When a light wave enters into a medium of different optical density,

Ben rushin is waiting at a stoplight. when it finally turns green, ben accelerated from rest at a rate of a 6.00 m/s2 for a time

jasenka [17]

In the 4.10 seconds that elapsed, Ben reaches a velocity of

$v_f=v_0+at\implies v_f=0\,\dfrac{\mathrm m}{\mathrm s}+\left(6.00\,\dfrac{\mathrm m}{\mathrm s^2}\right)(4.10\,\mathrm s)$

$\implies v_f=24.6\,\dfrac{\mathrm m}{\mathrm s}$

In this time, his displacement $\Delta x$ satisfies

${v_f}^2-{v_0}^2=2a\Delta x\implies\left(24.6\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(0\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(6.00\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x$

$\implies\Delta x=50.4\,\mathrm m$

4 0

3 years ago

1) A force of 157 N is applied to a box 25o above the horizontal. If it is applied over a distance of 14.5 m how much work is do

vfiekz [6]

#1

Work done is given by

$W = F.d cos\theta$

here we know that

F = 157 N

d = 14.5

$\theta = 25^0$

now from the above formula

$W = (157)(14.5)cos25 = 2063.2 J$

#2

By energy conservation

Initial total potential energy of egg = final total kinetic energy of egg

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

$v = \sqrt{2(9.8)(8.2)}$

$v = 12.7 m/s$

#3

By momentum conservation we know that

$m_1v_1 = (m_1 + m_2)v$

$215(9.5) = (215 + 19.7) v$

$v = \frac{215 (9.5)}{215 + 19.7}$

$v = 8.7 m/s$

#4

by momentum conservation we know that

$m_1v_1 + m_2v_2 = (m_1 + m_2) v$

let direction of velocity of Jay Z car motion is positive direction

now we will say

$1100(-20) + 1475v = (1100 + 1475)(7)$

$1475 v = 2575(7) + 1100(20)$

$v = 27.14 m/s$

Yes Jay Z is overspeeding as his speed is more than speed limit

#5

Tension force in the string will be centripetal force

so here we can say

$T = \frac{mv^2}{R}$

here we know that

m = 0.63 kg

v = 7.8 m/s

R = 0.23 m

$T = \frac{0.63 (7.8)^2}{0.23}$

$T = 166.6 N$

6 0

3 years ago

Bottlenose dolphins use echolocation pulses with a frequency of about 100 kHz, higher than the frequencies used by most bats. Wh

Alinara [238K]

Answer:

1. greater

2. direct

3. smaller

4. inverse

Explanation:

The speed of sound in water is greater than in air; hence for the same frequency the sound wavelength in water is greater than in air (for the given frequency the wavelength is in the direct proportion with the speed of sound).

To "see" an object via the echolocation creature needs to use sound with the wavelength smaller than the size of an object viewed.

That means to "see" objects of the same size dolphin and bat need to use ultrasound of the same wavelength, hence dolphin needs to use higher frequency (for the given speed of sound the wavelength is in inverse proportion with the frequency).

8 0

3 years ago

A projectile is fired from the ground (you can assume the initial height is the same as the ground) in a field so there are no o

Cerrena [4.2K]

Answer:

A) 112 m. B) 27.2 m C) 41.1 m/s i + 13.4 m/s j D) 43.2 m/s

Explanation:

A) Once fired, no external forces act on the projectile in the x-direction, so it keeps moving to the right at constant speed, which is the projection on the x-axis of the initial velocity vector:

v₀ₓ = v₀* cos 33º = 49 m/s* cos 33º = 41.1 m/s

In the y-direction, the component of the velocity can be found as the projection of v₀ on the y-axis, as follows:

v₀y = v₀* sin 33º = 49 m/s* sin 33º = 26.7 m/s

Both velocities are independent each other, as no one has a projection on the other.

In the vertical direction, the projectile is in free fall all time, under the influence of gravity , which accelerates it downward.

So, at any time, in the vertical direction, the velocity can be calculated as follows:

vfy = v₀y -g*t (same equation as for an object thrown upwards)

When the object is at its maximum height, the velocity, in the vertical direction, will be momentarily zero, so we can find the time when this happens as follows:

vfy= 0 ⇒ v₀y = g*t ⇒ t = v₀y / g = 26.7 m/s / 9.8 m/s² = 2.72 s

As the time is the same for both movements, we can replace this value in the expression for the displacement x at constant speed, as follows:

x = v₀ₓ* t = 41.1 m/s* 2.72 s = 112 m

B) Like above, as the time is the same for both movements, we can find the time for the instant that the projectile hit the wall, as follows:

x = v₀ₓ* t ⇒ 55. 8 m = 41.1 m/s * t

⇒ t = 55. 8 m / 41.1 m/s = 1.36 s

We can replace this value of t in the equation for the vertical displacement, as follows:

Δy = v₀y*t -1/2*g*t² = (26.7m/s*1.36s) - 1/2*9.8m/s²*(1.36s)² = 27.2 m

C) The velocity of the projectile, at any time, has two components, one horizontal and one vertical.

As explained above, x-component is constant, equal to v₀x:

vx = v₀x i = 41.1 m/s i

For vy, we can apply acceleration definition, using the value of v₀y and t that we have just found:

vfy = voy - g*t = 26.7 m/s - 9.8m/s*1.36 sec = 13.4 m/s

vfy = 13.4 m/s j

v = 41.1 m/s i + 13.4 m/s j

D) Finally, in order to get the speed of the projectile when it hit the wall, we need just to find the magnitude of the velocity, as we get the magnitude of any vector given its vertical and horizontal components:

v = √(41.1 m/s)² +(13.4 m/s)² =43.2 m/s

5 0

3 years ago

. A light bulb glows because of it’s resistance, and the brightness of the bulbincreases with the electrical power delivered to

sesenic [268]

Complete Question

The complete question is shown in the first uploaded image

Answer:

a

When the both bulb are in the circuit bulb B glows equally brighter to bulb A

This because the power delivered to the both bulb are equal

b

The bulb A on the right will glow brighter than the bulb A on the left due to the fact that the power supplied to bulb A on the right is higher than that gotten by bulb A on the left.

Explanation:

From the question we are been told that the two bulbs are identical

So their resistance denoted by R is the same

Considering the left circuit where the two bulbs are connected in series which mean that the same current is passing through them

$R_A =R_B =R$