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3 years ago

What is the formula for lithium and chlorine?

2 answers:

7 0

LiCl, because lithium (Li) has one positively charged ion (1+), and
chloride (Cl) has one negatively charged ion (1-), so they
cancel each other out.

aniked [119]3 years ago

3 0

The answer is licl you must. check it on chemistry study guide

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A small boy is playing with a ball on a stationary train. If he places the ball on the floor of the train, when the train starts

expeople1 [14]

A small boy is playing with a ball on a stationary train. If he places the ball on the floor of the train, when the train starts moving the ball moves toward the back of the train. This happened due to inertia

An object at rest remains at rest, or if in motion, remains in motion unless a net external force acts on it .

When a train starts moving forward, the ball placed on the floor tends to fall backward is an example of inertia of rest. Due to the reason that the lower part of the ball is in contact with the surface and rest of the part is not . As the train starts moving, its lower part gets the motion as the floor starts moving but the upper part will remain as it is as it is not in contact with the floor , hence do not attain any motion due to the inertia of rest simultaneously i.e. it tends to remain at the same place.

To learn more about inertia here :

brainly.com/question/11049261

#SPJ1

8 0

1 year ago

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th

NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

$v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2$

$v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)$

$v_{pf}=-31.4\ m/s$

$v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}$

$v_{gf} = 0.0988\ m/s$

4 0

3 years ago

A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.

Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

$\Delta K = \Delta E_{p} = q\Delta V$

<u>Where</u>:

K: is the kinetic energy

$E_{p}$ : is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV

$\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J$

Now, the kinetic energy of the particle as it moves through point B is:

$\Delta K = K_{f} - K_{i}$

$K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J$

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!

8 0

3 years ago

Consider two wheels with fixed hubs. The hub cannot move, but the wheel can rotate about it. The hubs are fixed to a stationary

kykrilka [37]

Answer:

Magnitude of force on wheel B is 4 N

Explanation:

Given that

$I=mr^2$

For wheel A

m= 1 kg

d= 1 m,r= 0.5 m

F=1 N

We know that

T= F x r

T=1 x 0.5 N.m

T= 0.5 N.m

T= I α

Where I is the moment of inertia and α is the angular acceleration

$I_A=1 \times 0.5^2\ kg.m^2$

$I_A=0.25\ kg.m^2$

T= I α

0.5= 0.25 α

$\alpha = 2\ rad/s^2$

For Wheel B

m= 1 kg

d= 2 m,r=1 m

$I_B=1 \times 1^2\ kg.m^2$

$I_B=1 \ kg.m^2$

Given that angular acceleration is same for both the wheel

$\alpha = 2\ rad/s^2$

T= I α

T= 1 x 2

T= 2 N.m

Lets force on wheel is F then

T = F x r

2 = F x 1

So F= 2 N

Magnitude of force on wheel B is 2 N

3 0

3 years ago

An object with a mass M = 250 g is at rest on a plane that makes an angle θ = 30 o above the horizontal. The coefficient of kine

liubo4ka [24]

Answer:

v = 79.2 m/s

Solution:

As per the question:

Mass of the object, m = 250 g = 0.250 kg

Angle, $\theta = 30^{\circ}$

Coefficient of kinetic friction, $\mu_{k} = 0.100$

Mass attached to the string, m = 0.200 kg

Distance, d = 30 cm = 0.03 m

Now,

The tension in the string is given by:

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = T$ (1)

Also

T = m(g + a)

Thus eqn (1) can be written as:

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = m(g - a)$

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = mg + ma$

$mg - Mgsin\theta - \mu_{k}Mgcos\theta = (M - m)a$

$a = \frac{0.2\times 9.8 - 0.250\times 9.8\times sin30^{\circ} - 0.1\times 0.250\times 9.8\times cos30^{\circ}}{0.250 - 0.200}$

$a = 10.45\ m/s^{2}$

Now, the speed is given by the third eqn of motion with initial velocity being zero:

$v^{2} = u^{2} + 2ad$

where

u = initial velocity = 0

Thus

$v = \sqrt{2ad}$

$v = \sqrt{2\times 10.45\times 0.03} = 0.792\ m/s$

3 0

3 years ago