Which landform is the farthest north?

A 12.0-g plastic ball is dropped from a height of 2.50 m. Just as it strikes the floor, it is moving at a speed of 3.20 m/s. How

nalin [4]

Answer:

0·233 J

Explanation:

Given

Mass of the ball = 0·012 kg

Initially the ball is at a height of 2·5 m

As initially the ball is dropped, it's initial velocity will be equal to 0

Therefore initially it has zero kinetic energy and has only potential energy

∴ Initially total mechanical energy of the ball = potential energy of the ball

Initial potential energy of the ball = m × g × h

where

m is the mass of the ball

g is the acceleration due to gravity

h is the height of the ball

∴ Potential energy = 0·012 × 9·8 × 2·5 = 0·294 J

Velocity of the ball after striking the floor = 3·2 m/s

After striking the floor, the total mechanical energy = kinetic energy just after striking the floor

Kinetic energy = 0·5 × m × v²

where m is the mass of the ball

v is the velocity of the ball

∴ Kinetic energy of the ball = 0·5 × 0·012 × 3·2² = 0·061 J

Mechanical energy that is lost = 0·294 - 0·061 = 0·233 J

∴ Mechanical energy that the ball lost during its fall = 0·233 J

6 0

2 years ago

The average distance from the sun to Pluto is approximately 6.10 × 109 km. How long does it take light from

Scorpion4ik [409]

$V= \frac{S}{t}$
$t= \frac{S}{V}$ <u />
$t= \frac{S}{c}$
$t= \frac{6.1*10^{12}}{299792458}$
$t=20347.4098071s$

It takes 20347.4098071s for light from the sun to reach Pluto.
The 6.1*10^9 is replaced by 6.1*10^12 on line 4 because we convert the distance from km to m.
c = speed of light. If a different value was given in the previous question then use that instead of the value I used to do the final calculation.

3 0

3 years ago

A 90kg person jumps from a 30m tower into a tub of water with a volume of 5m3 initially at 20°C. Assuming that all of the work d

Kryger [21]

Answer:

T₂ = 20.06 ° C

Explanation:

Given

P = 90 kg, T₁ = 20 ° C, h = 30 m, c = 1.82 kJ / Kg * ° C

Using the formula to determine the final temperature of the water

T₂ = T₁ * P * h / Eₐ * c

The work done of the person to the water

Eₐ = 1000 kg / m³ * 5 m³ * 9.8 m / s²

Eₐ = 49000 N

T₂ = 20 ° C +[ (90 kg * 30m) / (49000 N * 1.82) ]

T₂ = 20.06 ° C

4 0

3 years ago

A 39-cm-long vertical spring has one end fixed on the floor. Placing a 2.2 kg physics textbook on the spring compresses it to a

Inessa05 [86]

Answer:

The spring constant is 215.6 N/m.

Explanation:

Given that,

Distance = 39 cm

Compresses length = 29 cm

Mass = 2.2 kg

We need to calculate the distance

Using formula of distance

$x=l-l'$

Put the value into the formula

$x=39-29=10\ cm$

We need to calculate the spring constant

Using formula of restoring force

$F=kx$

$k=\dfrac{mg}{x}$

Where, F = force

x = distance

Put the value into the formula

$k=\dfrac{2.2\times9.8}{10\times10^{-2}}$

$k=215.6\ N/m$

Hence, The spring constant is 215.6 N/m.

7 0

2 years ago

A ray of light is moving from a material having a high indexof refraction into a material with a lower index of refraction.

luda_lava [24]

(a) Away from the normal

We can find the direction of bending of the ray of light by using Snell's equation:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where we have:

n1, n2: index of refraction of the first and second medium

$\theta_1, \theta_2$ ; angle that the incident and the refracted ray form with the normal to the surface

Here, the light ray moves from a material with high index of refraction to a material with lower index, so we have

$n_1 > n_2$

Re-arranging Snell's law we find

$sin \theta_2 = \frac{n_1}{n_2} sin \theta_1$

since we have

$\frac{n_1}{n_2}>1$

this implies

$sin \theta_2 > sin \theta_1\\\theta_2 > \theta_1$

so the ray of light bends away from the normal.

(b) The wavelength is greater in the second material (the one with lower index of refraction)

The wavelength of the light in a medium is given by

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda_0$ is the wavelength of the light in a vacuum

n is the refractive index

The equation can be rewritten as

$\lambda_0 = \lambda_1 n_1 = \lambda_2 n_2$

and again it can be rewritten as

$\lambda_2 = \frac{n_1}{n_2} \lambda_1$

where

$\lambda_1 = 600 nm\\\frac{n_1}{n_2}>1$

Therefore, we have that the wavelength in the second medium (the one with lower index of refraction) is longer than the wavelength in the first medium.

(c) The frequency remains the same

Wavelength and speed of a light ray depend on the medium in which the wave is travelling through, however the frequency does not depend on that, so it remains the same in the two mediums.

8 0

2 years ago