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Dmitrij [34]
3 years ago
11

X applies a force of 500 N to push Y 100 m across a road how much work does X do?​

Physics
1 answer:
dolphi86 [110]3 years ago
7 0

Answer:

from the formula of work done =Force *Distance 500*100=50000

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Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
BARSIC [14]

Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
2 years ago
20% Part (a) Use an "E Field Sensor" and move it along either equipotential. What can you say about the E field along an equipot
Alex

Answer:

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

Explanation:

As we know that the relation between electric field and electric potential is given as

\Delta E = -\frac{dV}{dr}

here if we say that potential is constant because electric field sensor is moving along equi-potential line.

Then we will say

V = constant

so we have

\Delta E = 0

so electric field will remain constant always in magnitude and always remains perpendicular to the surface

so we have

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

6 0
3 years ago
An electric fan is set on its HIGH setting. After the Low push button is depressed, the angular speed of the fan blades decrease
brilliants [131]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

<span>a=Δω/Δt
</span><span>a=2π*Δf/Δt
</span><span>a=2π*(f2-f1)/Δt
</span>
<span>f1=f2-a*Δt/2π
</span><span>f2=800/60 rev/sec
</span><span>a=-42 rad/sec^2
</span><span>Δt=1.75sec
</span><span>so f1=25 rev/sec
 f1=1500 rev/min</span>
8 0
3 years ago
A diffraction grating with 1000 lines per mm is used in a spectrometer to measure the wavelengths of light emitted from a gas di
frez [133]

Answer:

= 9.8°

Explanation:

Width of one slit (a₁ ) = 1 / 1000 mm=0.001 mm = 10⁻⁶ m.

width of one slit in case 2 (a₂ ) = 1/500 =2 x 10⁻⁶ m

angular position of fringe, Sinθ  = n λ /a

n is order of fringe , λ is wave length of light and a  is slit aperture

So Sinθ  ∝ 1 / a

Sin θ₁ /Sin θ₂ = a₂/a₁ ;

Sin20°/sinθ₂ = 2 / 1

sinθ₂ = Sin 20° / 2 = .342/2 = .171

θ₂ = 9.8 °

4 0
3 years ago
Need help please... thank u​
tangare [24]

Answer:

the answer from my side is both vertical and horizontal

4 0
2 years ago
Read 2 more answers
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