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Airida [17]
3 years ago
9

In order to start an object moving, you must first overcome the force of _________ friction.

Physics
2 answers:
navik [9.2K]3 years ago
7 0
It would be static friction which is what you have to overcome when an object is not in motion. When you move an object friction works against it like gravity and air resistance. I hope this helps!
Luda [366]3 years ago
6 0

Answer:

Static friction.

Explanation:

Static friction is defined as the force of friction  that keeps an object at rest position.

This friction must be overcome to start an object moving . If an object is in motion, it will experiences a force of friction called  kinetic friction.

If a force is applied which is of small magnitude to an object, the static friction will posses or applies an equal magnitude which is in opposite direction of applied force.

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Approximately 100-120 wolves
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3 years ago
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An object has a mass of 7 kg and is accelerating at 4 m/s2. How far would the object move if it took 168 J of work to move it?​
KIM [24]

Answer:

6m

Explanation:

use work =force×distance

6 0
3 years ago
Find the angle for the third-order maximum for 591 nm wavelength light falling on a diffraction grating having 1460 lines per ce
Marat540 [252]

Answer:

15.32°

Explanation:

We have given the wavelength \lambda =591nm=591\times 10^{-9}m

Diffraction grating is 1460 lines per cm

So  d=\frac{10^{-2}}{1460}=6.71\times 10^{-6}m (as 1 m=100 cm )

For maximum diffraction

dsin\Theta =m\lambda here m is order of diffraction

So 6.71\times 10^{-6}sin\Theta =3\times 591\times 10^{-9}

sin\Theta =0.264

\Theta =15.32^{\circ}

6 0
3 years ago
The mean time between collisions for electrons in room temperature copper is 2.5 x 10-14 s. What is the electron current in a 2
Darya [45]

Answer:

1.87 A

Explanation:

τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s

d = diameter of copper wire = 2 mm = 2 x 10⁻³ m

Area of cross-section of copper wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

E = magnitude of electric field = 0.01 V/m

e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of electron = 9.1 x 10⁻³¹ kg

n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³

i = magnitude of current

magnitude of current is given as

i = \frac{Ane^{2}\tau E}{m}

i = \frac{(3.14\times 10^{-6})(8.47\times 10^{28})(1.6\times 10^{-19})^{2}(2.5\times 10^{-14}) (0.01)}{(9.1\times 10^{-31})}

i  = 1.87 A

4 0
3 years ago
n electromagnetic wave in vacuum has an electric field amplitude of 611 V/m. Calculate the amplitude of the corresponding magnet
enot [183]

Answer:

The  corresponding  magnetic field is  

Explanation:

From the question we are told that

    The electric field amplitude is  E_o   =  611\  V/m

   

Generally the  magnetic  field amplitude is  mathematically represented as

              B_o  =  \frac{E_o }{c }

Where c is the speed of light with a constant value

         c = 3.0 *0^{8} \ m/s

So  

        B_o   =  \frac{611 }{3.0*10^{8}}

         B_o   =  2.0 4 *10^{-6} \  Vm^{-2} s

Since 1  T  is  equivalent to  V  m^{-2} \cdot  s

         B_o  =  2.0 4 *10^{-6} \ T

6 0
3 years ago
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