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Simora [160]
3 years ago
12

A planet orbits a star, in a year of length 2.35 x 107 s, in a nearly circular orbit of radius 3.49 x 1011 m. With respect to th

e star, determine (a) the angular speed of the planet, (b) the tangential speed of the planet, and (c) the magnitude of the planet's centripetal acceleration.
Physics
1 answer:
Naya [18.7K]3 years ago
7 0

Answer:

a)   w = 9.599 10⁴ rad / s , b)   v = 3.35 10¹⁶ m / s , c)    a = 3.22  10²¹ m / s²

Explanation:

For this exercise we must use the relation of angular kinematics

a) angular velocity, the distance remembered in orbit between time (period)

         w = 2π r / T

         w = 2 π 3.59 10¹¹ / 2.35 10⁷

         w = 9.599 10⁴ rad / s

b) linear and angular velocity are related by the equation

          v = w r

          v = 9,599 10⁴ 3.49 10¹¹

          v = 3.35 10¹⁶ m / s

c) the centripetal acceleration is

            a = v² / r = w² r

            a = (9,599 10⁴)²   3.49 10¹¹

            a = 3.22  10²¹ m / s²

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Determine the potential difference between the ends of the wire of resistance 5 Ω if 720 C passes through it per minute.
Strike441 [17]

Answer:

The potential difference between the ends of a wire is 60 volts.

Explanation:

It is given that,

Resistance, R = 5 ohms

Charge, q = 720 C

Time, t = 1 min = 60 s

We know that the charge flowing per unit charge is called current in the circuit. It is given by :

I = 12 A

Let V is the potential difference between the ends of a wire. It can be calculated using Ohm's law as :

V = IR

V = 60 Volts

So, the potential difference between the ends of a wire is 60 volts. Hence, this is the required solution.

8 0
2 years ago
150c passes through cell in 30 seconds. cell has a potential difference of 12v. what is current in the circuit
zzz [600]

The current in the circuit is 5 A

Explanation:

The intensity of current is given by the equation:

I=\frac{q}{t}

where

I is the current

q is the amount of charge passing through a given point of the circuit in a time interval of t

For the cell in this problem, we have

q = 150 C is the charge

t = 30 s is the time interval

Substituting into the equation, we f ind

I=\frac{150}{30}=5 A

Learn more about current:

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6 0
3 years ago
What mass of silver (in grams) is solidified when 749 joules of heat are released by a sample of molten silver at its freezing p
Anastasy [175]

heat released Q = 749 joules

heat of fusion of silver L = 109 J/g

Here phase of silver is changing from liquid to solid

so temperature will remain same

all heat will be released due to its phase change

and in this case we use Q=mL

where m is the mass of silver in gram

Q= mL

749 = m * 109

m = 749/109

m = 6.87 gram

4 0
3 years ago
Read 2 more answers
Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship jour
Elena-2011 [213]

Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = \frac{t}{\sqrt{1-\frac{v^2}{c^2} } }

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 10

t₁= 16.7  years

D ) Given t = 10 years , t₁ = ? v = .4c

\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 10

t₁= 10.9  years

E ) Given t = 20 years , t₁ = ? v = .8c

\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 20

t₁= 33.4   years

7 0
2 years ago
A cheetah, which is the fastest land mammal, can accelerate at 9 m/s from rest to in 31 m/s. How long does
Anna007 [38]

31 m/s ÷ 9 m/s² = 3.44 s

Time = Change in velocity divided (÷) by acceleration.

8 0
3 years ago
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