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3 years ago

The speed of light is about 300,000 km/s (3×105km/s). If Jupiter is 0.72 light hours from the Sun, how far is this?

Physics

2 answers:

Daniel [21]3 years ago

7 0

Answer:

7.77 x 10⁸ Km

Explanation:

given,

Speed of light = 3 x 10⁵ Km/s

= 3 x 10⁸ m/s

1 light hour = 1.079 x 10⁹ Km

now,

0.72 light hour = 0.72 x 1.079 x 10⁹ Km

= 0.7769 x 10⁹ Km

= 7.77 x 10⁸ Km

The Jupiter is 7.77 x 10⁸ Km far from sun.

frozen [14]3 years ago

7 0

Answer:

Explanation:

speed of light, v = 3 x 10^8 m/s

time = 0.72 hour

distance = velocity x time

distance = 3 x 10^8 x 0.72 x 3600

distance = 7776 x 10^8 m = 7.78 x 10^8 km

Thus, the distance is 7.78 x 10^8 km.

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An object moves on a trajectory given by Bold r left parenthesis t right parenthesis equals left angle 10 cosine 6 t comma 10 si

spayn [35]

Answer:

Explanation:

(r) = <10 cos 6t, 10 Sin 6t>

The distance traveled by the object is the magnitude of vector r.

The magnitude of vector r is given by

$r = \sqrt{(10 Cos 6t)^{2}+(10 Sin 6t)^{2}}$

$r =10 \sqrt{(Cos^{2} 6t)+(Sin^{2} 6t)$

r = 10

5 0

2 years ago

A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

Explanation:

$m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg$

$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

3 0

3 years ago

A light horizontal spring has a spring constant of 138 N/m. A 3.35 kg block is pressed against one end of the spring, compressin

BARSIC [14]

Answer:

$U_k = 0.113$

Explanation:

using the law of the conservation of energy:

$E_i -E_f=W_f$

$\frac{1}{2}Kx^2=NU_kd$

where K is the spring constant, x is the spring compression, N is the normal force of the block, $U_k$ is the coefficiet of kinetic friction and d is the distance.

Also, by laws of newton, N is calculated by:

N = mg

N = 3.35 kg * 9.81 m/s

N = 32.8635

So, Replacing values on the first equation, we get:

$\frac{1}{2}(138)(0.123)^2= (32.8635)U_k(0.281m)$

solving for $U_k$ :

$U_k = 0.113$

8 0

3 years ago

The interference pattern seen when light passes through narrow, closely spaced slits, is due to

Lelu [443]

<h2>Answer: Diffraction</h2><h2 />

Diffraction is a characteristic phenomenon that occurs in all types of waves .

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Note that the principal condition for the occurrence of this phenomena is that <u>the obstacle must be comparable in size (similar size) to the size of the wavelength. </u>

<u />

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3 years ago

block with of mass m is at rest on horizontal frictionless surface at time t=0. A force given by F=Bt+C is applied horizontally

Alexeev081 [22]

Answer:

$v_{2} =\frac{1}{2}$