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2 years ago

Which type of force pulls objects toward one another

Physics

2 answers:

Klio2033 [76]2 years ago

4 0

Gravity ALWAYS does that, and electrostatic force does it when two objects have opposite charges.

abruzzese [7]2 years ago

3 0

Gravitational force, magnetic force and electric force

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SVETLANKA909090 [29]

Electronic friction ^.^

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3 years ago

Read 2 more answers

The world’s largest wind turbine has blades that are 80 m long and makes 1 revolution every 5.7 seconds. What is the velocity fo

klemol [59]

Answer:

The free end of the blade has a tangential velocity of about 88.19 m/s

Explanation:

The angular velocity of the blades is $2 \pi /5.7\,\,rad/sec$

since the blades are 80 m long, then the tangential velocity of the free end of the blade is:

$v_{tan} \approx 88.19\,\,m/s$

3 0

3 years ago

1.<br> Kinetic energy is the energy of _____.

lorasvet [3.4K]

Explanation:

kinetic energy is energy that it possesses due to its motion.

7 0

3 years ago

A velocity selector in a mass spectrometer uses a 0.150 T magnetic field. (a) What electric field strength (in volts per meter)

Alekssandra [29.7K]

Answer:

The electric field strength is $6.6\times10^{5}\ V/m$

Explanation:

Given that,

Magnetic field = 0.150 T

Speed $v= 4.40\times10^{6}\ m/s$

We need to calculate the electric field strength

Using formula of velocity

$v=\dfrac{E}{B}$

$E=v\times B$

Where, v = speed

B = magnetic field

Put the value into the formula

$E=4.40\times10^{6}\times0.150$

$E=660000\ V/m$

$E=6.6\times10^{5}\ V/m$

Hence, The electric field strength is $6.6\times10^{5}\ V/m$

4 0

2 years ago

Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’2) mm apart and then falls on a screen 7

IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

$sin \theta=\frac{n \lambda}{d}$

where

n is the order of the maximum

$\lambda$ is the wavelength

$a$ is the distance between the slits

In this problem,

n = 5

$\lambda=525 nm =5.25\cdot 10^{-7} m$

$a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m$

So we find

$\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}$

And given the distance of the screen from the slits,

$D=75.0 cm = 0.75 m$

The distance of the 5th bright fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm$

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

$sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}$

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

$\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}$

And the distance of the 8th dark fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm$

5 0

3 years ago