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3 years ago

a motorcycle starting from rest has an acceleration of 2.6m/s how long does it take the motorcycle to travel a distance of 120

Physics

1 answer:

Setler79 [48]3 years ago

5 0

46.15 seconds to reach 120

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You are driving your car and the traffic light ahead turns red.You apply the breaks for 3s and the velocity of the car decreases

hammer [34]

Answer:

The car's displacement during this time is 25.65 meters.

Explanation:

Given that,

Final velocity of the car, v = 4.5 m/s

Deceleration of the car, $a=-2.7\ m/s^2$

Let u is the initial speed of the car. It is given by :

$v=u+at$

$u=v-at$

$u=4.5-(-2.7)\times 3$

u = 12.6 m/s

Let d is the car's displacement during this time. It can be calculated using second equation of motion as :

$d=ut+\dfrac{1}{2}at^2$

$d=12.6\times 3+\dfrac{1}{2}\times (-2.7)\times 3^2$

d = 25.65 meters

So, the car's displacement during this time is 25.65 meters. Hence, this is the required solution.

7 0

3 years ago

5) What is the net force caused by the moon acting on earth when the moon is 3.86x10^8 m away? The moon has a mass of 7.46x10^23

Dimas [21]

Answers:

5) $1.99(10)^{21} N$

6) $1.37(10)^{18} N$

7) $1.64(10)^{21} N$

8) $4.29(10)^{10} N$ more than Venus force of gravity on Pluto

Explanation:

According to Newton's law of Universal Gravitation, the force $F$ exerted between two bodies of masses $M$ and $m$ and separated by a distance $R$ is equal to the product of their masses and inversely proportional to the square of the distance:

$F=G\frac{Mm}{R^{2}}$ (1)

Where $G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

This is the equation we will use to solve each question in this problem.

<h3>5) Gravitational force between Earth and Moon</h3>

In this case we have:

$F_{earth-moon}$ is the gravitational force between Earth and Moon

$M=5.97(10)^{24} kg$ is the mass of the Earth

$m=7.46(10)^{23} kg$ is the mass of the Moon

$R=3.86(10)^{8} m$ is the distance between Earth and Moon

Solving:

$F_{earth-moon}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(5.97(10)^{24} kg)(7.46(10)^{23} kg)}{(3.86(10)^{8} m)^{2}}$ (2)

$F_{earth-moon}=1.99(10)^{21} N$ (3)

<h3>6) Gravitational force between Jupiter and Venus</h3>

Assuming for a moment that the planets are perfectly aligned and all are in the same orbital period, we can make a rough estimation of the distance between Jupiter and Venus, knowing the distance of each to the Sun:

distance between Sun and Jupiter - distance between Sun and Venus=distance between Jupiter and Venus= $R_{jupiter-venus}$ (4)

$R_{jupiter-venus}=778.3(10)^{9} m - 108(10)^{9} m=6.703(10)^{11} m$ (5)

Using this value in the Law of Universal Gravitation equation:

$F_{jupiter-venus}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.90(10)^{27} kg)(4.87(10)^{24} kg)}{(6.703(10)^{11} m)^{2}}$ (6)

$F_{jupiter-venus}=1.37(10)^{18} N$ (7)

<h3>7) Gravitational force between Saturn and Mars</h3>

Using the same assumption we made in the prior question:

distance between Sun and Saturn - distance between Sun and Mars=distance between Saturn and Mars= $R_{saturn-mars}$ (8)

$R_{saturn-mars}=1427(10)^{9} m - 227.9(10)^{9} m=227.9(10)^{9} m$ (9)

Using this value in the Law of Universal Gravitation equation:

$F_{saturn-mars}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.989(10)^{30} kg)(6.42(10)^{23} kg)}{(227.9(10)^{9} m)^{2}}$ (10)

$F_{saturn-mars}=1.64(10)^{21} N$ (11)

<h3>8) How much more is earths force of gravity on Pluto than Venus force of gravity on Pluto?</h3>

Firstly, we need to find $F_{earth-pluto}$ and then find $F_{venus-pluto}$ in order to find the difference.

<u>For $F_{earth-pluto}$ :</u>

$M=5.97(10)^{24} kg$ is the mass of the Earth

$m=1.46(10)^{22} kg$ is the mass of Pluto

$R_{earth-pluto}=5.7504(10)^{12} m$ is the distance between Earth and Pluto

$F_{earth-pluto}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(5.97(10)^{24} kg)(1.46(10)^{22} kg)}{(5.7504(10)^{12} m)^{2}}$ (12)

$F_{earth-pluto}=1.759(10)^{11} N$ (13) Force between Earth and Pluto

<u></u>

<u>For $F_{venus-pluto}$ :</u>

$M=4.87(10)^{24} kg$ is the mass of Venus

$m=1.46(10)^{22} kg$ is the mass of Pluto

$R_{venus-pluto}=5.792(10)^{12} m$ is the distance between Venus and Pluto

$F_{venus-pluto}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(4.87(10)^{24} kg)1.46(10)^{22} kg)}{(5.792(10)^{12} m)^{2}}$ (14)

$F_{venus-pluto}=1.33(10)^{11} N$ (15) Force between Venus and Pluto

Calculating the difference:

$F_{earth-pluto}-F_{venus-pluto}=1.759(10)^{11} N-1.33(10)^{11} N$

Finally:

$F_{earth-pluto}-F_{venus-pluto}=4.29(10)^{10} N$ (16)

Hence:

Earths force of gravity on Pluto is $4.29(10)^{10} N$ than Venus force of gravity on Pluto.

3 0

3 years ago

A camcorder has a power rating of 13 watts. If the output voltage from its battery is 6 volts, what current does it use?

leva [86]

Power = (voltage) x (current)

13 watts = (6 volts) x (current)

Divide each side by (6 volts):

Current = (13 watts) / (6 volts)

Current = (13/6) Amperes

<em>Current = 2.17 Amperes</em>

5 0

4 years ago

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and the bottom of a b

erma4kov [3.2K]

Answer:

$h = 269.6 m$

Explanation:

Pressure at the bottom of the building and at the top of the building must be related as

$P_{top} = P_{bottom} - \rho g h$

$P_{top} = 730 mm Hg$

$P_{bottom} = 755 mm Hg$

now we will have

$(755 \times 10^{-3})(13.6 \times 10^3)(9.81) = P_{bottom}$

$P_{bottom} = 1.007 \times 10^5 Pa$

$P_{top} = (730\times 10^{-3})(13.6 \times 10^3)(9.81)$

$P_{top} = 0.974 \times 10^5$

now we have

$(1.007 - 0.974)\times 10^5 = 1.25 (9.81) h$

$h = 269.6 m$

7 0

3 years ago

Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too gr

IgorLugansk [536]

(a) 273.9 V

The power rating of the resistor is given by

$P=\frac{V^2}{R}$

where

P is the power rating

V is the potential difference across the resistor

R is the resistance

If the maximum power rating is $P=5.0 W$ , and the resistance of the resistor is $R=15 k\Omega = 15000 \Omega$ , then we can find the maximum potential difference across the resistor by re-arranging the previous equation for V: