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irakobra [83]
3 years ago
5

If a car travels 30.0 mi. north for 30.0 min., 50.0 mi. east for 1.00 hour, and 30.0 mi. south for 30.0 min., what is the averag

e speed of the car over the total trip? 55 mph 55.0 mph 60.0 mph 60. mph
Physics
1 answer:
san4es73 [151]3 years ago
4 0
Note that Speed = Distance/Time.

1st trip:
Distance = 30 mi
Time = 30 min = 0.5 h
Speed = 30/0.5 = 60 mi/h

2nd trip:
Distance = 50 mi
Time = 1 hour
Speed = 50 mi/h

3rd trip:
Distance = 30 mi
Time = 30 min = 0.5 h
Speed = 30/0.5 = 60 mi/h

Average speed = (60 + 50 + 60)/3 = 56.7 mph
The closestanswer is 55 mph

Answer: 55 mph

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Armand hands the rock to Pierre, who is standing next to the trampoline. Explain how Armand moves, in terms of the forces acting
Nikolay [14]

Answer:

Armando's weight ,restored force created by the trampoline

a harmonic movement within the trampoline

Explanation:

In a trampoline we have two forces that actuate Armando's weight and the restored force created by the trampoline that depends on the deformation distance of the elastic canvas.

Amando's weight is vertical and directed towards the center of the Earth and has a constant value, this weight is balanced with the elastic force the springboard exerts on Armando in a vertical direction.

In general, when entering the trampoline, a small jump is made, this creates a speed that deforms the canvas until the speed is reduced to zero, at this point the elastic force is greater than the weight and the boy begins to climb, After the boy leaves the canvas he meets only the force of gravity and his speed decreases to zero and begins his fall.

In Summary Armando is in a harmonic movement within the trampoline

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3 years ago
An object is moving in a circle. If the radius of the object is doubled, and the period remains constant, the magnitude of the v
Anton [14]

Answer:

Twice

Explanation:

From the formula for velocity in a circle

V= 2πr/T

Where V is velocity

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T is period

We see that as r increases V increases so if r is doubled V becomes doubled

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The principles of magnetism apply everywhere on earth. What does this tell us about God and His character?
Bas_tet [7]

Answer:

God is omnipresent.

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This means God is everywhere and He works where ever we are in the world

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3 years ago
A 0.5-kilogram apple falls from a height of 2 meters to 1.50 meters. Ignoring frictional effects, what is the kinetic energy of
spin [16.1K]

The  final kinetic energy of the ball is 2.45 J

Explanation:

We can solve this problem by using the law of conservation of energy.

In absence of frictional effect, the mechanical energy of the apple must be conserved during the fall. So we can write:

U_i +K_i = U_f + K_f

where :

U_i is the initial potential energy, at the top

K_i is the initial kinetic energy, at the top

U_f is the final potential energy, at the bottom

K_f is the final kinetic energy, at the bottom

By explicing the potential energy, we can rewrite the equation as:

mgh_i + K_i = mgh_f + K_f

where:

m = 0.5 kg is the mass of the apple

g=9.8 m/s^2 is the acceleration of gravity

h_i = 2 m is the initial height

h_f=1.50 m is the final height

The initial kinetic energy is zero, since the ball starts from rest:

K_i = 0

Therefore we can solve the equation for K_f, the final kinetic energy of the ball:

K_f = mg(h_i-h_f)=(0.5)(9.8)(2-1.50)=2.45 J

Learn more about kinetic energy and potential energy:

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3 years ago
Read 2 more answers
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
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