When people don't clearly recognize the obstacle causing their frustration, they

How long will it take a basketball starting from rest to roll without slipping 2.6 m down an incline that makes an angle of 30.8

Romashka-Z-Leto [24]

Answer:

t = 1.02 s

Explanation:

friction = 0 N (no slipping)

downward force along the inclined plane = F = mgsin(30.8)

F = ma

⇒ a = 9.81 sin(30.8)

a = 5.02 m/s^2

now we have

displacement = s = 2.6 m

acceleration = a = 5.02m/s^2

initial velocity = Vi = 0m/s

time = t = ?

APPLYING THE 2ND EQUATION OF MOTION:

S = (Vi)(t)+0.5at^2

2.6 = 0(t) + 0.5 * 5.02 * t^2

t = 1.02 s

7 0

3 years ago

A ball of mass 3 kg is released at the top of a track, as shown in the image below. The top of the ramp is 1 metre higher than

aliya0001 [1]

Answer:

potential energy

Explanation:

3 0

3 years ago

PLZ help !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

vlada-n [284]

Answer:

The leftmost red square.

Explanation:

As the earth surface we are looking at is moving toward the right, the impact point of the wind would be left of the original appearing target.

6 0

3 years ago

A plane electromagnetic wave, with wavelength 4.1 m, travels in vacuum in the positive direction of an x axis. The electric fiel

marusya05 [52]

(a) $7.32\cdot 10^7 Hz$

The frequency of an electromagnetic waves is given by:

$f=\frac{c}{\lambda}$

where

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\lambda=4.1 m$ is the wavelength of the wave in the problem

Substituting into the equation, we find

$f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz$

(b) $4.60\cdot 10^8 rad/s$

The angular frequency of a wave is given by

$\omega = 2\pi f$

where

f is the frequency

For this wave,

$f=7.32\cdot 10^7 Hz$

So the angular frequency is

$\omega=2\pi(7.32\cdot 10^7 Hz)=4.60\cdot 10^8 rad/s$

(c) $1.53 m^{-1}$

The angular wave number of a wave is given by

$k=\frac{2\pi}{\lambda}$

where

$\lambda$ is the wavelength of the wave

For this wave, we have

$\lambda=4.1 m$

so the angular wave number is

$k=\frac{2\pi}{4.1 m}=1.53 m^{-1}$

(d) $1.03\cdot 10^{-6}T$

For an electromagnetic wave,

$E=cB$

where

E is the magnitude of the electric field component

c is the speed of light

B is the magnitude of the magnetic field component

For this wave,

E = 310 V/m

So we can re-arrange the equation to find B:

$B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T$

(e) z-axis

In an electromagnetic wave, the electric field and the magnetic field oscillate perpendicular to each other, and they both oscillate perpendicular to the direction of propagation of the wave. Therefore, we have:

- direction of propagation of the wave --> positive x axis

- direction of oscillation of electric field --> y axis

- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis

(f) 127.5 W/m^2

The time-averaged rate of energy flow of an electromagnetic wave is given by:

$I=\frac{E^2}{2\mu_0 c}$

where we have

E = 310 V/m is the amplitude of the electric field

$\mu_0$ is the vacuum permeability

c is the speed of light

Substituting into the formula,

$I=\frac{(310 V/m)^2}{2(4\pi\cdot 10^{-7} H/m) (3\cdot 10^8 m/s)}=127.5 W/m^2$

(g) $1.53\cdot 10^{-8} kg m/s$

For a surface that totally absorbs the wave, the rate at which momentum is transferred to the surface given by

$\frac{dp}{dt}=\frac{A}{c}$

where the <S> is the magnitude of the Poynting vector, given by

$=\frac{EB}{\mu_0}=\frac{(310 V/m)(1.03\cdot 10^{-6} T)}{4\pi \cdot 10^{-7}H/m}=254.2 W/m^2$

and where the surface is

A = 1.8 m^2

Substituting, we find

$\frac{dp}{dt}=\frac{(254.2 W/m^2)(1.8 m^2)}{3\cdot 10^8 m/s}=1.53\cdot 10^{-8} kg m/s$

(h) $8.47\cdot 10^{-7} N/m^2$

For a surface that totally absorbs the wave, the radiation pressure is given by

$p=\frac{}{c}$

where we have

$=254.2 W/m^2$

$c=3\cdot 10^8 m/s$

Substituting, we find

$p=\frac{254.2 W/m^2}{3\cdot 10^8 m/s}=8.47\cdot 10^{-7} N/m^2$

8 0

3 years ago

Three forces act on a statue. Force F⃗ 1F→1 (magnitude 45.0 NN) points in the +x-direction, Force F⃗ 2F→2 (magnitude 105 NN) poi

Trava [24]

Answer:

Resultant force = (232.93î + 246.10j) N

x-component of the resultant force = (+232.93î) N

y-component of the resultant force = (+246.1j) N

Explanation:

The net external force on the statue is equal to the resultant force on the statue.

And the resuphant force is a vector sum of all the other forces acting on the statue.

Force 1 = (45î) N

Force 2 = (105j) N

Force 3 = (235cos 36.9°)î + (235 sin 36.9°)j = (187.93î + 141.10j) N

Resultant force = (Force 1) + (Force 2) + (Force 3)

Resultant force = 45î + 105j + (187.93î + 141.10j) = (232.93î + 246.10j) N

Hope this helps!!!

5 0

3 years ago