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Natali5045456 [20]

1 year ago

7

As a space shuttle moves through the dilute ionized gas of Earth's ionosphere, the shuttle's potential is typically changed by -

1.1 V during one revolution. Assuming the shuttle is a conducting sphere of radius 10 m, estimate the amount of charge it collects.

1 answer:

Svetllana [295]1 year ago

8 0

Answer:

The amount of charge the space shuttle collects is -1.224nC

Explanation:

The magnitude of Electric potential is given as;

V = kq/r

where;

V is the electric potential in volts

k is coulomb's constant

r is the radius of the sphere or distance moved by the charge

given; V = -1.1 V, k = 8.99 x 10⁹ Nm²/C², r = 10m

Substituting this values in the above equation, we estimate the amount of charge space shuttle collects.

q = (V*r)/k

q = (-1.1 *10)/(8.99 x 10⁹ )

q = -1.224 X 10⁻⁹ C

q = -1.224nC

Therefore, the amount of charge the space shuttle collects is -1.224nC

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A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car p

DedPeter [7]

Answer:

a) $v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}$

b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

c) $a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }$

Explanation:

<em><u>The knowable variables are </u></em>

$d_{t}=25m$

$t_{1}=1.3 s$

$t_{2}=3.9 s$

$t_{3}=5.5 s$

Since the three traffic signs are <u>equally spaced</u>, the <u>distance between each sign is $\frac{25}{3} m$ </u>

a) $v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}$

b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

Since we know the velocity in two points and the time the car takes to pass the traffic signs

c) $a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }$

6 0

1 year ago

A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity

Likurg_2 [28]

Answer:

a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2 = 239.6 N,

b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm

Explanation:

Given that:

γ= 9.5 kN/m³ = 9500N/m3

b = 6 inches = 0.1524 m

t = 0.0013 mm

d = 2 inches = 0.0508 m

n = 1750 rpm

$H_{nom}=2hp=1491.4W$

L = 9 ft = 2.7432 m

Ks = 1.25

g = 9.81 m/s²

a)

$w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m$

$V=\frac{\pi d n}{60} =\pi *0.0508*1750/60=4.65 m/s$

$F_c=\frac{wV^2}{g}=1.88*4.65^2/9.81=4.15N$

$(F_1)_a=bF_aC_pC_v=0.1524*6000*0.7*1=640N$

$T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm$

$F_2=(F_1)_a-\frac{2T}{D}= 640-\frac{2*10.17}{0.0508} =239.6N$

$F_i=\frac{(F_1)_a+F_2}{2} -F_c=435.65N$

b)

$H_a=1491*1.25=1863.75W$

$n_f_s=\frac{H_a}{H_{nom}K_S }=1$

dip = $\frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm$

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How do you do projectile motion problems

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The key projectile motion is that gravity allows downward only

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2 years ago

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A certain satellite travels in an approximately circular orbit of radius 2.0 × 106 m with a period of 7 h 11 min. Calculate the

kap26 [50]

Answer: Mass of the planet, M= 8.53 x 10^8kg

Explanation:

Given Radius = 2.0 x 106m

Period T = 7h 11m

Using the third law of kepler's equation which states that the square of the orbital period of any planet is proportional to the cube of the semi-major axis of its orbit.

This is represented by the equation

T^2 = ( 4π^2/GM) R^3

Where T is the period in seconds

T = (7h x 60m + 11m)(60 sec)

= 25860 sec

G represents the gravitational constant

= 6.6 x 10^-11 N.m^2/kg^2 and M is the mass of the planet

Making M the subject of the formula,

M = (4π^2/G)*R^3/T^2

M = (4π^2/ 6.6 x10^-11)*(2×106m)^3(25860s)^2

Therefore Mass of the planet, M= 8.53 x 10^8kg

5 0

1 year ago

A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is 20 m above the ground below. A cannonball

OLga [1]

<u>Answer:</u>

Cannonball will be in flight before it hits the ground for 2.02 seconds

<u>Explanation:</u>

Initial height from ground = 20 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when s = 20 meter.

Substituting

$20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds$

So it will take 2.02 seconds to reach ground.

5 0

1 year ago