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3 years ago

How do I do balance equations?

Physics

1 answer:

san4es73 [151]3 years ago

4 0

Answer:

1. Count the atoms of each element in the reactants and the products.

2. Use coefficients; place them in front of the compounds as needed.

Explanation:

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A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is

IRINA_888 [86]

Answer:

s=4.44 m

Explanation:

Given that

Coefficient of the kinetic friction ,μ = 0.13

Initial velocity ,u= 3.4 m/s

Final velocity of the box ,v= 0 m/s

The acceleration due to friction force

a= - μ g

Now by putting the values in the above equation

a= - 0.13 x 10 ( take g= 10 m/s²)

a= - 1.3 m/s²

We know that

v²= u ² + 2 a s

s=distance

a=acceleration

v=final speed

u=initial speed

Now by putting the values in the above equation

0²= 3.4² - 2 x 1.3 x s

$s=\dfrac{3.4^2}{2\times 1.3}\ m$

s=4.44 m

The distance cover by box will be 4.44 m.

6 0

3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago

Water is returned from earth’s surface to the atmosphere by

Tems11 [23]

Answer:

Evaporation

Explanation:

Evaporation is a form of mass tranfer phenomena where by water are moved from the earth surface into the atmosphere as vapours,it is path of the water cycle a decription of the path moved by land water until it turns into rain, humidity,air and temperature are factors that influence evaporation though evaporation can happen at all temperature

4 0

3 years ago

Convert the value from meters/second to kilometers/hour. One kilometer is equal to 1,000 meters, and 1 hour is equal to

Paha777 [63]

Answer:

we must multiply the speed value by 3.6

Explanation:

To reduce a magnitude from one unit to another, it must be multiplied by the equivalence in the order necessary so that the units do not change.

To reduce speed from m / s to km / h

we write the speed and the two conversion factors

v [m / s] [1 km / 1000m] [3600 s / 1h

v 3600/1000

v 3.6

therefore we must multiply the speed value by 3.6

3 0

4 years ago

Read 2 more answers

Determine the speed of an object moving at 35 seconds over a distance at 472.5 meters

Jlenok [28]

Answer:

volume is base*length*height

v = b*l*h

472.5 = 7*3*length

(rearranging the equation)

length = 472.5/21

= 22.5

6 0

3 years ago

Read 2 more answers