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2 years ago

13

An object launched from Earth must attain a speed of 7,900 m/s to achieve a low orbit. What happens if the object’s maximum spee

d is less than 7,900 m/s?

1 answer:

Bingel [31]2 years ago

5 0

It will not achieve a state of low orbit and will not go into space.

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What is the potential energy of the system composed of the three charges q1, q3, and q4, when q1 is at point R? Define the poten

Ad libitum [116K]

Answer:

Incomplete question check attachment for complete question

Also it is given that

q1=-0.7uC

q4=-1.7uC

q3=-1.7uC

Also the distance are given as

a=2.2cm=0.022m

d2=3.6cm=0.036m

Explanation:

The potential energy due to point R is given as

The potential energy due to charge q1 and q3 plus the potential energy due to charge q4 and q1 plus the potential energy due to charge q3 and q4

So, let take it one after the other

Potential energy is give as

P.E=kq1q2/r

Therefore,

Potential energy due to charge q1 and q3

U¹³=kq1q3/r

To get the distance between charge q1 and q3, we will apply Pythagoras theorem

r=√(d2²+a²)

r=√(0.036²+0.022²)

r=0.0422m

k is a constant =9×10^9Nm²/C²

Then,

U¹³=kq1q3/r

U¹³=9×10^9×0.7×10^-6×1.7×10^-6/0.0422

U¹³=0.254J

Potential energy due to charge q1 and q4

U¹⁴=kq1q4/r

To get the distance between charge q1 and q4, we will apply Pythagoras theorem

r=√(d2²+a²)

r=√(0.036²+0.022²)

r=0.0422m

k is a constant =9×10^9Nm²/C²

Then,

U¹⁴=kq1q4/r

U¹⁴=9×10^9×0.7×10^-6×1.7×10^-6/0.0422

U¹⁴=0.254J

Potential energy due to charge q3 and q4

U³⁴=kq3q4/r

r=2a=2×0.022=0.044m

k is a constant =9×10^9Nm²/C²

Then,

U³⁴=kq3q4/r

U³⁴=9×10^9×1.7×10^-6×1.7×10^-6/0.044

U¹⁴=0.591J

Then, the total energy is

U= U¹³+ U¹⁴ + U³⁴

U=0.254+0.254+0.591

U=1.099J

Then also, the potential energy is zero because at infinity both U¹³ and U¹⁴ will have infinite potential because their distance apart will be infinite.

6 0

3 years ago

The magnetic field at the earth's surface can vary in response to solar activity. During one intense solar storm, the vertical c

Flauer [41]

Answer:

$EMF = 33880 Volts$

Explanation:

As per Faraday's law of Electromagnetic induction we know that

Rate of change in magnetic flux will induce EMF in the closed conducting loop

so we have

$\phi = B.A$

now we have

$A = (110 \times 10^3)(110 \times 10^3)$

$A = 1.21 \times 10^{10}$

now we have

$\phi = B(1.21 \times 10^{10})$

now the induced EMF through this loop is given as

$EMF = (\frac{dB}{dt})(1.21 \times 10^{10})$

$EMF = (2.8 \times 10^{-6})(1.21 \times 10^{10})$

$EMF = 33880 Volts$

5 0

3 years ago

Now assume that Eq. 6-14 gives the magnitude of the air drag force on the typical 20 kg stone, which presents to the wind a vert

chubhunter [2.5K]

Answer:

362.41 km/h

Explanation:

F = Force

m = Mass = 84 kg

g = Acceleration due to gravity = 9.81 m/s²

C = Drag coefficient = 0.8

ρ = Density of air = 1.21 kg/m³

A = Surface area = 0.04 m²

v = Terminal velocity

F = ma

$F=\frac{1}{2}\rho CAv^2\\\Rightarrow mg=\frac{1}{2}\rho CAv^2\\\Rightarrow v=\sqrt{2\frac{mg}{\rho CA}}\\\Rightarrow v=\sqrt{2\frac{20\times 9.81}{1.21\times 0.8\times 0.04}}\\\Rightarrow v=100.66924\ m/s$

Converting to km/h

$100.66924\times 3.6=362.41\ km/h$

The terminal velocity of the stone is 362.41 km/h

5 0

2 years ago

Read 2 more answers

a 230 kg roller coaster reaches the top of the steepest hill with a speed of 6.2 km/h. It then descends the hill, which is at an

igor_vitrenko [27]

Answer: 81.619 kJ

Explanation:

Given

Mass of roller coaster is $m=230\ kg$

It reaches the steepest hill with speed of $u=6.2\ km/h\ or \ 1.72\ m/s$

Hill to bottom is 51 m long with inclination of $45^{\circ}$

Height of the hill is $h=51\sin 45^{\circ}=36.06\ m$

Conserving energy to get kinetic energy at bottom

Energy at top=Energy at bottom

$\Rightarrow K_t+U_t=K_b+U_b\\\Rightarrow \dfrac{1}{2}mu^2+mgh=K_b+0\\\\\Rightarrow K_b=0.5\times 230\times 1.72^2+230\times 9.8\times 36.06\\\Rightarrow K_b=340.216+81,279.24\\\Rightarrow K_b=81,619.456\ J\\\Rightarrow K_b=81.619\ kJ$

8 0

2 years ago

What happens when calcium reacts with chlorine?

choli [55]

Answer: They create calcuim chloride, CaCl2

7 0

2 years ago