A light bulb has a resistance of 100 ohms. If a current of 1.2 amps is going through it, calculate

A 12.0 cm object is 9.0 cm from a convex mirror that has a focal length of -4.5 cm. What is the distance of the image from the m

Rasek [7]

Answer:

- 3 cm

Explanation:

From the mirror formula;

1/f = 1/v + 1/u ; where f is the focal length, v is the image distance, and u is the object distance.

1/-4.5 = 1/9 + 1/v

1/v = -1/4.5 - 1/9

= -1/3

Therefore;

v = -3 cm

Hence;

Image distance is - 3cm

5 0

2 years ago

4. According to Newton third law what is the action force? The swimmer pushes against the water the water pushes back on the swi

marshall27 [118]

A.The swimmer pushes the water
C. the walls force against the ball

3 0

2 years ago

Read 2 more answers

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

2 years ago

A student says that a speed of 50 m/s is faster than a speed of 140 km/h because the number is bigger. What would you say to the

ladessa [460]

The units are not consistent - 1 m/s is not the same as 1 km/h.

First thing to do would be to convert from one unit of speed to the other, say km/h to m/s. There are 1000 meters (m) for every kilometer (km) and 3600 seconds (s) for every hour (h), so

$1\,\dfrac{\mathrm{km}}{\mathrm h}\cdot\dfrac{10^3\,\mathrm m}{\mathrm{km}}\cdot\dfrac{\mathrm h}{3600\,\mathrm s}\approx0.278\,\dfrac{\mathrm m}{\mathrm s}$

So in fact 1 km/h is about 4 times slower than 1 m/s.

4 0

2 years ago

When the inferior angle of the scapula moves in the superior and lateral direction, the motion is called:

kkurt [141]

The motion of the inferior angle of the scapula in the superior and lateral direction is called upward rotation.

<h3>What are the possible motions of the scapula?</h3>

The scapula or the shoulder blade has about six different types of motion it undergoes.

The six ways of movement of the scapula are:

protraction,
retraction,
elevation,
depression,
upward rotation, and
downward rotation

When the inferior angle of the scapula moves in the superior and lateral direction, the motion is called upward rotation.

Learn more about scapula motion at: brainly.com/question/16868917

#SPJ12

8 0

1 year ago