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Masja [62]
3 years ago
11

5.

Physics
1 answer:
Alex787 [66]3 years ago
6 0
C. A hammer pushes a nail
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As a block of mass 42 kilograms drops from the edge of a 40-meter-high cliff it experiences a loss of energy due to air resistan
JulsSmile [24]

Answer:

<em>The block hits the ground at 27.9 m/s</em>

Explanation:

<u>Gravitational Potential Energy (GPE)</u>

It's the energy stored in an object because of its height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or 9.8 m/s^2.

When the block is at the edge of the cliff it has potential energy that can be transformed into any other type of energy as it starts falling to the ground.

The GPE of the block of mass m=42 Kg at h=40 m is:

U = 42*9.8*40

U = 16,464 J

The block loses 81 J due to air resistance, thus the energy stored when it hits the ground is 16,464 J - 81 J = 16,383 J.

This energy is stored as kinetic energy, whose formula is:

\displaystyle K=\frac{1}{2}mv^2

Solving for v:

\displaystyle v=\sqrt{\frac{2K}{m}}

\displaystyle v=\sqrt{\frac{2*16,383 }{42}}

v=\sqrt{780.143}

v = 27.9 m/s

The block hits the ground at 27.9 m/s

5 0
3 years ago
Matter that has a fixed composition is a ________. Which is formed by_______.
Alenkasestr [34]
A) Pure substances, Chemical changes

8 0
4 years ago
In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is created on a screen that is 2.00 m a
balu736 [363]

Answer:

The value is  \lambda  = 214.3  \ nm

Explanation:

From the question we are told that

   The  slit separation is  d =  3.00 * 10^{-5} m

    The  distance of the screen is  D =   2.00\ m

    The  order of fringe is  n  =  7

    The path difference is  y =  10.0 \ cm  =  0.1 \  m

    Generally the path difference is mathematically represented as

      y =  \frac{n *  \lambda  *  D}{ d}

=>   0.1 =  \frac{7 *  \lambda  * 2.00 }{ 3.00 * 10^{-5}}

=>  \lambda  =  \frac{0.1 *3.00 * 10^{-5} }{7 * 2.00 }

=>   \lambda  =  \frac{0.1 *3.00 * 10^{-5} }{7 * 2.00 }

=>   \lambda  = 2.143 *10^{-7} \  m    

=>    \lambda  = 214.3  \ nm

6 0
3 years ago
At a pressure of one atmosphere methane boils at −161°C and freezes at −182.5°C. Consider a temperature scale where the boiling
lawyer [7]

Answer:

-421.627906977\^{\circ}M

Explanation:

Absolute zero in the Kelvin scale = -273.15 °C

Temperature difference in the boiling and freezing point

-161-(-182.5)=21.5\ ^{\circ}C

Change in 21.5°C = Change of 100°C in M

1^{\circ}C=\dfrac{100}{21.5}=4.6511627907\ ^{\circ}M

Difference in the Celsius scale

-182.5-(-273.15)=90.65\ ^{\circ}C

Change in 90.65\ ^{\circ}C corresponds to

4.6511627907\times 90.65=421.627906977\ ^{\circ}M

Therefore the absolute zero in the Methane scale is

-421.627906977\ ^{\circ}M

3 0
3 years ago
A mass on a string of unknown length oscillates as a pendulum with a period of 1.8 s. What is the period in the following situat
Maslowich

Answer:

Using

Period ( P) is given as

P~√(L/g).

a) since mass has no effect on the period of a pendulum. So, the period will remain 1.8seconds

b) using the formula above ,period varies with the square root of the length. Thus , when the length doubles, the period is multiplied by √2. So, the period is 1.8s*√2 = 2.54s

c) in this case, the period is multiplied by √(1/2).

1.8√(1/2)=1.27s.

d) amplitude of the pendulum doesn't affect the period (unless itsvery high, so, the period is still 1.8s

7 0
4 years ago
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