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2 years ago

- The gravitational force on a 352 kg satellite is -2100 N. Find its

Physics

1 answer:

Dimas [21]2 years ago

6 0

Answer:

8173.8 km

Explanation:

Fg=G(M*m/r^2)

r=sqrt(G*M*m/Fg)

G=6.6743*10^-11

M(earth)=5.972*10^24kg

M(object)=352kg

Fg=-2100N

r=sqrt((6.6743*10^-11)*(5.972*10^24kg)*(352)/-2100)

r=8173808m

r=8173.8km

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2 years ago

A cinder cone is formed from relatively ________ viscosity magma with a ________ gas content. low; high high; high low; low high

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2 years ago

Monochromatic light is incident on (and perpendicular to) two slits separated by 0.200 mm, which causes an interference pattern

marusya05 [52]

Answer:

I = 0.636*Imax

Explanation:

(a) To find the fraction of the maximum intensity at a distance y from the central maximum you use the following formula:

$I=I_{max}cos^2(\frac{\pi d}{\lambda L}y)$ (1)

I: intensity of light

Imax: maximum intensity of light

d: separation between slits = 0.200mm = 0.200 *10^-3 m

L: distance from the screen = 613cm = 0.613 m

y: distance to the central peak of the interference pattern

λ: wavelength of light = 656.3 nm = 656.3 *10^-9 m

You replace the values of all variables in the equation (1):

$I=I_{max}cos^2(\frac{\pi (0.200*10^{-3}m)}{(656.3*10^{-9}m)(0.613m)}0.600m)\\\\I=I_{max}cos^2(937.06)=0.636I_{max}$

Hence, the fraction of the maximum intensity is I = 0.636*Imax

6 0

2 years ago

In a nuclear reactor, each atom of uranium (of atomic mass 235 u) releases about 200 MeV when it fissions. What is the change in

Ket [755]

Answer:

0.002372187708 kg

Explanation:

Each atom of Uranium 235 releases 200 MeV = 200×10⁶×1.60218×10⁻¹⁹

= 200×1.60218×10⁻¹³ Joule

Number of atoms in a 2.6 kg sample mass = (2.6/0.235)×6.02214076×10²³

⇒Number of atoms in a 2.6 kg sample mass = 66.627×10²³ atoms

Change in energy = Change in mass / (speed of light)²

ΔE = Δmc²

⇒200×1.60218×10⁻¹³×66.627×10²³ = Δm×(3×10⁸)²

⇒Δm = 200×1.60218×10⁻¹³×66.627×10²³/(3×10⁸)²

⇒Δm = 2372.187708×10⁻⁶ kg

∴Change in mass = 0.002372187708 kg

7 0

2 years ago

A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi

Vlad [161]

Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block $m = 6.60$ kg

Initial speed of block $v _{i} = 1.56$ $\frac{m}{s}$

Final speed of block $v_{f} = 0$ $\frac{m}{s}$

Coefficient of kinetic friction $\mu _{k} = 0.62$

Ramp inclined at angle $\theta =$ 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

$\mu _{k} mgd \cos 28.4 = \Delta K - \Delta U$

Where $\Delta U = mg d\sin 28.4$

$\mu _{k} mgd \cos 28.4 = \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4$

$\mu _{k} mgd \cos 28.4 +mgd\sin 28.4 = \frac{1}{2}mv_{i} ^{2}$

$d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}$

$d = 0.122$ m

Therefore, the distance travel by block before coming to rest is 0.122 m

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3 years ago