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Alex777 [14]
3 years ago
10

A proton accelerates from rest in a uniform electric field of 630 N/C. At one later moment, its speed is 1.50 Mm/s (nonrelativis

tic because v is much less than the speed of light). (a) Find the acceleration of the proton.
Physics
1 answer:
Vlad1618 [11]3 years ago
8 0

Answer:

the acceleration of the proton is 6.025 x 10¹⁰ m/s².

Explanation:

Given;

magnitude of electric field, E = 630 N/C

final speed of the proton, v = 1.5 M m/s = 1.5 x 10⁶ m/s

charge of proton, Q = 1.6 x 10⁻¹⁹ C

mass of proton, m = 1.673 x 10⁻²⁷ kg

The force experienced by the proton is calculated as;

F = ma = EQ\\\\a = \frac{EQ}{m} \\\\a = \frac{(630)(1.6\times 10^{-19})}{1.673 \times 10^{-27}} \\\\a = 6.025 \times 10^{10} \ m/s^2

Therefore, the acceleration of the proton is 6.025 x 10¹⁰ m/s².

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A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from
Ulleksa [173]

Answer:

1270.64\ \text{J}

Explanation:

m = Mass of object = \dfrac{mg}{g}

mg = Weight of object = 20 N

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

\theta = Angle of slope = 30^{\circ}

f = Force of friction

fd = Work done against friction

The force balance of the system is

\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}

The work done against friction is 1270.64\ \text{J}.

8 0
3 years ago
Miswer
quester [9]

Answer: 4.0

Explanation:

5 0
3 years ago
Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen
kakasveta [241]

Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

\lambda = 436.1 \ nm

or,

  =436.1\times 10^{-9} \ m

Distance,

d = 0.31 \ mm

or,

  =0.31\times 10^{-3} \ m

Distance between the 1st and 2nd dark fringes,

(y_2-y_1) = 6\times 10^{-3} \ m

As we know,

⇒ \frac{d}{L} (y_2-y_1) = \lambda

or,

⇒ L=\frac{d(y_2-y_1)}{\lambda}

By substituting the values, we get

       =\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}

       =\frac{0.31\times 6\times 10^3}{436.1}

       =\frac{1860}{436.1}

       =4.26 \ m

3 0
3 years ago
An astronaut lands on a new, recently discovered planet in a different star system. The astronaut measures the acceleration due
Nezavi [6.7K]

Answer:

The radius of the new planet is ~2.04 * 10⁶ m, or 2,041,752 m.

Explanation:

We can use Newton's Law of Universal Gravitation:

  • \displaystyle F_g=G\frac{Mm}{r^2}

Let's look at Newton's 2nd Law:

  • F=ma

We can set these equations equal to each other:

  • \displaystyle G\frac{Mm}{r^2} =ma

The mass of the second mass (astronaut) cancels out. We are left with:

  • \displaystyle G\frac{M}{r^2} =a

We are solving for the radius of the new planet, so we can rearrange the equation:

  • \displaystyle r=\sqrt{\frac{GM}{a} }

Substitute in our known values given in the problem (<u><em>G = 6.67 * 10⁻¹¹ </em></u><em> ; </em><u><em>M = 7.5 * 10²³</em></u><em> ; </em><u><em>a = 12</em></u>).

  • \displaystyle r =\sqrt{\frac{(6.67\times 10^{-11})(7.5 \times 10^{23}}{12} }
  • r=2.04 \times 10^6

The radius of the new planet is ~2.04 * 10⁶ m.

3 0
2 years ago
When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam
iris [78.8K]

Answer:

Explanation:

Given:

U1 = 1.6 m/s

U2 = -1.1 m/s

M1 = 1850 kg

M2 = 1400 kg

V1 = 0.27 m/s

Using momentum- collision equation,

M1U1 + M2U2 = M1V1 + M2V2

1850 × 1.6 - 1400 × 1.1 = 1850 × 0.27 + 1400 × V2

1420 = 499.5 + 1400V2

V2 = 0.6575 m/s

B.

KE = 1/2 × MV^2

KEa1 + KEa2 = KEb1 + KEb2

Delta KE = KE2 - KE1

KEa1 = 2368 J

KEb1 = 847 J

KEa2 = 67.433 J

KEb2 = 302.6 J

KE1 = KEa1 + KEb1

= 3215 J

KE2 = 370.033 J

Delta KE = -2845 J.

5 0
3 years ago
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