How do you calculate Friction force without the coefficient of friction

Which of these types of rock form near earths surface? Select all that apply.

OlgaM077 [116]

A. Igneous

Extrusive igneous rocks cool and solidify quicker than intrusive igneous rocks. They are formed by the cooling of molten magma on the earth's surface.

8 0

3 years ago

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Two identical masses are attached to two identical springs that hang vertically. The two masses are pulled down and released, bu

Alex17521 [72]

Answer:

Lets take mass of A = m

Mass of B = m

Lets take spring constant = K

Displacement of mass A = x

Displacement of mass B= x'

Given that

x ' > x

a)

We know that time period of spring mass system given as

$T= 2\pi\sqrt{\dfrac{m}{K}}$

Time does not depends on the displacement.

So time will be same for both.

b)

We know that acceleration given as

a= ω² .x

ω=natural frequency ( ω² = m K)

Here x ' > x

So Acceleration of mass B is greater than mass A.

c)

Velocity at equilibrium position

V= ω .x

Here x ' > x

So velocity of mass B is greater than mass A.

d)

As we know that at equilibrium point acceleration of the mass is zero.So both the mass have same acceleration and the value of this acceleration will be zero m/s².

3 0

3 years ago

At a pressure of one atmosphere oxygen boils at −182.9°C and freezes at −218.3°C. Consider a temperature scale where the boiling

Gelneren [198K]

Answer: -254.51°O

Explanation:

Ok, in our scale, we have:

-182.9°C corresponds to 100° O

-218.3°C corresponds to 0°

Then we can find the slope of this relation as:

S = (100° - 0°)/(-182.9°C - (-218.3°C)) = 2.82°O/°C

So we can have the linear relationship between the scales is:

Y = (2.82°O/°C)*X + B

in this relation, X is the temperature in Celcius and Y is the temperature in the new scale.

And we know that when X = -182.9°C, we must have Y = 0°O

then:

0 = (2.82°O/°C)*(-182.9°C) + B

B = ( 2.82°O/°C*189.9°C) = 515.778°O.

now, we want to find the 0 K in this scale, and we know that:

0 K = -273.15°C

So we can use X = -273.15°C in our previous equation and get:

Y = (2.82°O/°C)*(-273.15°C) + 515.778°O = -254.51°O

5 0

3 years ago

It is weigh-in time for the local under 85 kg rugby team. The bathroom scale used to assess eligibility can be described by Hook

Grace [21]

15 m/n is the answer

4 0

3 years ago

(1 point) find the half-life (in hours) of a radioactive substance that is reduced by 1010 percent in 9595 hours.

Bad White [126]

The decay function is of the form
$N(t) = N_{0} \, e^{-kt}$
where
N₀ = initial amount
k = decay constant
t = hours

The material decays by 10% in 95 hours. Therefore
$0.9N_{0} = N_{0} \, e^{-95k} \\\\ -95k = ln(0.9) \\\\ k= \frac{ln(0.9)}{-95}=0.001109$

The time for the half life is given by
$0.5N_{0} = N_{0} \, e^{-0.001109t} \\\\ -0.0001109t = ln(0.5) \\\\ t = \frac{ln(0.5)}{-0.001109} = 625 \, h$

Answer: The half life is 625 hours

8 0

3 years ago