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2 years ago

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The gravitational field of m1 is denoted by g1. Enter an expression for the gravitational field g1 at position la in terms of m1

, la, and the gravitational constant G.

1 answer:

Andrei [34K]2 years ago

4 0

Answer:

The expression of gravitational field due to mass $m_!$ at a distance $l_a$

Explanation:

We have given mass is $m_1$

Distance of the point where we have to find the gravitational field is $l_a$

Gravitational constant G

We have to find the gravitational filed

Gravitational field is given by $g=\frac{Gm_1}{l_a^2}$

This will be the expression of gravitational field due to mass $m_!$ at a distance $l_a$

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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

A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio

dlinn [17]

Answer:

a = 4.72 m/s²

Explanation:

given,

mass of the box (m)= 6 Kg

angle of inclination (θ) = 39°

coefficient of kinetic friction (μ) = 0.19

magnitude of acceleration = ?

box is sliding downward so,

F - f = m a

f is the friction force

m g sinθ - μ N = ma

m g sinθ - μ m g cos θ = ma

a = g sinθ - μ g cos θ

a = 9.8 x sin 39° - 0.19 x 9.8 x cos 39°

a = 4.72 m/s²

the magnitude of acceleration of the box down the slope is a = 4.72 m/s²

3 0

2 years ago

Calculate the current through a 10.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V

Kipish [7]

Answer:

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

Explanation:

Given:

Length = l = 10 meter

$Radius = r = 0.321\ mm =0.321\times 10^{-3}\ meter$

$Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter$

V = 12 Volt

To Find:

Current, I =?

Solution:

Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

$R=\dfrac{\rho\times l}{A}$

Where,

R = Resistance

l = length

A = Area of cross section = πr²

$\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter$

Substituting the values we get

$R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}$

$R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}$

$R=\dfrac{1\times 10^{2}}{3.23}$

$R=30.95\ ohm$

Now by Ohm's Law,

$V= I\times R$

Substituting the values we get

$I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere$

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

4 0

2 years ago

The sun transfers energy to the earth by radiation at a rate of approximately 1.00 kW per square meter of surface.

Mashutka [201]

Answer:

1320336992.2512 m²

1320.33 kilometers or 509.79 miles

Explanation:

Energy transferred by the sun

$W=0.24\times 1\times 10^3=240\ W/m^2$

Energy required by the United States is $1\times 10^{19}\ J/yr$ (assumed)

Power

$P=\frac{E}{t}\\\Rightarrow P=\frac{1\times 10^{19}}{365.25\times 24\times 3600}\\\Rightarrow P=316880878140.2895\ W$

Area

$A=\frac{P}{W}\\\Rightarrow A=\frac{316880878140.2895}{240}\\\Rightarrow A=132033699.2512\ m^2$

Area of the solar collector would be 1320336992.2512 m²

Converting to km²

$1\ m^2=\frac{1}{1000\times 1000}\ km^2$

$1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1000\times 1000}\ km^2=1320.33\ km^2$

Converting to mi²

$1\ m^2=\frac{1}{1609.34\times 1609.34}\ mi^2$

$1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1609.34\times 1609.34}\ mi^2=509.79\ mi^2$

Each side of the square would be 1320.33 kilometers or 509.79 miles

4 0

2 years ago

This means that the speed at which the bullet travels across Earth's surface (its magnitude of horizontal velocity) does not aff

Dmitry_Shevchenko [17]

Answer: the speed at which it falls toward the Earth.

Explanation:

A bullet travelling across Earth's surface with some horizontal velocity is classical example of projectile motion.

Projectile motion is an idealization of the motion under the action of gravity neglecting the influence of the air (no drag force nor friction).

This kind of motion is the result of two independent motions: vertical motion and horizontal motion.

The observed net velocity is the vectorial sum of the vertical and horizontal velocities.

The horizontal velocity is constant, since there is not any force acting in the horizontal axis. Thi is, the object, following the first Law of Newton (inertia law) tends to continue in uniform rectilinear movement (with zero acceleration).

The vertical velocity, this is the velocity at which the bullet falls toward the Earth, is influenced (accelerated) by the action of the gravity of the Earth. So, the vertical velocity is accelerated by the pull of the Earth.

Vertical and horizontal velocities are independent of each other, which means that the speed or the magnitude of the horizontal velocity does not affect the speed at which an object (the bullet) falls toward the Earth.

6 0

2 years ago