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Burka [1]
3 years ago
11

What's a Blackhole ?​

Physics
1 answer:
Vlad1618 [11]3 years ago
7 0

Answer:

A black hole is a region of spacetime where gravity is so strong that nothing—no particles or even electromagnetic radiation such as light—can escape from it. The theory of general relativity predicts that a sufficiently compact mass can deform spacetime to form a black hole.

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When an object moves, where does the energy come from?
Eddi Din [679]
Ans: Kinetic and potential energies are found in all objects. If an object is moving, it is said to have kinetic energy (KE). Potential energy (PE) is energy that is "stored" because of the position and/or arrangement of the object. The classic example of potential energy is to pick up a brick.
5 0
3 years ago
How would you set it up?
Firdavs [7]

Answer:

THE LAST OPTION

Explanation:

5 0
4 years ago
The speed of sound in air is 345 m/s. A tuning fork vibrates above the open end of a sound resonance tube. If sound waves have w
anzhelika [568]

Answer:

594.8 Hz

Explanation:

Parameters given:

Speed of sound, v = 345 m/s

Wavelength = 58 cm = 0.58 m

Speed of a wave is given as:

Speed = wavelength * frequency

Therefore:

Frequency = Speed/Wavelength

Frequency = 345/0.58

Frequency = 594.8 Hz

8 0
3 years ago
A 15 kg block is on a ramp which is inclined at 20o above the horizontal. It is connected by a string to a 19 kg mass which hang
12345 [234]

Answer:

The magnitude of the acceleration of the 19 kg block is 1.414 m/s²

Explanation:

From Newton's second law of motion;

F_{Net} = ma

where;

m is the mass of the objects involved, kg

a is the acceleration of the object, m/s²

different forces on the block and string

⇒force due to 15 kg block

=mgcosθ = 15×9.8×cos20 = 15×9.8×0.9396

= 138.12 N

⇒Tensional Force on 19 kg mass:

T = mg = 19×9.8 = 186.2 N

F_{Net} = T-mg = a(m_1+m_2)

186.2 - 138.12 = a(15+19)

48.08 =  a(34)

a = 48.08/34

a = 1.414 m/s²

Therefore, the magnitude of the acceleration of the 19 kg block is 1.414 m/s²

7 0
4 years ago
A bar-magnet with magnetic moment 2.5 Am^2 is placed in a homogeneous magnetic field (of 0.1 T that is directed along the z-axis
Angelina_Jolie [31]

Answer:

1)

Force on bar magnet  = 0

Torque on bar magnet = 0

2)

Force on bar magnet  = 0

Torque on bar magnet = 0.177 Nm

3)

Force on bar magnet  = 0

Torque on bar magnet = 0.25 Nm

Explanation:

Part 1)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

\tau = MBsin\theta

when bar magnet is inclined along z axis along magnetic field

then we will have

\tau = MBsin0 = 0

Part 2)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

\tau = MBsin\theta

when bar magnet is pointing 45 degree with z axis then we will have

\tau = MBsin45

\tau = (2.5)(0.1)sin45

\tau = 0.177 Nm

Part 3)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

\tau = MBsin\theta

when bar magnet is pointing 90 degree with z axis then we will have

\tau = MBsin90

\tau = (2.5)(0.1)sin90

\tau = 0.25 Nm

8 0
4 years ago
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