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Alexus [3.1K]
3 years ago
7

Which of the following is equivalent to Planks constant

Physics
1 answer:
Alinara [238K]3 years ago
8 0

Answer:

6.62607004 x 10^(-34)m²kg/s

Explanation:

This is the constant that shows the value of energy of a photon in relation to it's frequency.

Please let me know if you want this explained further!

Thanks!

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An airplane weighing 11,000 N climbs to a
Gennadij [26K]

The power in horsepower is 40.1 hp

Explanation:

We start by calculating the work done by the airplane during the climb, which is equal to its change in gravitational potential energy:

W=(mg)\Delta h

where

mg = 11,000 N is the weight of the airplane

\Delta h = 1.6 km = 1600 m is the change in height

Substituting,

W=(11,000)(1600)=17.6\cdot 10^6 J

Now we can calculate the power delivered, which is given by

P=\frac{W}{t}

where

W=17.6\cdot 10^6 J is the work done

t=9.8 min \cdot 60 = 588 s is the time taken

Substituting,

P=\frac{17.6\cdot 10^6 J}{588}=2.99\cdot 10^4 W

Finally, we can convert the power into horsepower (hp), keeping in mind that

1 hp = 746 W

Therefore,

P=\frac{2.99\cdot 10^4}{746}=40.1 hp

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0
3 years ago
The 2.6-kg collar is released from rest at A and slides down the inclined fixed rod in the vertical plane. The coefficient of ki
neonofarm [45]

Answer:

The attached diagram explains the system,

Sum of Fy = 0

N=9.81

N - mgCos60 = 0

F= ukN= (0.53)(9.81) =

F= 5.12 N

So

F.d= 1/2(mv.v) - mgdsin60

-5.12*0.5 = 0.5*v^2 - 2*(9.81)*(0.5*sin60)

(a)  v = 2.436 m/s

For deflection

-F.x = 1/2(mv.v) - mgxsin60 + 1/2 (k*x*x)

by solving for with values of v, m, g, F, k

800x^2 - 11.87 x - 5.938 = 0

by solving the quadratic equation

x = 0.093, -0.079

(b) x = 0.093 m

correct Answer is 0.093m

Explanation:

3 0
3 years ago
A 2200-kg auto moving northward at 12.0 m/s runs into a 3800-kg truck which is also moving northward, but at 5.00 m/s. If the ve
Nookie1986 [14]

Answer:

7.56 m/s

Explanation:

8 0
3 years ago
When a gas is rapidly compressed (say, by pushing down a piston) its temperature increases. When a gas expands against a piston,
shusha [124]

Answer:

Explained in explanation

Explanation:

The first law of thermodynamics states that the change in internal energy of a system(ΔU) is equal to the sum of the net heat transfer into the system(Q) and the net work done on the system(W). In equation, this law is;

ΔU = Q + W

Now, when there's gas inside a container with a movable piston that's tightly fitting, we will assume that the piston can move up and down thereby compressing the gas or allowing the gas to expand against it.

Now these gas molecules inside the container possess kinetic energy. Thus, the internal energy(U) of the system is simply the sum of all the kinetic energies of the individual gas molecules present in the container.

Therefore, if the temperature(T) of the gas increases, then the speed and internal energy(U) of the gas molecules will also increase. In the same way, if the temperature of the gas decreases, the speed and internal energy of the gas molecules would also decrease.

Now, back to the question, when the piston is pushed down, it does work on the gas and the gas does negative work on the piston. Thus, the gas will be get compressed to a smaller space, and thereby making the gas molecules to hit the piston at a faster rate. Thus, there is a decrease in speed and as we saw earlier that when there is a decrease in speed, it means temperature has decreased.

Whereas, when the piston is moved up, the gas does positive work on the piston and the speed of the gas molecules will increase. Like I said earlier that increase in speed means increase in temperature.

4 0
3 years ago
Two spherical asteroids have the same radius R. Asteroid 1 hasmass M and asteroid 2 has mas 2M. The two asteroids are releasedfr
gtnhenbr [62]

Answer:

 v_1 =\sqrt{\dfrac{16GM}{15R}}

 v_2 =\sqrt{\dfrac{4GM}{15R}}

Explanation:

given,

mass of asteroid 1 = M

mass of asteroid 2 = 2M

radius of two asteroid = R

Distance between the asteroid = 10 R

Speed of the asteroid before collision = ?

using conservation of momentum

M u + 2M u' = M v₁ + 2 M v₂

initial speed of asteroid is equal to zero

0 = v₁ + 2 v₂

v₁ = -2 v₂

using conservation of momentum

initial potential energy is converted into potential energy and the kinetic energy of both the asteroids.

 \dfrac{GM(2M)}{10R}=\dfrac{GM(2M)}{2R}+\dfrac{1}{2}Mv_1^2 + \dfrac{1}{2}(2M)v_2^2

 \dfrac{GM(2M)}{10R}-\dfrac{GM(2M)}{2R}=\dfrac{1}{2}M(-2v_2)^2 + \dfrac{1}{2}(2M)v_2^2

 6v_2^2 = \dfrac{8GM}{5R}

 v_2 =\sqrt{\dfrac{4GM}{15R}}

now,

 v_1 =-2\sqrt{\dfrac{4GM}{15R}}

 v_1 =\sqrt{\dfrac{16GM}{15R}}

hence, the velocity of asteroid are

 v_1 =\sqrt{\dfrac{16GM}{15R}}

 v_2 =\sqrt{\dfrac{4GM}{15R}}

6 0
2 years ago
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