A car is at x1 = 15m at time t1 = 5 secs & later seen at x2=20m at time t2 = 10secs. find the average velocity of the car.

What is the acceleration of a proton moving with a speed of 6.5 m/s at right angles to a magnetic field of 1.5 T?

Brilliant_brown [7]

Answer:

The acceleration of the proton is 9.353 x 10⁸ m/s²

Explanation:

Given;

speed of the proton, u = 6.5 m/s

magnetic field strength, B = 1.5 T

The force of the proton is given by;

F = ma = qvB(sin90°)

ma = qvB

where;

m is mass of the proton, = 1.67 x 10⁻²⁷ kg

charge of the proton, q = 1.602 x 10⁻¹⁹ C

The acceleration of the proton is given by;

$a = \frac{qvB}{m}\\\\a = \frac{(1.602*10^{-19})(6.5)(1.5)}{1.67*10^{-27}}\\\\a = 9.353*10^8 \ m/s^2$

Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²

4 0

2 years ago

A drop of oil of volume 10m it spread out on water to make a circular firm of radius 10m calculate the tickness of the firm

Effectus [21]

Answer:

h = 3.1 cm

Explanation:

Given that,

The volume of a oil drop, V = 10 m

Radius, r = 10 m

We need to find the thickness of the film. The film is in the form of a cylinder whose volume is as follows :

$V=\pi r^2 h\\\\h=\dfrac{V}{\pi r^2}\\\\h=\dfrac{10}{\pi \times 10^2}\\\\h=0.031\ m\\\\h=3.1\ cm$

So, the thickness of the film is equal to 3.1 cm.

8 0

2 years ago

What is the SI unit of pressure? A. g/cm3 B. the pascal C. the newton D. m/s2

sesenic [268]

The basic definition of pressure is force/area and the scientific community defined that as the Pascal (Pa).

6 0

3 years ago

Read 2 more answers

In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a

matrenka [14]

Answer:

Approximately $\displaystyle\rm \left[ \begin{array}{c}\rm191\; m\\\rm-191\; m\end{array}\right]$ .

Explanation:

Consider this $45^{\circ}$ slope and the trajectory of the skier in a cartesian plane. Since the problem is asking for the displacement vector relative to the point of "lift off", let that particular point be the origin $(0, 0)$ .

Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at $45^{\circ}$ to the x-axis. Since this slope is connected to the ramp, it should also go through the origin. Based on these conditions, this line should be represented as $y = -x$ .

Convert the initial speed of this diver to SI units:

$\displaystyle v = \rm 110\; km\cdot h^{-1} = 110 \times \frac{1}{3.6} = 30.556\; m\cdot s^{-1}$ .

The question assumes that the skier is in a free-fall motion. In other words, the skier travels with a constant horizontal velocity and accelerates downwards at $g$ ( $g \approx \rm -9.81\; m\cdot s^{-2}$ near the surface of the earth.) At $t$ seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:

$x$ -coordinate: $30.556t$ meters (constant velocity;)
$y$ -coordinate: $\displaystyle -\frac{1}{2}g\cdot t^{2} = -\frac{9.81}{2}\cdot t^{2}$ meters (constant acceleration with an initial vertical velocity of zero.)

To eliminate $t$ from this expression, solve the equation between $t$ and $x$ for $t$ . That is: express $t$ as a function of $x$ .

$x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}$ .

Replace the $t$ in the equation of $y$ with this expression:

$\begin{aligned} y = &-\frac{9.81}{2}\cdot t^{2}\\ &= -\frac{9.81}{2} \cdot \left(\frac{x}{30.556}\right)^{2}\\&= -0.0052535\;x^{2}\end{aligned}$ .

Plot the two functions:

$y = -x$ ,
$\displaystyle y= -0.0052535\;x^{2}$ ,

and look for their intersection. Refer to the diagram attached.

Alternatively, equate the two expressions of $y$ (right-hand side of the equation, the part where $y$ is expressed as a function of $x$ .)

$-0.0052535\;x^{2} = -x$ ,

$\implies x = 190.35$ .

The value of $y$ can be found by evaluating either equation at this particular $x$ -value: $x = 190.35$ .

$y = -190.35$ .

The position vector of a point $(x, y)$ on a cartesian plane is $\displaystyle \left[\begin{array}{l}x \\ y\end{array}\right]$ . The coordinates of this skier is approximately $(190.35, -190.35)$ . The position vector of this skier will be $\displaystyle\rm \left[ \begin{array}{c}\rm191\\\rm-191\end{array}\right]$ . Keep in mind that both numbers in this vectors are in meters.