Label each formula in the chemical quation below as either a reactant of a product.

A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When

natita [175]

Answer:

a) $\mu_{k} = 0.704$ , b) $R = 0.312\,m$

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s$

$\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}$

$\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}$

$\mu_{k} = 0.704$

b) The speed of the block is determined by using the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = x\cdot \sqrt{\frac{k}{m} }$

$v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }$

$v \approx 9.829\,\frac{m}{s}$

The radius of the circular loop is:

$\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}$

$\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}$

$R = 0.312\,m$

5 0

3 years ago

The half-life of a certain isotope is 15 minutes. How much of a 400 g sample will remain after 90 minutes?12.5 g6.25 g26.7 g66.7

Virty [35]

There is a total of 6 half lives that need to take place.

ONE HALF LIFE = 200

TWO HALF LIFES = 100

THREE HALF LIFES = 50

FOUR HALF LIFES = 25

FIVE HALF LIFES = 12.5

SIX HALF LIFES = 6.25

The answer is 6.25g

4 0

3 years ago

A velocity selector has a magnetic field of magnitude 0.22 T perpendicular to an electric field of magnitude 0.51 MV/m.

ohaa [14]

Answer:

speed of a particle = 2.31 × $10^{6}$ m/s

energy of proton required = 27.77 KeV

energy of electron required = 15.171 eV

Explanation:

given data

magnetic field of magnitude = 0.22 T

electric field of magnitude = 0.51 MV/m

to find out

speed of a particle and energy must protons have to pass through undeflected and energy must electrons have to pass through undeflected

solution

we know that force due to magnetic and electric field is express as

force due to magnetic field B = qvB ..............1

and force due to electric field E = qE .....................2

so without deflection force due to magnetic field = force due to electric field

so here qvB = qE

and V = $\frac{E}{B}$ ...................3

put here value

V = $\frac{0.51*10^6}{0.22}$

speed of a particle = 2.31 × $10^{6}$ m/s

and

now energy of proton required will be here as

energy of proton required = mass of proton × $\frac{V^2}{2}$

put here value

energy of proton required = 1.67 × $10^{-27}$ × $\frac{(2.31*10^6)^2}{2}$

energy of proton required = 4.45 × $10^{-15}$ J

energy of proton required = 4.45 × $10^{-15}$ J ÷ (1.602 × $10^{-19}$

energy of proton required = 27777.777 eV

energy of proton required = 27.77 KeV

and

now we get here energy of electron required that is

energy of electron required = mass of electron × $\frac{V^2}{2}$

put here value

energy of electron required = 9.11 × $10^{-31}$ × $\frac{(2.31*10^6)^2}{2}$

energy of electron required = 24.305× $10^{-19}$ J

energy of electron required = 24.305 × $10^{-19}$ J ÷ (1.602 × $10^{-19}$

energy of electron required = 15.171 eV

5 0

4 years ago

Read 2 more answers

PLS HELP ME ASAPPPP. 50 POINTSSS

dybincka [34]

Answer:

1.8 kj

Explanation:

Explanation:

A substance's specific heat tells you how much heat is required to increase the mass of

1 g

of that substance by

1

∘

C

.

The equation that establishes a rel;ationship between heat absorbed and change in temperature looks like this

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed

m

- the mass of the sample

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

You have all the information needed to find the amount of heat required to increase the temperature of your sample of mercury by that many degrees Celsius, so just rearange the above equation and solve for

q

=

250.0

g

⋅

0.14

J

g

∘

C

⋅

(

62

−

10

)

∘

C

=

1820 J

I'll leave the answer rounded to two sig figs and expressed in kilojoules

q

=

1.8 kJ

Answer Explanation:

A substance's specific heat tells you how much heat is required to increase the mass of

1 g

of that substance by

1

∘

C

.

The equation that establishes a rel;ationship between heat absorbed and change in temperature looks like this

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed

m

- the mass of the sample

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

You have all the information needed to find the amount of heat required to increase the temperature of your sample of mercury by that many degrees Celsius, so just rearange the above equation and solve for

q

=

250.0

g

⋅

0.14

J

g

∘

C

⋅

(

62

−

10

)

∘

C

=

1820 J

I'll leave the answer rounded to two sig figs and expressed in kilojoules

q

=

1.8 kJ

Answer link

4 0

3 years ago

Two states of matter are described below.

miv72 [106K]

State X = neon; State Y = crayon

5 0

3 years ago

Read 2 more answers