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3 years ago

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Newton’s first law says that a body will remain in a state of rest, or move in a straight line at a constant speed, unless it is

acted upon by an outside force. What is acting upon the Hubble Space Telescope (HST) to keep it in the curved path of its orbit around the Earth?

1 answer:

dangina [55]3 years ago

7 0

Answer:

Gravity

Explanation:

The Hubble is continually attracted to the earth due to the action of gravity. Therefore, it is thanks to gravity that the space telescope is kept in orbit. Without him, the direct motion generated by inertia would take him out of course. Gravity slows it down and keeps it in the curved path of its orbit around the Earth.

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Identify the forces acting on motor vehicle in straight line motion on a horizontal surface

tigry1 [53]

Friction, engine thrust, normal reaction and weight, these are the forces acting on motor when it moves in a straight line

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3 years ago

In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x=

swat32

In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x=20.0 cm to x=60.0cm.

<h3>Where is the electric potential, when the particle moved?</h3>

The charge field system's electric potential energy rose. The particle experiences an electric force that is directed against the x-axis. It is pushed uphill by an outside force, which raises the potential energy.

When a charge to be moved against an applied electric field, electric potential energy is needed. A charge must be moved through a stronger electric field with more energy than it would require to carry it via a weaker electric field.

In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x=20.0 cm to x=60.0cm.

The electric potential energy of the charge field system:

(a) increase
(b) remain constant
(c) decrease
(d) change unpredictably

The correct option is a).

To learn more about electric potential, refer to:

brainly.com/question/21808222

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2 years ago

What is the wavelength (in nm) of radiation that has an energy content of 1.0x10^2 kJ/mol?

hammer [34]

<span>Energy = h nu, where nu is the frequency
h = 6.63 x 10^-34 J-s, Planck's constant
So nu = E/h = 1 x 10^5 J /h = 0.15 x 10^29 / s
nu lambda = c, the speed of light.
lambda = wavelength = c / nu =3 x 10^8 / 0.15 x 10^29 = 20 x 10^-21 m.

this can possibly be a gamma ray. Gamma rays are very penetrating. It's both matter and an energy. They are electromagnetic radiation that results from a radioactive material.

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7 0

3 years ago

Read 2 more answers

The Sun delivers an average power of 1150 W/m2 to the top of the Earth’s atmosphere. The permeability of free space is 4π × 10−7

My name is Ann [436]

Answer:

E=930.84 N/C

Explanation:

Given that

I = 1150 W/m²

μ = 4Π x 10⁻⁷

C = 2.999 x 10⁸ m/s

E= C B

C=speed of light

B=Magnetic filed ,E=Electric filed

Power P = I A

A=Area=4πr² ,I=Intensity

$I=\dfrac{CB^2}{2\mu_0}$

$I=\dfrac{CE^2}{2\mu_0 C^2}$

$E=\sqrt{{2I\mu_0 C}}$

$E=\sqrt{{2\times 1150\times 4\pi \times 10^{-7}(2.99792\times 10^8)}}$

E=930.84 N/C

Therefore answer is 930.84 N/C

4 0

3 years ago

A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the sa

aleksandr82 [10.1K]

Answer: The value of the celsius temperature of the cube is 472.2°c.

Explanation:

The expression for the power radiated is as follows;

$P=A\epsilon\sigma T^{4}$

Here, A is the area, $\sigma$ is the stefan's constant, $\epsilon$ is the emissivity and T is the temperature.

It is given in the problem that A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the same emissivity as the sphere.

Then the expression for the radiated power for the cube and the sphere can be expressed as;

$A_{1}\epsilon \e\sigma T_{1}^{4}=A_{2}\epsilon \e\sigma T_{2}^{4}$

Here, $A_{1}$ is the area of the sphere, $A_{2}$ is the area of the cube, $T_{1}$ is the temperature of the sphere and $T_{2}$ is the temperature of the cube.

The radiated powers and emissivity of the cube and the sphere are same.

$A_{1}T_{1}^{4}=A_{2}T_{2}^{4}$

The area of the sphere is $A_{1}=4\pi \times r^{2}$ .

Here, r is the radius of the sphere.

The area of the cube is $A_{2}=6\times a^{2}$ .

Here, a is the edge of the cube.

Put $A_{1}=4\pi \times r^{2}$ and $A_{2}=6\times a^{2}$ .

$T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}$ ....(1)

The masses and the densities of the sphere and the cube are same. Then the volumes are also same.

$V_{1}=V_{2}$

Here, $V_{1},V_{1}$ are the volumes of the sphere and the cube.

$\frac{4}{3}\pi r^{3}=a^{3}$

$\frac{r}{a}=(\frac{3}{4\pi })^{\frac{1}{3}}$

Put this value in the equation (1).

$T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}$ $T_{2}=T_{1}(\frac{2\pi }{3}\times ((\frac{3}{4\pi })^{\frac{1}{3}})^{2})^{\frac{1}{4}}$

Put T_{1}=500°c.

$T_{2}=(500)(\frac{2\pi }{3}\times (\frac{3}{4\pi })^{\frac{2}{3}})^{\frac{1}{4}}$

$T_{2}=472.2^{\circ}c$

Therefore, the value of the celsius temperature of the cube is 472.7°c.

5 0

3 years ago