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umka2103 [35]
3 years ago
6

How does the doppler effect affect the velocity from a sourse that is approaching you?

Physics
1 answer:
Firlakuza [10]3 years ago
3 0
The apparent change in the frequency of a sound caused by the motion of either The Listener or the source of the sound
You might be interested in
A baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg. What’s the baseballs velocity?
Mashcka [7]

Answer:

<em>v=40 m/s south</em>

Explanation:

<u>Momentum </u>

It's a physical magnitude that measures the product of the mass by the velocity of a particle. Its units in the International System is kg.m/s and the formula is

p=m.v

Where m is the mass and v the velocity of the particle. If we wanted to solve for v, we have

\displaystyle v=\frac{p}{m}

The baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg, thus

\displaystyle v=\frac{6}{0.15}=40\ m/s

The velocity is directed to the south

3 0
3 years ago
Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11
tiny-mole [99]

Answer:

The line charge density is 1.59\times10^{-4}\ C/m

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

V=EA

V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2

\lambda=\dfrac{V\times2\epsilon_{0}}{r}

Where, r = radius

V = potential difference

Put the value into the formula

\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}

\lambda=1.59\times10^{-4}\ C/m

Hence, The line charge density is 1.59\times10^{-4}\ C/m

4 0
3 years ago
Two separate disks are connected by a belt traveling at 5m/s. Disk 1 has a mass of 10kg and radius of 35cm. Disk 2 has a mass of
pantera1 [17]

Explanation:

Given that,

Linear speed of both disks is 5 m/s

Mass of disk 1 is 10 kg

Radius of disk 1 is 35 cm or 0.35 m

Mass of disk 2 is 3 kg

Radius of disk 2 is 7 cm or 0.07 m

(a) The angular velocity of disk 1 is :

v=r_1\omega_1\\\\\omega_1=\dfrac{v}{r_1}\\\\\omega_1=\dfrac{5}{0.35}\\\\\omega_1=14.28\ rad/s

(b) The angular velocity of disk 2 is :

v=r_2\omega_2\\\\\omega_2=\dfrac{v}{r_2}\\\\\omega_2=\dfrac{5}{0.07}\\\\\omega_2=71.42\ rad/s

(c) The moment of inertia for the two disk system is given by :

I=I_1+I_2\\\\I=\dfrac{1}{2}m_1r_1^2+\dfrac{1}{2}m_2r_2^2\\\\I=\dfrac{1}{2}(m_1r_1^2+m_2r_2^2)\\\\I=\dfrac{1}{2}\times (10\times (0.35)^2+3\times (0.07)^2)\\\\I=0.619\ kg-m^2

Hence, this is the required solution.

6 0
3 years ago
Two metal plates 15mm apart have a potential difference of 750v between them. The force on a small charged sphere placed between
Romashka [77]

Answer:

50,000 V/m

Explanation:

The electric field between two charged metal plates is uniform.

The relationship between potential difference and electric field strength for a uniform field is given by the equation

\Delta V=Ed

where

\Delta V is the potential difference

E is the magnitude of the electric field

d is the  distance between the plates

In this problem, we have:

\Delta V=750 V is the potential difference between the plates

d = 15 mm = 0.015 m is the distance between the plates

Therefore, rearranging the equation we find the strength of the electric field:

E=\frac{\Delta V}{d}=\frac{750}{0.015}=50,000 V/m

6 0
3 years ago
If it requires 5.0 j of work to stretch a particular spring by 1.8 cm from its equilibrium length, how much more work will be re
ANEK [815]
15.277.. j. I did the problem using a proportion. an additional 3.7m to the current 1.8 cm=5.5cm.

Therefore, 5.0 j/1.8cm=x/5.5cm
5 0
3 years ago
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