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3 years ago

A sidereal day is the time it takes for

Physics

1 answer:

ch4aika [34]3 years ago

8 0

Answer:

B. the stars to come back to the same positions in the sky.

Explanation:

In fact, the solar day is equivalent to more than a rotation, because when the point has turned completely, it is not, as it should, in the same position with respect to the Sun.

The reason for this is that while performing the rotation, the Earth simultaneously moved following its orbit around the Sun.

When the reference point completed its rotation, the Earth already moved almost 2,500,000 km., So that to see the Sun again it will be necessary to turn a little more.

Solar day is more than a rotation. The sidereal or sidereal day, commonly used by astronomers, is also based on the rotation of the Earth; but in this case a distant star is taken as a reference (sidereal comes from the Latin sidus which means "star").

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A 50.0-kg block is being pulled up a 13.0° slope by a force of 250 N which is parallel to the slope, but the block does not slid

Elan Coil [88]

Weight of block, Wb = mass*gravity = 50*9.8 = 490 N

Since block is being pulled up by a 13-degree slope

Therefore, Force which is acting parallel to the slop:

F p =490 Sin $13^{0}$ = 110.2N

Force which is acting perpendicular to the slope:

Fv =490 Cos $13^{0}$ = 477.4 N

Net force can be given as follows:

F n = (250 - 110.2 - 0.2*477.4) N

Fn=44.3N

Now acceleration is given by the ratio of force to mass

a = Fn/m

=44.3/50 = 0.89 ms^-2

8 0

3 years ago

How long would it take light from the sun to reach mercury?

BARSIC [14]

About 499.0 seconds.....

7 0

4 years ago

A proton has a speed of 3.50 Ã 105 m/s when at a point where the potential is +100 V. Later, itâs at a point where the potential

ruslelena [56]

Answer:

(a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

Explanation:

Given that,

Speed $v= 3.50\times10^{5}\ m/s$

Initial potential V=100 V

Final potential = 150 V

(a). We need to calculate the change in the protons electric potential

Potential energy of the proton is

$U=qV=eV$

Using conservation of energy

$K_{i}+U_{i}=K_{f}+U_{f}$

$\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}$

$]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})$

$\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V$

$\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}$

Put the value into the formula

$\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}$

$\Delta V=639.2=0.639\ kV$

(b). We need to calculate the change in the potential energy of the proton

Using formula of potential energy

$\Delta U=q\Delta V$

Put the value into the formula

$\Delta U=1.6\times10^{-19}\times639.2$

$\Delta U=1.022\times10^{-16}\ J$

(c). We need to calculate the work done on the proton

Using formula of work done

$\Delta U=-W$

$W=q(V_{2}-V_{1})$

$W=-1.6\times10^{-19}(150-100)$

$W=-8\times10^{-18}\ J$

Hence, (a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

5 0

3 years ago

Young _____ stars would most likely be found in a giant molecular cloud.

marysya [2.9K]

The correct answer is letter A. red. Young red stars would most likely be found in a giant molecular cloud. Red stars are cooler than the sun, with surface temperatures of 3,500 K for a bright red star

6 0

3 years ago

Which quantity may be calculated directly using Newton's second law of motion?

xxMikexx [17]

Answer:

Weight

Explanation:

The second law of motion by Newton can be used to determine weight.

4 0

2 years ago