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3 years ago

The study of charges in motion and their

Physics

1 answer:

vredina [299]3 years ago

4 0

Answer:

The study of charges in motion and their interaction with magnetic fields is known as electromagnetism.

Any material with an unequal number of electrons and or protons could generally be termed as ion.

Explanation:

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A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c

stealth61 [152]

Answer :

(a) The acceleration of the car is, $-2m/s^2$

(b) The distance covered by the car is, 150 m

Explanation :

By the 1st equation of motion,

$v=u+at$ ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = ?

Now put all the given values in the above equation 1, we get:

$5m/s=25m/s+a\times (10s)$

$a=-2m/s^2$

The acceleration of the car is, $-2m/s^2$

By the 2nd equation of motion,

$s=ut+\frac{1}{2}at^2$ ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = $-2m/s^2$

Now put all the given values in the above equation 2, we get:

$s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2$

By solving the term, we get:

$s=150m$

The distance covered by the car is, 150 m

8 0

3 years ago

565,900 seconds into days

alukav5142 [94]

Answer:

Explanation:

4 0

3 years ago

Calculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of (a) 100 V, (b) 1.0 kV

forsale [732]

Answer:

(a) $\lambda=1.227\ A$

(b) $\lambda=0.388\ A$

Explanation:

Given that,

(a) An electron accelerated from rest through a potential difference of 100 V. The De Broglie wavelength in terms of potential difference is given by :

$\lambda=\dfrac{h}{\sqrt{2meV} }$

Where

m and e are the mass of and charge on an electron

On solving,

$\lambda=\dfrac{12.27}{\sqrt{V} }\ A$

V = 100 V

$\lambda=\dfrac{12.27}{\sqrt{100} }\ A$

$\lambda=1.227\ A$

(b) V = 1 kV = 1000 V

$\lambda=\dfrac{12.27}{\sqrt{V} }\ A$

$\lambda=\dfrac{12.27}{\sqrt{1000} }\ A$

$\lambda=0.388\ A$

$\lambda=\dfrac{12.27}{\sqrt{10^5} }\ A$

$\lambda=0.038\ A$

Hence, this is the required solution.

7 0

3 years ago

In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 4.8 m/s at an

tensa zangetsu [6.8K]

The equation to be used here is the trajectory of a projectile as written below:

y = xtanθ +/- gx²/2v²(cosθ)²
where
y is the vertical distance
x is the horizontal distance
θ is the angle of trajectory or launch angle
g is 9.81 m/s²
v is the initial velcity

Since the angle is below horizontal, let's use the minus equation. Substituting the values:

- 0.8 m = xtan15° - (9.81 m/s²)x²/2(4.8 m/s)²(cos15°)²
Solving for x,
x = 2.549 m

However, we only take half of this distance because it was specified that the distance asked before bouncing. Hence, the horizontal distance is equal to 1.27 m.

5 0

3 years ago

Which of these best describes how the kinetic energy of carbon dioxide molecules change as the carbon dioxide is heated?

spin [16.1K]

B.
carbon dioxide molecules have more energy; therefore, the kinetic energy increases

4 0

3 years ago

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