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3 years ago

Any help please, physics is not my strong suit?

Physics

2 answers:

frez [133]3 years ago

5 0

Answer:

2000mg = 2g

5L = 5000mL

16cm = 160mm

Mashcka [7]3 years ago

5 0

1) kg

2) m

3) g

4) mL

5) mm

6) L

7) km

8) cm

9) mg

10) 2g

11) 104000m

12) 4.8m

13) 5600g

14) 8cm

15) 5000mL

16) 0.198kg

17) 0.075L

18) 0.15m

19) 560cm

20) 16mm

21) 25km

22) 6500mg

23) 66mm

24) 0.12g

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During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

lord [1]

Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

Solution:

As per the question:

Mass of Earth, $M_{e} = 5.972\times 10^{24}\ kg$

Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

Distance between the earth and the sun, $R_{es} = 1.5\times 10^{11}\ m$

Distance between the sun and the moon, $R_{sm} = 1.5\times 10^{11}\ m$

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

$F_{se} = 3.521\times 10^{20}\ N$

7 0

3 years ago

Expressing Experimental Error If the accepted value of π is 3.1416, what are the fractional error and the percent error of the e

LUCKY_DIMON [66]

Answer:

1) Δx = 0.16 m / s² , Δx = 9.72-9.8 = 0.08 m / s²

2) e% = 1.63% , , e% = 0.816%

3) 1e% = 0.04

Explanation:

The fractional error in a quantity is the absolute error between the accepted value of the quantity and the percentage error is the fractional error per 100

In this case, it is not indicated which is the measured experimental value of pi, suppose that a value of 3.142 is measured

fractional error

e = (ax - ax_average) / x_average

e = (3.142 - 3.1416) / 3.1416

e = 1.27 10⁻⁴

the percentage error is

% = e 100

% = 1.27 10-4 100

% = 1.27 10⁻²%

1) Δx = 9.96-9.8 = 0.16 m / s²

Δx = 9.72-9.8 = 0.08 m / s²

2) the percentage difference

x = 9.96 m / s2

e% = (9.96 - 9.80) / 9.80 100

e% = 1.63%

x = 9.72 m / s2

e% = (9.72 -9.80) / 9.80 100

e% = 0.816%

3) the mean value is

x_average = (x1 + x2) / 2

x_average = (9.96 + 9.72) / 2

x_average = 9.84 m / s2

e% = (9.84 - 9.80) / 9.80 100

1e% = 0.04

7 0

3 years ago

A ball is revolving horizontally in a circle of radius 1.8 m, and is held by a rigid, massless rod. The mass of the ball is 0.1

OLga [1]

Answer:

$\omega=15 \ rad/s$

Explanation:

given,

radius of circle = 1.8 m

mass of the ball = 0.1 Kg

linear speed of the ball = 27 m/s

angular velocity of orbit = ?

$v = r \omega$

ω is the angular speed of the circle

$\omega=\dfrac{v}{r}$

$\omega=\dfrac{27}{1.8}$

$\omega=15 \ rad/s$

the angular velocity of the orbit is $\omega=15 \ rad/s$

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3 years ago

A prediction or educated guess that can be tested is ?

Ksivusya [100]

An educated guess is a “hypothesis”

7 0

3 years ago

What is the resistance of a conductor

Ivenika [448]

Opposition for the current flow

Explanation:

Resistance is defined as the ability pf the conductor to oppose the flow of electrons through it. It is usually represented in Ohm.
Resistance is an important parameter in a circuit design and it should be properly designed.
Resistance of a conductor depends upon the material nature, length of the conductor and area of cross section.
In conductor, Resistance is low. It is due to that charges flow through the conductor freely.

7 0

3 years ago