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3 years ago

What are the ranks in order on your profile in Brainly

Physics

2 answers:

velikii [3]3 years ago

8 0

Answer:

All I know is that first you're Beginner second you're Ambitious and third you're Virtuoso

Explanation:

Zarrin [17]3 years ago

8 0

Beginner

Helping Hand

Ambitious

Virtuoso

Expert

Ace

Genius

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A 5.00-g bullet is fired into a 900-g block of wood suspended as a ballistic pendulum. The combined mass swings up to a height o

Thepotemich [5.8K]

Answer:

The value is $KE_b =0.710 \ J$

Explanation:

From the question we are told that

The mass of the bullet is $m_b = 5.00 \ g = 0.005 \ kg$

The mass of the wood is $m_w = 900 \ g = 0.90\ kg$

The height attained by the combined mass is $h = 8.0 \ cm = 0.08 \ m$

Generally according to the law of energy conservation

$KE _b = PE_c$

Here $KE_b$ is the kinetic energy of the bullet before collision.

and $PE_c$ is the potential energy of the combined mass of bullet and wood at the height h which is mathematically represented as

$PE_m = [m_b + m_w] * g * h$

$KE_b =PE_c = [0.005 + 0.90] * 9.8 *0.08$

=> $KE_b =0.710 \ J$

3 0

3 years ago

An air-track glider attached to a spring oscillates between the 14.0 cm mark and the 71.0 cm mark on the track. The glider compl

horsena [70]

Answer:

A = 2,8333 s

Explanation:

El periodo es definido como el tiene que toma de dar una oscilación.

En este caso realiza varias osicilacion por lo cual debemos encontrar el promedio del perdono.

T = t/n

calculemos

A = 34,0/ 12,0

A = 2,8333 s

6 0

2 years ago

A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250

Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30$

coefficient of static friction $\mu =0.35$

coefficient of kinetic friction $\mu _k=0.25$

distance traveled $d=77.3 cm$

spring constant $k=35 N/m$

work done by gravity+work done by friction=Energy stored in Spring

$mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}$

$mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}$

$0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}$

$x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}$

$x=0.234 m$

$x=23.4 cm$

6 0

3 years ago

What is an elastic collision?

Anit [1.1K]

Answer:

A collision in which both total momentum and total kinetic energy are conserved

Explanation:

In classical physics, we have two types of collisions:

- Elastic collision: elastic collision is a collision in which both the total momentum of the objects involved and the total kinetic energy of the objects involved are conserved

- Inelastic collision: in an inelastic collision, the total momentum of the objects involved is conserved, while the total kinetic energy is not. In this type of collisions, part of the total kinetic energy is converted into heat or other forms of energy due to the presence of frictional forces. When the objects stick together after the collision, the collisions is called 'perfectly inelastic collision'

6 0

3 years ago

The mean time between collisions for electrons in room temperature copper is 2.5 x 10-14 s. What is the electron current in a 2

Darya [45]

Answer:

1.87 A

Explanation:

τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s

d = diameter of copper wire = 2 mm = 2 x 10⁻³ m

Area of cross-section of copper wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

E = magnitude of electric field = 0.01 V/m

e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of electron = 9.1 x 10⁻³¹ kg

n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³

$i$ = magnitude of current

magnitude of current is given as

$i = \frac{Ane^{2}\tau E}{m}$

$i = \frac{(3.14\times 10^{-6})(8.47\times 10^{28})(1.6\times 10^{-19})^{2}(2.5\times 10^{-14}) (0.01)}{(9.1\times 10^{-31})}$

$i$ = 1.87 A

4 0

2 years ago