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zmey [24]
2 years ago
13

"The greater the height of an object, the *BLANK* its gravitational potential energy."

Physics
2 answers:
Brut [27]2 years ago
6 0
The answer is greater
-Dominant- [34]2 years ago
3 0

Answer:

the greater the height of an object the *greater* its gravitational potential energy

greater

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Accelerating charges radiate electromagnetic waves. calculate the wavelength of radiation produced by a proton in a cyclotron wi
levacccp [35]

B = magnetic field in the cyclotron = 0.400 T

q = magnitude of charge on a proton = 1.6 x 10⁻¹⁹ C

m = mass of the proton = 1.67 x 10⁻²⁷ kg

f = frequency of revolution of proton in the cyclotron = ?

v = speed of electromagnetic waves = 3 x 10⁸ m/s

λ = wavelength of electromagnetic wave = ?

Frequency of revolution of proton in the cyclotron is given as

f = qB/(2πm)

inserting the values

f = (1.6 x 10⁻¹⁹)(0.400)/(2 (3.14) (1.67 x 10⁻²⁷))

f = 6.1 x 10⁶ Hz

wavelength of electromagnetic wave is given as

λ = v/f

λ = (3 x 10⁸)/(6.1 x 10⁶)

λ = 49.2 m

7 0
2 years ago
Read 2 more answers
Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a
Nat2105 [25]

Answer:

Explanation:

Given

diameter d=20 \mu m

density \rho =1300 kg/m^3

frequency \nu =150 Hz

Length of silk strand L=14 cm

Velocity in the string is as follows

\nu =\sqrt{\frac{T}{\mu }}

The expression for Fundamental Frequency

f=\frac{\nu }{2l}

f=\frac{1}{2l}\times \sqrt{\frac{T}{\mu }}

f=\frac{1}{2l}\times \sqrt{\frac{T}{\frac{m}{l}}}

f=\frac{1}{2l}\times \sqrt{\frac{Tl}{\rho V}}

Squaring

f^2=\frac{1}{4l^2}\times \frac{Tl}{\rho V}

T=4\rho \cdot A\cdot l^2\cdot f^2

T=4\times 1300\times \frac{\pi }{4}(20\times 10^{-6})^2\times (0.14)^2\times 150^2

T=7.2\times 10^{-4} N

8 0
3 years ago
A rock is attached to the left end of a uniform meter stick that has the same mass as the rock. How far from the left end of the
kotykmax [81]

Answer:

M₂ = M  then L₂ = L

M₂> M  then L₂ = \frac{M}{M_{2}} L

Explanation:

This is a static equilibrium exercise, to solve it we must fix a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive

          ∑ τ = 0

           

The mass of the rock is M and placed at a distance, L the mass of the rod M₁, is considered to be placed in its center of mass, which by uniform e is in its geometric center (x = 0) and the triangular mass M₂, with a distance L₂

The triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.

         M L + M₁ 0 - m₂ L₂ = 0

         M L - m₂ L₂ = 0

         L₂ = \frac{M}{M_{2}} L

From this answer we have several possibilities

* if the two masses are equal then L₂ = L

* If the masses are different, with M₂> M then L₂ = \frac{M}{M_{2}} L

6 0
2 years ago
An electron in a mercury atom changes from energy level b to a higher energy level when the
emmainna [20.7K]
The energy is 3.06 electronvolts,  E = 3.06eV

1eV = 1.6 * 10^-19 J

3.06 eV = 3.06* 1.6 * 10^-19 J = 4.896 * 10^-19 J


4 0
2 years ago
a football player kicks a ball with a mass of 0.42kg. The average acceleration of the football was 14.8 m/s2. How much force did
Zarrin [17]

Answer:

6.216 N

Explanation:

As for Newton's second law of motion

F=ma

where F= the acting force

           m=subjected mass

           a= the acceleration

applying F=ma to the football

F=m*a

 =0.42*14.8

 =6.216 N

6.216 N of a force is supplied to the ball

5 0
2 years ago
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