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3 years ago

What affect does friction have on motion

Physics

1 answer:

satela [25.4K]3 years ago

3 0

Answer: friction reduces the speed during motion

Explanation:

The more the friction, the lesser the speed during motion

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2 (a) What is the distance from the Sun to Earth in terms of solar radii? Earth radii?

ohaa [14]

a) Distance Earth-Sun is 215.5 solar radii and 23,548 Earth radii

Mercury: 1.7 days

Mars: 6.5 days

Jupiter: 21.7 days

Uranus: 86.8 days

Neptune: 130.2 days

c) $1.3\cdot 10^6$ Earths can fit inside the Sun

Explanation:

The distance between the Sun and the Earth is 150 millions km, so

$d=150\cdot 10^6 km = 1.50\cdot 10^{11} m$

The solar radius is

$r_s = 6.96\cdot 10^5 km = 6.96\cdot 10^8 m$

Therefore the distance Earth-Sun in solar radii is

$d_s = \frac{d}{r_s}=\frac{1.50\cdot 10^{11}}{6.96\cdot 10^8}=215.5$

The Earth radius is

$r_e = 6.37\cdot 10^6 m$

Therefore the distance Earth-Sun in Earth radii is

$d_e=\frac{d}{r_e}=\frac{1.50\cdot 10^{11}}{6.37\cdot 10^6}=23,548$

The speed of the solar wind is

$v=400 km/s = 4\cdot 10^5 m/s$

The value of 1 AU (Astronomical Unit) is

$1 AU = 1.50\cdot 10^{11}m$ (distance Earth-Sun)

The distance between the Sun and Mercury is:

$d=0.4 AU \cdot 1.50\cdot 10^{11}=6.0\cdot 10^{10} m$

So the time taken by a parcel of solar wind to reach Mercury is:

$t=\frac{d}{v}=\frac{6.0\cdot 10^{10}}{4.0\cdot 10^5}=150,000 s$

Converting into days ( $1d=86400 s$ ),

$t=\frac{150,000}{86400}=1.7 d$

The distance between the Sun and Mars is:

$d=1.5 AU \cdot 1.50\cdot 10^{11}=2.25\cdot 10^{11} m$

So the time taken by a parcel of solar wind to reach Mars is:

$t=\frac{d}{v}=\frac{2.25\cdot 10^{11}}{4.0\cdot 10^5}=562,500 s$

Converting into days ( $1d=86400 s$ ),

$t=\frac{562,500}{86400}=6.5 d$

The distance between the Sun and Jupiter is:

$d=5 AU \cdot 1.50\cdot 10^{11}=7.5\cdot 10^{11} m$

So the time taken by a parcel of solar wind to reach Jupiter is:

$t=\frac{d}{v}=\frac{7.5\cdot 10^{11}}{4.0\cdot 10^5}=1.88 \cdot 10^6 s$

Converting into days ( $1d=86400 s$ ),

$t=\frac{1.88\cdot 10^6}{86400}=21.7 d$

The distance between the Sun and Uranus is:

$d=20 AU \cdot 1.50\cdot 10^{11}=3.0\cdot 10^{12} m$

So the time taken by a parcel of solar wind to reach Uranus is:

$t=\frac{d}{v}=\frac{3.0\cdot 10^{12}}{4.0\cdot 10^5}=7.5 \cdot 10^6 s$

Converting into days ( $1d=86400 s$ ),

$t=\frac{7.5\cdot 10^6}{86400}=86.8 d$

The distance between the Sun and Neptune is:

$d=30 AU \cdot 1.50\cdot 10^{11}=4.5\cdot 10^{12} m$

So the time taken by a parcel of solar wind to reach Neptune is:

$t=\frac{d}{v}=\frac{4.5\cdot 10^{12}}{4.0\cdot 10^5}=11.3 \cdot 10^6 s$

Converting into days ( $1d=86400 s$ ),

$t=\frac{11.3\cdot 10^6}{86400}=130.2 d$

As we said in part a), we have:

$r_e = 6.37\cdot 10^6 m$ (radius of the Earth)

$r_s=6.96\cdot 10^8 m$ (radius of the Sun)

So the volume of the Earth can be calculated as:

$V_e=\frac{4}{3}\pi r_e^3 = \frac{4}{3}\pi (6.37\cdot 10^6)^3=1.08\cdot 10^{21} m^3$

While the volume of the Sun is

$V_s=\frac{4}{3}\pi r_s^3 = \frac{4}{3}\pi (6.96\cdot 10^8)^3=1.41\cdot 10^{27} m^3$

Therefore, the number of Earths that could fit inside the Sun is:

$\frac{V_s}{V_e}=\frac{1.41\cdot 10^{27}}{1.08\cdot 10^{21}}=1.3\cdot 10^6$

Learn more about the Solar System:

brainly.com/question/2887352

brainly.com/question/10934170

#LearnwithBrainly

6 0

3 years ago

You drop a rock off the top of a building. It takes 4.5 s to hit the ground. How tall is

madam [21]

Since the rock as been dropped off the building you can assume that initial velocity equals 0m/s.

t=4.5s
acceleration due to gravity=9.8m/s^2

s=ut+1/2at^2

Since u=0, s=1/2at^2

s=1/2(9.8)(4.5)^
s=99.225m

4 0

3 years ago

Determina el trabajo realizado al desplazar un bloque 3 m sobre una superficie horizontal, si se desprecia la fricción y la fuer

bija089 [108]

Answer: El trabajo realizado al desplazar el bloque es: 75 N.m

Datos:

Fuerza= F= 25 N

Distancia= d= 3 m

Explicación:

El trabajo es el producto de la fuerza ejercida sobre un cuerpo con el desplazamiento producido por la fuerza. Para que exista trabajo, la dirección de la fuerza debe ser igual a la dirección del desplazamiento.

Trabajo= Fuerza x Desplazamiento

Reemplazando los datos:

T= 25 N * 3 m

T= 75 N.m

5 0

2 years ago

Which involves more work in the scientific sense: moving the boxes and the furniture down from the second floor or up to the fif

Advocard [28]

Answer:

moving the boxes and the furniture up to the fifth floor.

Explanation:

The work done when moving an object up or down is equal to its change of gravitational potential energy:

$W=mg \Delta h$

where

m is the mass of the object

g is the gravitational acceleration

$\Delta h$ is the change in height of the object

We have two opposite situations:

1) when an object is moved upward, $\Delta h$ is positive, so the work done W is positive as well: this means that we need to do work in order to move the object, because we have to "win" the effect of gravity, which pulls the object downward

2) when an object is moved downward, $\Delta h$ is negative, so the work done W is negative as well: this means that we do not need to do work on the object, because it is already done by gravity, which pulls the object downward.

Therefore, more work is done when we are moving the boxes and the furniture up to the fifth floor.

8 0

2 years ago

An object has an initial velocity of 15 m/s. How long must it accelerate at a constant rate of 3. 0 m/s² before its final veloci

Lera25 [3.4K]

Answer: 5 sec

Explanation:

30 = 15 + (3*t)

so, t=(30-15)÷3

= 5 sec

4 0

1 year ago