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levacccp [35]
3 years ago
8

John is traveling north at 20 meters/second and his friend Betty is traveling south at 20 meters/second. If north is the positiv

e direction, what are John and Betty's speeds?
Physics
2 answers:
Tems11 [23]3 years ago
6 0

Explanation:

John is traveling north at 20 meters/second and his friend Betty is traveling south at 20 meters/second.

We know that the total distance covered divided by total time taken is called speed of an object. It is a scalar quantity i.e. it have only magnitude. While velocity is a vector quantity.

In this case, north direction is considered as positive while south direction is considered as negative.

So, John's speed is 20 meters/second and Betty's speed is 20 meters/second as speed is a scalar quantity.

DanielleElmas [232]3 years ago
3 0
John's velocity is 20 m/s north.
Betty's velocity is 20 m/s south.
Their speeds are both 20 m/s.
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In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.2 106 m/s.
Murrr4er [49]

The central force acting on the electron as it revolves in a circular orbit is 9.52 \times 10^{-8} \ N.

The given parameters;

  • <em>speed of electron, v = 2.2 x 10⁶ m/s</em>
  • <em>radius of the circle, r = 4.63 x 10⁻¹¹ m</em>

<em />

The central force acting on the electron as it revolves in a circular orbit is calculated as follows;

F = \frac{M_e v^2}{r} \\\\

where;

M_e is mass of electron = 9.11 x 10⁻³¹ kg

F = \frac{(9.11 \times 10^{-31}) \times(2.2\times 10^6)^2 }{4.63 \times 10^{-11}} \\\\F = 9.52 \times 10^{-8} \ N

Thus, the central force acting on the electron as it revolves in a circular orbit is 9.52 \times 10^{-8} \ N.

Learn more about centripetal force here:brainly.com/question/20905151

8 0
3 years ago
Which of these belongs to a tropic level with the most energy?
MrRa [10]

Answer:

Berries is the correct answer because it is the produce in your pyramid and as each living thing is devoured by another there is less energy. For instance the berry has the most energy because it’s energy has just come from the sun. But then an insect eats it and consumes most of its energy but some energy is released into the atmosphere. Then a rodent eats the bug and consumes its energy but yet again some energy is released into the atmosphere. So each time there is less and less energy. Does that help any?

Explanation:

it’s energy has just come from the sun. But then an insect eats it and consumes most of its energy but some energy is released into the atmosphere.

8 0
3 years ago
A tiger runs at 58 km/h [S]. What is the displacement of the tiger in 38 s?
Artyom0805 [142]
58 K/h = 58000/3600= 16.1 m/s
In 38 s displacement is 38x16.1= 612.2 m
3 0
3 years ago
Two capacitors of capacitances 25 µF and 50 µF are connected in series with a 33-V battery. How much energy is stored in the 25-
torisob [31]

Answer:

6.05×10⁻³ J

Explanation:

Note: Two capacitors connected in series behaves like two resistors connected in parallel.

Using

1/Ct = 1/C1+1/C2

Ct = (C1×C2)/(C1+C2)............................ Equation 1

Where Ct = combined capacitance of the two capacitor, C1 = Capacitance of the first capacitor, C2 = capacitance of the second capacitor.

Given: C1 = 25 µF, C2 = 50 µF

Substitute into equation 1

Ct = (25×50)/(25+50)

Ct = 1250/75

Ct = 16.67 µF.

Using

Q = CV.................... Equation 2

Where Q = Charge, V = Voltage.

Given: V = 33 V, C = 16.67 µF = 16.67×10⁻⁶ F

Substitute into equation 2

Q = 33(16.67×10⁻⁶)

Q = 5.5×10⁻⁴ C.

Since both capacitors are connected in series, the same amount of charge flows through them.

Using,

E = 1/2Q²/C.................. Equation 3

Where E = Energy stored in the 25-µF capacitor

Given: Q =5.5×10⁻⁴ C, C = 25 µF = 25×10⁻⁶ F

Substitute into equation 3

E = 1/2(5.5×10⁻⁴)²/ 25×10⁻⁶

E = 6.05×10⁻³ J.

5 0
3 years ago
In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting a
IRISSAK [1]

A. 4.64\cdot 10^{11}m

The orbital speed of the clumps of matter around the black hole is equal to the ratio between the circumference of the orbit and the period of revolution:

v=\frac{2\pi r}{T}

where we have:

v=30,000 km/s = 3\cdot 10^7 m/s is the orbital speed

r is the orbital radius

T=27 h \cdot 3600 =97,200 s is the orbital period

Solving for r, we find the distance of the clumps of matter from the centre of the black hole:

r=\frac{vT}{2\pi}=\frac{(3\cdot 10^7 m/s)(97200 s)}{2\pi}=4.64\cdot 10^{11}m

B. 6.26\cdot 10^{36}kg, 3.13\cdot 10^6 M_s

The gravitational force between the black hole and the clumps of matter provides the centripetal force that keeps the matter in circular motion:

m\frac{v^2}{r}=\frac{GMm}{r^2}

where

m is the mass of the clumps of matter

G is the gravitational constant

M is the mass of the black hole

Solving the formula for M, we find the mass of the black hole:

M=\frac{v^2 r}{G}=\frac{(3\cdot 10^7 m/s)^2(4.64\cdot 10^{11} m)}{6.67\cdot 10^{-11}}=6.26\cdot 10^{36}kg

and considering the value of the solar mass

M_s = 2\cdot 10^{30}kg

the mass of the black hole as a multiple of our sun's mass is

M=\frac{6.26\cdot 10^{36} kg}{2\cdot 10^{30} kg}=3.13\cdot 10^6 M_s

C. 9.28\cdot 10^9 m

The radius of the event horizon is equal to the Schwarzschild radius of the black hole, which is given by

R=\frac{2MG}{c^2}

where M is the mass of the black hole and c is the speed of light.

Substituting numbers into the formula, we find

R=\frac{6.26\cdot 10^{36} kg)(6.67\cdot 10^{-11})}{(3\cdot 10^8 m/s)^2}=9.28\cdot 10^9 m

8 0
3 years ago
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