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3 years ago

Sequence eye light enters the eye

Physics

1 answer:

Burka [1]3 years ago

8 0

Answer:

Well it is the courtnes

Explanation:

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In order to prevent injury in a car crash, it is recommended that you _______. Increase the initial velocity of the collision.

Vlad1618 [11]

Increase the change in momentum of the collision

5 0

3 years ago

Starting from rest, a 2.3x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts

Harman [31]

Answer:

3.13 m/s

Explanation:

From the question,

Since the flea spring started from rest,

Ek = W................... Equation 1

Where Ek = Kinetic Energy of the flea spring, W = work done on the flea spring.

But,

Ek = 1/2mv²............ Equation 2

Where m = mass of the flea spring, v = flea's speed when it leaves the ground.

substitute equation 2 into equation 1

1/2mv² = W.................... Equation 3

make v the subject of the equation

v = √(2W/m)................. Equation 4

Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg

Substitute into equation 4

v = √[2×3.6×10⁻⁴ )/2.3×10⁻⁴]

v = 7.2/2.3

v = 3.13 m/s

Hence the flea's speed when it leaves the ground = 3.13 m/s

4 0

3 years ago

Find the acceleration if a 32.5 N force is used on an object that has a mass of 128.6 kg.

Sholpan [36]

Answer:

Acceleration=3.95

Explanation:Use the formula a=m/f

a=128.6/32.5

a=3.95

3 0

3 years ago

At a particular instant, a hot air balloon is 100 m in the air and descending at a constant speed of 2.0 m/s. at this exact inst

rewona [7]

Answer:

86.4 m horizontal from landing spot

Explanation:

Find out how long before the ball hits the ground

vertical speed of ball = -2 m/s gravity = - 9.81 m/s^2

find time to hit ground from 100 m

( height will be zero when it hits the ground)

0 = 100 - 2 t - 1/2 ( 9.81) t^2

use Quadratic Formula to find t = 4.32 seconds

horizontal speed of ball = 20 m/s

in 4.32 seconds it will travel horizontally 20 m/s * 4.32 s = 86.4 m

3 0

2 years ago

A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the

zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

$V = \frac{\pi d^2}{4}h$

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

$V_0 = \frac{\pi d^2}{4}H$

$V_f = \frac{\pi d^2}{4}h$

We know as well by definition that

$1gal = 3.785*10^{-3}m^3$

Then we have for the statement that

$V_f = V_0 -1gal$

$V_f = V_0 - 3.785*10^{-3}$

Replacing the previous data

$\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}$

$\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}$

Solving to get h,

$h = 1.99963m$

Therefore the change is

$\Delta h = H-h$

$\Delta h = 2- 1.99963$

$\Delta h = 3.7*10^{-4}m=0.37mm$

Therefore te change in the height of the water in the tank is 0.37mm

4 0

3 years ago