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Assoli18 [71]
3 years ago
12

How is energy transferred whenhitting a nail?​

Physics
1 answer:
Juli2301 [7.4K]3 years ago
7 0

Answer:

kinetic energy

kinetic energy

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Given: Saturated air changes temperature by 0.5°C/100 m. The air is completely saturated at the dew point. The dew point has bee
erma4kov [3.2K]

Answer:

Explanation:

Given

saturated air temperature by 0.5^{\circ}C/100 m

Dew point temperature is given by t=2^{\circ}C

Dew point is defined as the temperature after which air no longer to uphold the water vapor fuse with it and some water vapor may condense to a liquid.

air continues to rise for 1400 m

i.e. change in temperature would be \Delta t =\frac{0.5}{100}\times 1400=7^{\circ}C

Final temperature t_f

t_f+\Delta t=t

t_f=2-7=-5^{\circ}C

3 0
4 years ago
What is the velocity of an object with a mass of 4 kg and a momentum of 24 kg.m/s
Eva8 [605]

Answer:

6m/s

Explanation:

momentum = mass × change in velocity

∆p =m(v)

24 = 4(v)

V =>24/4 = 6m/s

5 0
3 years ago
How would you get your friend to change his/her mind from using PED's?​
kvasek [131]

Answer:

Watch funny videos and start eating a lot of candy

6 0
3 years ago
A window washer drops a brush from a scaffold on a tall office building. What is the speed of the falling brush after 3.28 s? (N
valina [46]

Answer:

The speed of washer will be 32.144 m/s

Explanation:

Given that

time t= 3.28 s

Acceleration due to gravity

 g=9.8\ \frac{m}{s^2}

Here acceleration  is constant so we cam apply the motion equation

we know that

v= u + at

At initial condition  u =0 m/s

v= u + at

v= 0 + 9.8 x 3.28

v= 32.144 m/s

So the speed of washer will be 32.144 m/s

6 0
3 years ago
Read 2 more answers
The trajectory of a rock ejected from the Kilauea volcano, with a velocity of magnitude 6.4 m/s and at angle 2.2 degrees above t
Vika [28.1K]

Answer:

v = 8.03 m / s

Explanation:

This is a missile throwing exercise.

          y = y₀ + v_{oy}  t - ½ g t²

     

indicates that y = -1.2 m and the initial velocity is

         v_{oy} = v₀ sin θ

         v_{oy} = 6.4 sin 2.2

         v_{oy} = 0.2457 m / s

we substitute in the equation

         -1.2 = 0.2457 t - ½ 9.8 t²

          4.9 t² - 0.2457 t - 1.2 = 0

          t² - 0.05014 t -0.2449 = 0

we solve the quadratic equation

          t = [0.05014 ±√ (0.05014² + 4 0.2449)] / 2

          t = [0.05014 ± 0.9910] / 2

          t₁ = 0.5206 s

          t₂ = -0.47 s

time must be a positive magnitude therefore the correct answer is

          t = 0.5206 s

with this time we can calculate the vertical speed when the rock hits the ground

         v_{y} = v_{oy} - gt

         v_{y} = 0.2457 - 9.8 0.5206

         v_{y} = -4.856 m / s

the negative sign indicates that the speed is down

horizontal velocity is constant, due to no acceleration

         vₓ = v₀ₓ = v₀ cos 2,2

         v₀ₓ = 6.4 cos 2.2

         v₀ₓ = 6.395 m / s

therefore let's use Pythagoras' theorem to find the velocity

         v = √ (vₓ² + v_{y}^{2})

         v = √ (6,395² + 4,856²)

         v = 8.03 m / s

the direction can be found with trigonometry

        tan θ = v_{y} / vₓ

        θ = tan⁻¹ (-4,856 / 6,395)

        θ = - 37

the negative sign indicates that it is half clockwise from the x axis

   

4 0
4 years ago
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