A gecko crawls vertically up and down a wall. Its motion is shown on the following graph of vertical position yyy vs. time ttt.

PLZ SOMEONEE HELPP I’LL MARK BRANLIESTTTT

7nadin3 [17]

Answer:

I'm pretty sure it's 37.5 joules of energy

Explanation:

hope this helps!

8 0

2 years ago

(See picture) may I have help!!?

algol13

Picture is blurry…. try re uploading it

6 0

1 year ago

Two boats - Boat A and Boat B - are anchored a distance of 24 meters apart. The incoming water waves force the boats to oscillat

ozzi

Answer:

wavelength = 24 m

Period = 10 s

f = 0.1 Hz

Amplitude = 4 m

Explanation:

Wavelength:

Since the boats are at crest and trough, respectively at the same time. Hence, the horizontal distance between them is the wavelength of the wave:

wavelength = 24 m

Period:

The period is given as:

$Period = \frac{time}{no.\ of\ cycles} \\\\Period = \frac{10\ s}{1}\\\\$

Period = 10 s

Frequency:

The frequency is given as:

$f = \frac{1}{time\ period}\\\\f = \frac{1}{10\ s}\\\\$

f = 0.1 Hz

Amplitude:

Amplitude will be half the distance between extreme points, that is, crest and trough:

Amplitude = 8 m/2

Amplitude = 4 m

5 0

2 years ago

A long cylindrical insulating shell has an inner radius of a = 1.41 m and an outer radius of b = 1.67 m. The shell has a constan

Natasha2012 [34]

Answer:

a. E = 122.4 N/C

b. E = 58.2 N/C

c. E = 0

Explanation:

The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.

In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.

A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.

$E2\pi rh = \frac{\lambda V}{\epsilon_0} = \frac{\lambda \pi (b^2 - a^2)h}{\epsilon_0}\\E2\pi (1.97)h = \frac{(5.3\times 10^{-9})\pi(1.67^2 - 1.41^2)h}{\epsilon_0}\\E = 122.4~N/C$

B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.

$E2\pi rh = \frac{\lambda V_{enc}}{\epsilon_0} = \frac{\lambda \pi (r^2 - a^2}h{\epsilon_0}\\E2\pi (1.51)h = \frac{5.3\times 10^{-9})\pi(1.51^2 - 1.41^2)h}{\epsilon_0}\\E = 58.2~N/C$

C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.

6 0

2 years ago

6. Two blocks are released from rest at the same height. Block A slides down a steeper ramp than Block B. Both ramps are frictio

Stolb23 [73]

Answer:

a. the work done by the gravitational force on Block A is less than the work done by the gravitational force on Block B.

b. the speed of Block A is equal to the speed of Block B.

c. the momentum of Block A is less than the momentum of Block B.

Explanation:

a. The work done by the gravitational force is equal to:

w = m*g*h

where m is mass, g is the standard gravitational acceleration and h is height. Given that both blocks are released from rest at the same height, then, the bigger the mass, the bigger the work done.

b. With ramps frictionless, the final speed of the blocs is:

v = √(2*g*h)

which is independent of the mass of the blocks.

c. The momentum is calculated as follows:

momentum = m*v

Given that both bocks has the same speed, then, the bigger the mass, the bigger the momentum.

8 0

2 years ago