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Kruka [31]
3 years ago
5

Marc was told his dinner would be ready at 18:00. He left his house at 12:00 and travelled in his car at an average speed of 45m

ph to his mum's house 300 miles away. Did Marc make it home in time for dinner
Physics
1 answer:
Tema [17]3 years ago
5 0

Answer:

No

Explanation:

<u>No, Marc did not make it on time for dinner.</u>

The time duration between 12:00 and 18:00 is 6 hours.

Total distance that needs to be covered from his house to his mum's house and back = 300 miles x 2 = 600 miles

Time that would be needed to cover 600 miles to his mum's house at an average speed of 45 mph = distance/speed

    = 600/45

            = 13.33 hours

<em>The required time to cover the distance by Marc is approximately 13.33 hours. If he leaves the house by 12:00, he will definitely not get back 6 hours after. Hence, he could not have made it for dinner.</em>

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As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff ideal sprin
shepuryov [24]

Answer:

The magnitude of force must you apply to hold the platform in this position = 888.89 N

Explanation:

Given that :

Workdone (W) = 80.0 J

length x = 0.180 m

The equation  for this  work done by the spring is expressed as:

W = \frac{1}{2}k_{eq}x^2

Making the spring constant k_{eq} the subject of the formula; we have:

k_{eq} = \frac{2W}{x^2}

Substituting our given values, we have:

k_{eq} = \frac{2*80}{0.180^2}

k_{eq} = 4938.27 \ N.m^{-1}

The magnitude of the force that must be apply  to the hold platform in this position is given by the formula :

F = k_{eq}x

F = 4938.27*0.180

F = 888.89 N

6 0
3 years ago
Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 k
Harrizon [31]

Explanation:

Given that,

(a) Speed, v=6.66\times 10^6\ m/s

Mass of the electron, m_e=9.11\times 10^{-31}\ kg

Mass of the proton, m_p=1.67\times 10^{-27}\ kg

The wavelength of the electron is given by :

\lambda_e=\dfrac{h}{m_ev}

\lambda_e=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 6.66\times 10^6}

\lambda_e=1.09\times 10^{-10}\ m

The wavelength of the proton is given by :

\lambda_p=\dfrac{h}{m_p v}

\lambda_p=\dfrac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 6.66\times 10^6}

\lambda_p=5.96\times 10^{-14}\ m

(b) Kinetic energy, K=1.71\times 10^{-15}\ J

The relation between the kinetic energy and the wavelength is given by :

\lambda_e=\dfrac{h}{\sqrt{2m_eK}}

\lambda_e=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.71\times 10^{-15}}}

\lambda_e=1.18\times 10^{-11}\ m

\lambda_p=\dfrac{h}{\sqrt{2m_pK}}

\lambda_p=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.71\times 10^{-15}}}

\lambda_p=2.77\times 10^{-13}\ m

Hence, this is the required solution.

6 0
3 years ago
A 2-kW electric heater takes 15 min to boil a quantity of water. If this is done once a day and power costs 10 cents per kWh, wh
frozen [14]

Answer:

<h2> $1.50</h2>

Explanation:

Given data

power P= 2 kW

time t= 15 min to hours = 15/60= 1/4 h

cost of power consumption per kWh= 10 cent = $0.1

We are expected to compute the cost of operating the heater for 30 days

but let us computer the energy consumption for one day

Energy of heater  for one day= 2* 1/4 = 0.5 kWh

the cost of operating the heater for 30 days= 0.5*0.1*30= $1.50

<u><em>Hence it will cost  $1.50 for 30 days operation</em></u>

4 0
3 years ago
Leandra is baking cookies , and they are ready to come out of the oven . Before she takes them out of the oven , she puts her mi
ch4aika [34]

Answer: Leandra puts on her mittens because if you do not you will burn your self, due to extremely high temperatures.

Explanation:

8 0
3 years ago
Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated
Artemon [7]

Answer: Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN

Explanation: To find the answer we need to know more about the Newton's law of gravitation.

<h3>What is Newton's law of gravitation?</h3>
  • Gravitation is the force of attraction between any two bodies.
  • Every body in the universe attracts every other body with a force.
  • This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between these two masses.
  • Mathematically we can expressed it as,

                        F=\frac{GMm}{r^2} \\where, G=6.67*10^-11Nm^2kg^-2

<h3>How to solve the problem?</h3>
  • Here, we have given with the data's,

                      M=8.22*10^9kg\\m=1.38*10^8 kg\\r=1.43*10^3m

  • Thus, the force of attraction between these two bodies will be,

               F=6.67*10^-11*\frac{8.22*10^9*1.38*10^8}{1.43*10^3} =52.9kN

Thus, if two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg and, If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN.

Learn more about the Newton's law of gravitation here:

brainly.com/question/28045318

#SPJ4

6 0
1 year ago
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