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2 years ago

7

Do the data for the first part of the experiment support or refute the first hypothesis? Be sure to explain your answer and incl

ude how the variables changed in the first part of the experiment.

2 answers:

Sphinxa [80]2 years ago

5 0

The data for the first part of the experiment support the first hypothesis. As the force applied to the cart increased, the acceleration of the cart increased. Since the increase in the applied force caused the increase in the cart's acceleration, force and acceleration are directly proportional to each other, which is in accordance with Newton's second law.

klio [65]2 years ago

3 0

no experiment shown here ?

You might be interested in

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th

qaws [65]

Answer:

The distance between both cars is 990 m

Explanation:

The equations for the position and the velocity of an object moving in a straight line are as follows:

x = x0 + v0 * t + 1/2 * a * t²

v = v0 + a * t

where:

x = position of the car at time "t"

x0 = initial position

v0 = initial speed

t = time

a = acceleration

v = velocity

First let´s find how much time it takes the driver to come to stop (v = 0). We will consider the origin of the reference system as the point at which the driver realizes she must stop. Then x0 = 0

With the equation of velocity, we can obtain the acceleration and replace it in the equation of position, knowing that the position will be 250 m at that time.

v = v0 + a*t

v-v0 / t = a

0 m/s - 71.0 m/s / t =a

-71.0 m/s / t = a

Replacing in the equation for position:

x = v0* t +1/2 * a * t²

250 m = 71.0 m/s * t + 1/2 *(-71.0 m/s / t) * t²

250 m = 71.0 m/s * t - 1/2 * 71.0 m/s * t

250m = 1/2 * 71.0m/s *t

<u>t = 2 * 250 m / 71.0 m/s = 7.04 s</u>

It takes the driver 7.04 s to stop.

Then, we can calculate how much time it took the driver to reach her previous speed. The procedure is the same as before:

v = v0 + a*t

v-v0 / t = a now v0 = 0 and v = 71.0 m/s

(71.0 m/s - 0 m/s) / t = a

71.0 m/s / t =a

Replacing in the position equation:

x = v0* t +1/2 * a * t²

390 m = 0 m/s * t + 1/2 * 71.0 m/s / t * t² (In this case, the initial position is in the pit, then x0 = 0 because it took 390 m from the pit to reach the initial speed).

390m * 2 / 71.0 m/s = t

<u>t = 11.0 s</u>

In total, it took the driver 11.0s + 5.00 s + 7.04 s = 23.0 s to stop and to reach the initial speed again.

In that time, the Mercedes traveled the following distance:

x = v * t = 71.0 m/s * 23.0 s = 1.63 x 10³ m

The Thunderbird traveled in that time 390 m + 250 m = 640 m.

The distance between the two will be then:

<u>distance between both cars = 1.63 x 10³ m - 640 m = 990 m. </u>

3 0

2 years ago

The wavelength of green light is 550 nm.

SIZIF [17.4K]

Answer:

(a) momentum of photon is 1.205 x 10⁻²⁷ kgm/s

velocity of electron is 1323.88 m/s

momentum of the electron is 1.205 x 10⁻²⁷ kgm/s

(b) momentum of photon is 1.506 x 10⁻²⁷ kgm/s

velocity of electron is 1654.85 m/s

momentum of the electron is 1.506 x 10⁻²⁷ kgm/s

(c) The momentum of the photon is equal to the momentum of the electron

Explanation:

(a)

wavelength of green light, λ = 550 nm

momentum of photon is given by;

$p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{550*10^{-9}}\\\\p = 1.205 *10^{-27} \ kg.m/s$

velocity of electron is given by;

$P = \frac{h}{\lambda} \\\\mv = \frac{h}{\lambda}\\\\v = \frac{h}{m \lambda}\\\\v = \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(550*10^{-9})}\\\\v = 1323.88 \ m/s$

momentum of the electron is given by;

p = mv

p = (9.1 x 10⁻³¹) (1323.88)

p = 1.205 x 10⁻²⁷ kgm/s

(b)

wavelength of red light, λ = 440 nm

momentum of photon is given by;

$p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{440*10^{-9}}\\\\p = 1.506 *10^{-27} \ kg.m/s$

velocity of electron is given by;

$v = \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(440*10^{-9})}\\\\v = 1654.85 \ m/s$

momentum of the electron is given by;

p = mv

p = (9.1 x 10⁻³¹) (1654.85)

p = 1.506 x 10⁻²⁷ kgm/s

(c) The momentum of the photon is equal to the momentum of the electron.

7 0

2 years ago

Daria was swimming in a friend’s pool yesterday, when she saw that a fly had landed in the water about 5 feet away from her. She

jasenka [17]

Answer:

Daria probably suffers from Entomophobia.

5 0

2 years ago

A 2.0-m-tall man is 5.0 m from the converging lens of a camera. His image appears on a detector that is 50 mm behind the lens. H

ladessa [460]

Answer:

20 cm

Explanation:

We can solve the problem by using the magnification equation:

$M=\frac{h_i}{h_o}=-\frac{q}{p}$

where

$h_i$ is the size of the image

$h_o = 2.0 m$ is the height of the real object (the man)

$q=50 mm =0.050 m$ is the distance of the image from the lens

$p = 5.0 m$ is the distance of the object (the man) from the lens

Solving the formula for $h_i$ , we find

$h_i = -\frac{q}{p}h_o=-\frac{0.050 m}{5.0 m}(2.0 m)=-0.02 m = -20 cm$

And the negative sign means the image is inverted.

6 0

2 years ago

A measuring cylinder contains 60cm3 of oil at 0 celcius. When a piece of ice was roped into the cylinder it sank completely in o

mariarad [96]

Answer:

$S_i=\frac{9}{10} =0.9$

Explanation:

Given:

volume of oil in the cylinder, $V_o=60\ cm^2$

volume of the oil level when the ice is immersed, $V=90\ cm^3$

the volume level of oil when the ice melted, $V'=87\ cm^3$

<u>Now, therefore the volume of ice:</u>

$V_i=V-V_o$

$V_i=90-60$

$V_i=30\ cm^3$

<u>Now the volume of water:</u>

$V_w=V'-V_o$

$V_w=87-60$

$V_w=27\ cm^3$

As we know that the relative density is the ratio of density of the substance to the density of water.

<u>So, the relative density of ice:</u>

$S_i=\frac{\rho_i}{\rho_w}$ .....................(1)

as we know that density is given as:

$\rm \rho=\frac{mass}{volume}$

now eq. (1)

$S_i=\frac{m}{V_{i}}\div \frac{m}{V_w}$

where, m = mass of the water or the ice which remains constant in any phase

$S_i=\frac{V_w}{V_i}$

$S_i=\frac{27}{30}$

$S_i=\frac{9}{10} =0.9$

7 0

2 years ago