N. How much work does Joey do?

Physics
1 answer:
stich3 [128]3 years ago
8 0

Answer:

300 Joules

Explanation:

Just use the Work formula:

W = F . D

W = 10 . 30

W = 300 Joules

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A particle is constrained to move round a circle radius 382400km and makes a single revolution in 27.3 days. (i). Find the veloc
andreyandreev [35.5K]

The velocity and acceleration of the particle moving round the circle is;

<em><u>Velocity = 162.12 m/s</u></em>

<em><u>Velocity = 162.12 m/sAcceleration = 6.873 × 10^(-5) m/s²</u></em>

We are given;

Radius of circle; 382400 km = 382400000 m

Time; t = 27.3 days = 27.3 × 86400 s = 2358720 s

Now, formula for velocity is;

Velocity = distance/time

Thus;

I) velocity = 382400000/2358720

Velocity = 162.12 m/s

II) Acceleration is centripetal acceleration and is given by the formula;

a = v²/r

a = 162.12²/382400000

a = 6.873 × 10^(-5) m/s²

Read more at; brainly.com/question/12199398

5 0
3 years ago
What collides and creates a movement of heat called conduction?
lutik1710 [3]
The answer of this is C!!!
6 0
3 years ago
Alarge plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 MPa m1/2. If the plate is expos
Korvikt [17]

Answer:

minimum length of a surface crack is 18.3 mm

Explanation:

Given data

plane strain fracture toughness K = 82.4 MPa m1/2

stress σ = 345 MPa

Y = 1

to find out

the minimum length of a surface crack

solution

we will calculate length by this formula

length = 1/π ( K / σ Y)²

put all value

length = 1/π ( K / σ Y)²

length = 1/π ( 82.4 10^{3/2} / 345× 1)²

length = 18.3 mm

minimum length of a surface crack is 18.3 mm

4 0
3 years ago
What is the source of the radioactivity?
Nataly_w [17]
The Earth itself is a source of terrestrial radiation. Radioactive materials including uranium, thorium, and radium exist naturally in soil and rock. Essentially all air contains radon<span> , which is responsible for most of the dose that Americans receive each year from natural background sources.</span>
4 0
3 years ago
In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.
leonid [27]

Answer:

(a): F_e = 8.202\times 10^{-8}\ \rm N.

(b): F_g = 3.6125\times 10^{-47}\ \rm N.

(c): \dfrac{F_e}{F_g}=2.27\times 10^{39}.

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053\times 10^{-9} m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges q_1 and q_2 respectively is given by

F_e = \dfrac{k|q_1||q_2|}{r^2}

where,

  • k = Coulomb's constant = 9\times 10^9\ \rm Nm^2/C^2.
  • r = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, q_1 = +1.6\times 10^{-19}\ C.

The charge on the electron, q_2 = -1.6\times 10^{-19}\ C.

These two are separated by the distance, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.

Part (b):

The gravitational force of attraction between two objects of masses m_1 and m_1 respectively is given by

F_g = \dfrac{Gm_1m_2}{r^2}.

where,

  • G = Universal Gravitational constant = 6.67\times 10^{-11}\ \rm Nm^2/kg^2.
  • r = distance of separation between the masses.

For the given system,

The mass of proton, m_1 = 1.67\times 10^{-27}\ kg.

The mass of the electron, m_2 = 9.11\times 10^{-31}\ kg.

Distance between the two, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.

The ratio \dfrac{F_e}{F_g}:

\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.

6 0
3 years ago
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