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3 years ago

5

If a car is accelerating downhill under a net force 3674 N , what additional force would cause the car to have a constant veloci

ty

1 answer:

Semenov [28]3 years ago

6 0

Answer:

For the car to move with constant velocity the additional force required is $F__{dg} }= -3674 \ N$

Explanation:

From the question we are told that

The net force of the car is $F_{net} = 3674 \ N$

Generally the total force acting on the car is the net force plus the force due to gravity acting in direction of the car (Let denote it as $F__{dg}$ )

So the total force acting on the car is mathematically represented as

$F_{net} + F__{dg}} = F$

Here this F representing the total force can be mathematically represented as

$F = ma$

Now for constant velocity to be attained, the acceleration of the car will be zero

So at constant velocity

$F = m * 0$

=> $F = 0 \ N$

So

$F_{net} + F__{dg}} = 0$

=> $F__{dg} }= -F_{net}$

=> $F__{dg} }= -3674 \ N$

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7. Which of the following has the most total mechanical energy?

maria [59]

Answer:

An airplane flying

Explanation:

Both a bird and airplane fly, but the airplane is using more mechanical energy to run and power the plane, causing it to fly.

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3 years ago

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the

Rama09 [41]

1) $1.11\cdot 10^{-7} J$

The capacitance of a parallel-plate capacitor is given by:

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

$r=\frac{2.0 cm}{2}=1.0 cm=0.01 m$

so the area is

$A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2$

While the separation between the plates is

$d=0.50 mm=5\cdot 10^{-4} m$

So the capacitance is

$C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F$

And now we can find the energy stored,which is given by:

$U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J$

2) 0.71 J/m^3

The magnitude of the electric field is given by

$E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m$

and the energy density of the electric field is given by

$u=\frac{1}{2}\epsilon_0 E^2$

and using

$E=4\cdot 10^5 V/m$ , we find

$u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3$

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4 years ago

High angular momentum leads to faster rotation. why does faster rotation tend to lead to a spiral galaxy, rather than an ellipti

Lesechka [4]

High angular momentum leads to faster rotation, because faster rotation leads to collisions among particles that cause the gas to settle into a spiral disk, rather than a more spread out cloud.

What is the difference between an elliptical galaxy and a spiral galaxy?

Elliptical galaxies are ellipsoids with no clearly visible internal structure. Spiral galaxies have a very dense nucleus and a region of stars bulging outwards from the disks and, therefore, called the central bulge. Elliptical galaxies also have dense centers, but they do not protrude from the body of the galaxy.

Galaxies rotate, or spin around a central axis, and because of something called density waves, helps galaxies to become spiral in shape.

To learn more about Spiral galaxy here

brainly.com/question/27697600

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2 years ago

According to Lenz's law, an induced current in the direction depicted in the picture would be created when: a. An external downw

LenKa [72]

Answer:

C. An external downward field is created or an external downward field is removed

Explanation:

As we can see that from the attached figure that the induced current would be counter clockwise. So the field occur because of induced current i.e. out of page. This represent that the current is induced in order to rise the flux out of the direction of the page

Therefore because of the external field, the field out of page & flux would be reducing or the external upward field is eliminated

So option C is correct

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3 years ago

The lens of a telescope has a diameter of 25 cm. You are using it to look at two stars that are 2 × 1017 m away from you and 6 ×

vagabundo [1.1K]

Answer:

a)It is NOT possible using this telescope, to see the two stars as separate stars

b) $d_{min} =28.466m$

Explanation:

From the question we are told that:

Diameter of lens, $d = 25 cm \approx 0.25 m$

Distance from both star $D_f= 2*10^{17} m$

Distance between both stars $D_b= 6*10^9 m$

Wavelength of light $\lambda =700 nm \approx 700*10^-9 m$

Generally the equation for angle subtended by the two stars at the lens is mathematically given by

$\theta=\frac{D_f}{D_b}$

$\theta=\frac{6*10^9}{2*10^{17}}$

$\theta=3.0*10^{-8} rad$

Generally the equation for minimum angular separation of two object is mathematically given by

$\theta_{min} = 1.22*\lambda/d$

$\theta_{min}= \frac{1.22*700*10^-9}{0.25}$

$\theta_{min}= 3.416*10^-^6 rad$

Therefore

$\theta < \theta_{min}$

$3.0*10^{-8} rad< 3.416*10^-^6 rad$

It is NOT possible using this telescope, to see the two stars as separate stars

b)

Generally the equation for minimum diameter of the lens is mathematically given by

$d_{min} =\frac{ 1.22*\lambda}{\theta}$

$d_{min} =\frac{ 1.22*700*10^{-9}}{3*10^{-8}}$

$d_{min} =28.466m$

3 0

3 years ago