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2 years ago

Which of these is emitted during beta decay ?

Physics

2 answers:

satela [25.4K]2 years ago

4 0

Answer:

C. a small charged particle.

Explanation:

typically beta radiation emits an electron which is a small negativity charged particle.

hope it helps. :)

Vesna [10]2 years ago

3 0

Answer:

C) a small charged particle

Explanation:

AP3X

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As a car moves along the road, the distance traveled each second is measured and recorded. What is the

scoundrel [369]

If the car moves along the distance it will be 16 of the line graph where is independent of the graph

8 0

2 years ago

A physical quantity X is connected from X = ab2/C. Calculate percentage error in X, when percentage error in a,b,c are 4,2 and 3

Hoochie [10]

Answer: 11%

Explanation:

Given that

X = ab^2/C. Calculate percentage error in X, when percentage error in a,b,c are 4,2 and 3 respectively.

Percentage error of b = 2%

Percentage error of b^2 = 2 × 2 = 4

When you are calculating for percentage error that involves multiplication and division, you will always add up the percentage error values.

Percentage error of X will be;

Percentage error of a + percentage error of b^2 + percentage error of c

Substitute for all these values

4 + 4 + 3 = 11%

Therefore, percentage error of X is 11%

3 0

3 years ago

Four point charges are individually brought from infinity and placed at the corners of a square. Each charge has the identical v

Brut [27]

To solve this problem we will apply the concept of voltage given by Coulomb's laws. From there we will define the charges and the distance, and we will obtain the total value of the potential difference in the system.

The length of diagonal is given as

$l = 2a$

The distance of the center of the square from each of the corners is

$r = \frac{2a}{2}= a$

The potential electric at the center due to each cornet charge is

$V_1 = \frac{kQ_1}{r_1}$

$V_2 = \frac{kQ_2}{r_2}$

$V_3 = \frac{kQ_3}{r_3}$

$V_4 = \frac{kQ_4}{r_4}$

The total electric potential at the center of the given square is

$V = V_1+V_2+V_3+V_4$

$V = \frac{kQ_1}{r_1}+ \frac{kQ_2}{r_2}+\frac{kQ_3}{r_3}+\frac{kQ_4}{r_4}$

Al the charges are equal, and the distance are equal to a, then

$V = \frac{kQ}{a}+ \frac{kQ}{a}+\frac{kQ}{a}+\frac{kQ}{a}$

$V = \frac{4kQ}{a}$

Therefore the correct option is E.

3 0

3 years ago

An ac source is connected to a resistor R = 75Ω, an inductor L = 0.01 H, and a capacitor C = 4 μF. What is the phase difference

zvonat [6]

Answer:

135°.

Explanation:

R = 75 ohm, L = 0.01 H, C = 4 micro F = 4 x 10^-6 F

Frequency is equal to the half of resonant frequency.

Let f0 be the resonant frequency.

$f_{0}=\frac{1}{2\pi \sqrt{LC}}$

$f_{0}=\frac{1}{2\times 3.14 \sqrt{0.01\times 4\times 10^{-6}}}$

f0 = 796.2 Hz

f = f0 / 2 = 398.1 Hz

So, XL = 2 x 3.14 x f x L = 2 x 3.14 x 398.1 x 0.01 = 25 ohm

$X_{c}=\frac{1}{2\pi fC}$

Xc = 100 ohm

$tan\phi = \frac{X_{L}-_{X_{C}}}{R}$

tan Ф = (25 - 100) / 75 = - 1

Ф = 135°

Thus, the phase difference is 135°.

4 0

3 years ago

What is the direction of a vector with an x component of -12 m and a y component of 21 m?

Svet_ta [14]

Answer:

$Direction= 119.74^\circ$

Explanation:

<u>Displacement Vector</u>

The displacement, as every vector, has a magnitude r and a direction angle θ measured from the positive x-axis.

If we know the x-y components of the displacement, the magnitude and angle can be calculated by the equations:

$r=\sqrt{x^2+y^2}$

$\displaystyle \tan\theta=\frac{y}{x}$

The coordinates of the given vector are x=-12 m, y=21 m, thus:

$\displaystyle \tan\theta=\frac{21}{-12}=-1.75$

$\theta=tan^{-1}(-1.75)=-60.26^\circ$

Since the vector lies in the second quadrant, we add 180° to find the correct direction:

$\boxed{Direction=-60.26^\circ+180^\circ= 119.74^\circ}$

4 0

3 years ago