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masya89 [10]
2 years ago
10

If Earth were twice as far from the sun, the force of gravity attracting the Earth to the Sun would be

Physics
2 answers:
zalisa [80]2 years ago
6 0
If Earth was twice as far from the sun, the force of gravity attracting the Earth to the sun would be only one-quarter as strong. The correct answer will be C. 
sveticcg [70]2 years ago
3 0
It would be half as strong
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An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s
max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

4 0
2 years ago
Calculate the kinetic energy of a motorcycle of mass 60kg travelling at a velocity of 40km/h​
ELEN [110]

Answer:

1848.15J

Explanation:

KE =1/2 mv^2

Mass = 60kg, velocity =40km/h =11.11m/s

Hence

KE =30 x(11.1)^2 /2 = 1848.15J

5 0
2 years ago
The van of Hans and Frans is stuck on slippery ice, so they must get out and move it by hand. Hans pushes
Irina18 [472]

Answer: a= ff+fh/m

Explanation: bc khan academy said it was d. a=ff +fh/m

4 0
2 years ago
if you apply a Force of F1 to area A1 on one side of a hydraulic jack, and the second side of the jack has an area that is twice
Firlakuza [10]

Answer:

2F_{1}

Explanation:

F₁ = Force on one side of the jack

A₁ = Area of cross-section of one side of the jack

F₂ = Force on second side of the jack

A₂ = Area of cross-section of second side of the jack = 2 A₁

Using pascal's law

\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{A_{_{2}}}

\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{2A_{_{1}}}

F_{1}= \frac{F_{2}}{2}\\

F_{2}= 2F_{1}

7 0
2 years ago
a crane lifts a 35000 N steel girder a distance of 25 m in 45 s. How much power did the crane require to lift the girder
Angelina_Jolie [31]
Hello.

The formula for Power is Work divided by Time; however, we do not have our value for Work - yet.
To find for the Work inputted, we need to use its formula: Force * Distance.

Let's multiply our Force by our Distance. Remember that our Force is always  measured in Newtons (N), and our Distance is measured by Meters (M).
35,000 * 25 = 875,000 J (Unit for Work is "J" or "Joules")

Now that we have the value for Work, let's apply it to our Power formula.
P = 875,000 / 45; 19,444.44~

The Power required to lift the girder is 1944.44~ W (Unit for Power is "W" or "Watts").

I hope this helps!
7 0
2 years ago
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