Work is done if you carry a plant across a room at constant velocity. True False

how do i.... Find the angle between each pair of vectors below. (1 point each) a. a = 2x + 3y a = 4x + 2y

Shalnov [3]

Recall that the dot product between two vectors $\mathbf a$ and $\mathbf b$ satisfies

$\mathbf a\cdot\mathbf b=\|\mathbf a\|\|\mathbf b\|\cos\theta$

where $\|\mathbf x\|$ denotes the norm of a vector $\mathbf x$ , and $\theta$ is the angle between the two vectors.

So if $\mathbf a=(2,3)$ and $\mathbf b=(4,2)$ , then

$(2,3)\cdot(4,2)=\|(2,3)\|\|(4,2)\|\cos\theta$

$\implies2\cdot4+3\cdot2=\sqrt{2^2+3^2}\sqrt{4^2+2^2}\cos\theta$

$\implies\cos\theta=\dfrac{14}{\sqrt{13}\sqrt{20}}$

$\implies\theta\approx29.7^\circ$

6 0

3 years ago

A photon detector captures a photon with an energy of 4.29 ✕ 10−19 J. What is the wavelength, in nanometers, of the photon?

serious [3.7K]

Answer : The wavelength of photon is, $4.63\times 10^{2}nm$

Explanation : Given,

Energy of photon = $4.29\times 10^{-19}J$

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

$\nu$ = frequency of photon

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength of photon = ?

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$4.29\times 10^{-19}J=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{\lambda}$

$\lambda=4.63\times 10^{-7}m=4.63\times 10^{-7}\times 10^9nm=4.63\times 10^{2}nm$

Conversion used : $1nm=10^{-9}m$

Therefore, the wavelength of photon is, $4.63\times 10^{2}nm$

6 0

3 years ago

When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the

Taya2010 [7]

Answer:

The angular velocity is 15.37 rad/s

Solution:

As per the question:

$\theta = 54.6^{\circ}$

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v = $\sqrt{\frac{gx}{sin2\theta}}$

v = $\sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}$

v = $\sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s$

Now,

The angular velocity can be calculated as:

$v = \omega R$

Thus

$\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s$

8 0

3 years ago

Two billiard balls (each with mass equal to 170 g) collide head-on along the same line. Billiard ball A originally traveled east

barxatty [35]

Answer:

lucky mauld mauldgomary was an british poet...

5 0

3 years ago

A common cylindrical copper wire used in a lab is 841 m long. Find the radius (in mm) of a wire necessary to have 0.5 Ohms of re

slava [35]

The relationship between the resistance R of a wire and its resistivity $\rho$ is given by
$R= \frac{\rho L}{A}$
where L is the length of the wire and A is its cross sectional area.

In the problem, we have $R=0.5 \Omega$ , $\rho = 1.68 \cdot 10^{-8} \Omega m$ and $L=841 m$ . So we can solve the find the area A:
$A= \frac{\rho L}{R}=2.83 \cdot 10^{-5} m^2$

For a cylindrical wire, the cross sectional area is given by
$A= \pi r^2$
where r is the radius. We know the value of the area A, so now we can find the radius of the wire:
$r= \sqrt{ \frac{A}{\pi} }= \sqrt{ \frac{2.83 \cdot 10^{-5}m^2}{\pi} }=0.003 m=3.0 mm$

3 0

3 years ago