What forces are equal and opposite sides

If a wave had a period of 0.077 sec, it had a frequency of ?

Likurg_2 [28]

Answer: $12.98 Hz$

Explanation:

The frequency $f$ of a wave is given by the following equation:

$f=\frac{1}{P}$

Where:

$P=0.077 s$ is the period

Hence:

$f=\frac{1}{0.077 s}$

$f=12.98 s^{-1}=12.98 Hz$

7 0

3 years ago

A car accelerates uniformly from rest and

miv72 [106K]

Answer:

29.75 revolutions

Explanation:

The kinematic formula for distance, given a uniform acceleration a and an initial velocity v₀, is

$d=v_0t+\frac{1}{2}at^2$

This car is starting from rest, so v₀ = 0 m/s. Additionally, we have a = 9.2/9.7 m/s² and t = 9.7 s. Plugging these values into our equation:

$d=0t+\frac{1}{2}\left(\frac{9.2}{9.7}\right)(9.7)^2\\d=\frac{1}{2}(9.2)(9.7)\\d=4.6(9.7)\\d=44.62$

So, the car has travelled 44.62 m in 9.7 seconds - we want to know how many of the tire's <em>circumferences</em> fit into that distance, so we'll first have to calculate that circumference. The formula for the circumference of a circle given its diameter is $c=\pi{d}$ , which in this case is 47.8π cm, or, using π ≈ 3.14, 47.8(3.14) = 150.092 cm.

Before we divide the distance travelled by the circumference, we need to make sure we're using the same units. 1 m = 100 cm, so 105.092 cm ≈ 1.5 m. Dividing 44.62 m by this value, we find the number of revs is

$44.62/1.5\approx29.75$ revolutions

7 0

3 years ago

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an elect

klio [65]

Complete Question

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.4 cm?

Answer:

The magnetic field strength is $B= 0.0048 T$

Explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

$\Delta V q = \frac{1}{2}mv^2$

Making v the subject

$v = \sqrt{[\frac{2 \Delta V * q }{m}] }$

Where m is the mass of electron

v is the velocity of electron

q charge on electron

$\Delta V$ is the potential difference

Substituting values

$v = \sqrt{\frac{2 * 5.9 *10^3 * 1.60218*10^{-19} }{9.10939 *10^{-31]} }$ f

$= 4.5556 *10^ {7} m/s$

For the electron to move in a circular path the magnetic force[ $F = B q v$ ] must be equal to the centripetal force[ $\frac{mv^2}{r}$ ] and this is mathematically represented as

$Bqv = \frac{mv^2}{r}$

making B the subject

$B = \frac{mv}{rq}$

r is the radius with a value = 5.4cm = $= \frac{5.4}{100} = 5.4*10^{-2} m$

Substituting values

$B = \frac{9.1039 *10^{-31} * 4.556 *10^7}{5.4*10^-2 * 1.60218*10^{-19}}$

$= 0.0048 T$

7 0

4 years ago

What element is left over after the nuclear us reaction splits into different elements?

SVEN [57.7K]

Answer: A Radium

Explanation:

Thorium-232 is an alpha-emitting radionuclide, which decays to radium-228, which is a beta emitter with a half-life of about six years.

7 0

3 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$