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3 years ago

I need help solving the equation 2Na(s)+Cl2(g)>2NaCl(s)

Physics

1 answer:

IrinaK [193]3 years ago

3 0

it's a chemical eaction equation ... you don't solve them ... you bslance them .... it is balanced

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Examine the weather map and locate this symbol.

DochEvi [55]

The right answer is A

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2 years ago

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3 years ago

A 20 kg crate initially at rest on a horizontal floor requires a 80 N horizontal force to set it in motion. Find the coefficient

e-lub [12.9K]

Answer:

The coefficient of static friction between the crate and the floor is 0.41

Explanation:

Friction Force

When an object is moving and encounters friction in the air or rough surfaces, it loses acceleration and velocity because the friction force opposes motion.

The friction force when an object is moving on a horizontal surface is calculated by:

$Fr=\mu N$ [1]

Where $\mu$ is the coefficient of static or kinetics friction and N is the normal force.

If no forces other then the weight and the normal are acting upon the y-direction, then the weight and the normal are equal in magnitude:

N = W = m.g

The crate of m=20 Kg has a weight of:

W = 20*9.8

W = 196 N

The normal force is also N=196 N

We can find the coefficient of static friction by solving [1] for $\mu$ :

$\displaystyle \mu=\frac{Fr}{N}$

The friction force is equal to the minimum force required to start moving the object on the floor, thus Fr=80 N and:

$\displaystyle \mu=\frac{80}{196}$

$\mu=0.41$

The coefficient of static friction between the crate and the floor is 0.41

7 0

2 years ago

Recall specific heat of water is 4186 j/kg/C. Find the specific heat of sample.

Paraphin [41]

Answer:

Shown by explanation;

Explanation:

The heat of the sample = mass ×specific heat capacity of the sample × temperature change(∆T)

Assumption;I assume the mass of the samples are : 109g and 192g

∆T= 30.1-21=8.9°c.

The heat of the samples are for 109g are:

0.109 × 4186 × 8.9 =4060.84J

For 0.192g are;

∆T= 67-30.1-=36.9°c

0.192 × 4186×36.9=29656.97J

5 0

2 years ago

The overhang beam is subjected to the uniform distributed load having an intensity of w = 50 kn/m. determine the maximum shear s

alex41 [277]

Given:

Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.

We need to determine the maximum shear stress developed in the beam:

τ = F/A

Assuming the area of the beam is 100 m^2 with a length of 10 m.

τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2

8 0

3 years ago