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3 years ago

Joe and Matt fight over the last piece of sushi. Joe uses his chopsticks to

Physics

1 answer:

dimaraw [331]3 years ago

4 0

Answer:

Matt

Explanation:

because he has more force

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How much power is required to light a lightbulb at 100V of voltage when the lightbulb has a resistance of 500 Ohms?

salantis [7]

Answer:

Power = 20 Watts

Explanation:

Given the following data;

Voltage = 100 V

Resistance = 500 Ohms

To find the power that is required to light a lightbulb;

Mathematically, power can be calculated using the formula;

$Power = \frac {Voltage^{2}}{resistance}$

Substituting into the formula, we have;

$Power = \frac {100^{2}}{500}$

$Power = \frac {10000}{500}$

Power = 20 Watts

3 0

3 years ago

A petrol tanker ha 2800kg when empty and hold 30m3 0f petrol when full. The denity of petrol i 740kg/m3. Calculate the ma of the

White raven [17]

The mass of the tanker with petrol is 250000 N.

We are given that,

Mass of tanker= m =2800 kg

Volume of petrol= v =30 m³

Density of petrol= d =740 kg/m³

Thus , mass , density and volume relation can be given as,

density= mass/ Volume

Mass = Density × Volume

Mass = 740× 30

Mass = 22200 kg

The mass of the petrol is 22200 kg.

Total mass of tanker with petrol = Mass of petrol + Mass of tanker

Total mass of tanker with petrol= 22200+ 2800 kg

Total mass of tanker with petrol= 25000 kg

Total weight of the tanker with petrol = Mass of tanker full of petrol× g

Where, weight = m × g ,(g =10m/s²)

Total weight of the tanker with petrol= 25000×10 = 250000 N

Therefore, the mass of petrol, total mass of tanker with petrol and weight of tanker with petrol would be 22200 kg, 25000 kg and 250000 N.

To know more about mass

brainly.com/question/24100921

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4 0

1 year ago

A small metal object is tied to the end of a string and whirled round a circular path of radius 30cm. The object makes 20 oscill

LUCKY_DIMON [66]

Answer:

(i) The angular speed of the small metal object is 25.133 rad/s

(ii) The linear speed of the small metal object is 7.54 m/s.

Explanation:

Given;

radius of the circular path, r = 30 cm = 0.3 m

number of revolutions, n = 20

time of motion, t = 5 s

(i) The angular speed of the small metal object is calculated as;

$\omega = \frac{20 \ rev}{5 \ s} \times \frac{2 \pi \ rad}{1 \ rev} = \frac{40\pi \ rad}{5 \ s} = 8\pi \ rad/s = 25.133 \ rad/s$

(ii) The linear speed of the small metal object is calculated as;

$v = \omega r\\\\v = 25.133 \ rad/s \ \times \ 0.3 \ m\\\\v = 7.54 \ m/s$

6 0

3 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

7 0

3 years ago

A mechanic changing the spark plugs in a car notes that the instruction manual calls for a torque with a magnitude of

Papessa [141]

The magnitude (in N) of the force she must exert on the wrench is 150.1 N.

<h3>Force exerted by the wrench</h3>

The force exerted by the wrench is calculated using torque formula as follows;

torque, τ = F x r x sinθ

where;

F is the applied force
r is the perpendicular distance if force applied

F = τ /(r sinθ)

F = (39) / (0.3 sin 60)

F = 150.1 N

Thus, the magnitude (in N) of the force she must exert on the wrench is 150.1 N.

Learn more about torque here: brainly.com/question/14839816

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5 0

2 years ago