In a long-jump competition, Connor jumped 160 centimetres.

Mathematics

1 answer:

xeze [42]2 years ago

6 0

Answer:

290 cm

Step-by-step explanation:

There are 100 centimeters in a meter so 1.3m*100 is 130cm

130+160cm is 290cm.

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What is an equation of the line that passes through the point ( − 3 , − 1 ) and is perpendicular to the line x − 2 y =6

Tems11 [23]

Answer: y = -2x - 7

Step-by-step explanation:

The equation for a perpenducular line has a slope that is the negative inverse of the reference line, in this case x - 2y = 6.

Let's put this equation into standard slope-intercept form, y = mx+b, where m is the slope and b the y-intercept (the value of y when x is 0).

x - 2y = 6

- 2y = 6 - x

y = (1/2) x -3

The slope is (1/2), so a perpendicular line will have a slope of -2.

y = -2x + b

We can find b by using the given point ((-3,-1), a known solution):

-1 = -2*(-3) + b

-1 = 6 + b

b = -7

The perpendicular equation is y = -2x - 7

y = -2x

5 0

3 years ago

Electrons execute 1.4*10^6 revolutions per sec in a magnetic field of flux density

seropon [69]

The charge of the electrons is $8.02 \times 10^{-19} \ C$ .

The given parameters:

Number of revolution of the electron, N = 1.4 x 10⁶ rev per sec
Flux density, B = 10⁻⁵ T
Mass of electron, m = 9.11 x 10⁻³¹ kg

The angular speed of the electron is calculated as follows;

$\omega = 2\pi N\\\\\omega = 2\pi \ \frac{rad}{1 \ rev} \times 1.4 \times 10^6 \ rev/s\\\\\omega = 8.798 \times 10^6 \ rad/s$

The charge of the electrons is calculated as follows;

$F_c= m\omega ^2 r\\\\F_{B}= qvB = q\omega r B \\\\F_c = F_B\\\\m\omega ^2 r = q\omega r B \\\\q = \frac{m\omega ^2 r}{\omega r B} \\\\q = \frac{m \omega }{B} \\\\q = \frac{9.11 \times 10^{-31} \times 8.798 \times 10^6}{10^{-5}} \\\\q = 8.02 \times 10^{-19} \ C$