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3 years ago

What materials electricians should avoid wearing.

Physics

2 answers:

Nady [450]3 years ago

7 0

It believe its fire repellant clothing.

Rainbow [258]3 years ago

4 0

Arc flash PPE

Coveralls

Fire replant clothing

vest

safety gloves and footwear

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YOU READING THIS PLEASE GIVE ME A EASY 7TH GRADE TOPIC NO VOLCANIOS

Kipish [7]

Photothysenis. or fossils they are very easy

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3 years ago

Read 2 more answers

) Un círculo de 120 cm de radio gira a 600 rpm. Calcula: a) su velocidad angular

DIA [1.3K]

Responder:

20πrads ^ -1; 24πrads ^ -1; 0,1 seg; 10 Hz

Explicación:

Dado lo siguiente:

Radio (r) del círculo = 120 cm

600 revoluciones por minuto en radianes por segundo

(600 / min) * (2π rad / 1 rev) * (1min / 60seg)

(1200πrad / 60sec) = 20π rad ^ -1

Velocidad angular (w) = 20πrads ^ -1

Velocidad lineal = radio (r) * velocidad angular (w)

Velocidad lineal = (120/100) * 20πrad

Velocidad lineal = 1.2 * 20πrads ^ -1 = 24πrads ^ -1

C.) Período (T):

T = 2π / w = 2π / 20π = 0.1 seg

D.) Frecuencia (f):

f = 1 / T = 1 / 0.1

1 / 0,1 = 10 Hz

5 0

3 years ago

Brian has a medical condition called cone dystrophy, which is a disorder of the cone cells in the eye. What symptoms would you e

icang [17]

Answer:

Poor Color Vision
Sensitivity to Bright Light
Vision Loss

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3 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

2 years ago

A football player with a mass of 88 kg and a speed of 2.0 m/s collides head-on with a player from the opposing team whose mass i

Ket [755]

Answer:

Speed of another player, v₂ = 1.47 m/s

Explanation:

It is given that,

Mass of football player, m₁ = 88 kg

Speed of player, v₁ = 2 m/s

Mass of player of opposing team, m₂ = 120 kg

The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :

$m_1v_1+m_2v_2=(m_1+m_2)V$

V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.

$v_2=-\dfrac{m_1v_1}{m_2}$

$v_2=-\dfrac{88\ kg\times 2\ m/s}{120\ kg}$

$v_2=-1.47\ m/s$

So, the speed of another player is 1.47 m/s. Hence, this is the required solution.

7 0

3 years ago