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3 years ago

Score:

Physics

2 answers:

iren2701 [21]3 years ago

7 0

Answer:

Kepler's 1st law of planetary motion is called the law of ELLIPSES.

Explanation:

Kepler's 1st law of planetary motion can be referred to as the law of ellipses.

An ellipse is a form of an oval shape.

Kepler's 1st law of planetary motion states that every planet moves or rotates around the sun in a ellipses or elliptical orbit.

It is important to know that the sun is the focus or focal point of this ellipse.

Kepler's 1st law of planetary motion also serve as an indicator to show us s that when planet Earth rotates round the sun in an elliptical orbit, the distance between the Earth and Sun changes.

mylen [45]3 years ago

3 0

Kepler's 1st law is the law of ellipses

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A car is traveling at 96km/hr. what is the acceleration of a car traveling a distance of 100m and come to rest?

8090 [49]

Answer:

Explanation:

v² = u² + 2as

v = 0

u = 96 / 3.6 = 26.7 m/s

0² = 26.7² + 2a100

a = -3.5555555... ≈ -3.6 m/s²

the negative sign indicated the acceleration vector opposes the (assumed positive) initial velocity vector direction.

6 0

3 years ago

The electric potential difference across the membrane of a body cell is 0.070 V (higher on the outside than on the inside). The

12345 [234]

Answer:

The electric field is $8.75 \times 10^{6}~v~m^{-1}$ and the ditection is from outer to inner side of the membrane.

Explanation:

We know the electric field ( $\vec{E}$ ) is given by $\vec{E} = - \nabla V$ , 'V' being the potential.

In 1-D, it can be written as

$E=\dfrac{V}{d}$

where 'd' is the separation of space in between the potential difference is created.

Given, $V = 0.070~V~$ and the thickness of the cell membrane is $d = 8.0 \times 10^{-9}~m$ .

Therefore the created electric field through the cell membrane is

$E = \dfrac{0.07~V}{8 \times 10^{-9}~m} = 8.75 \times 10^{6}~m~s^{-1}$

5 0

3 years ago

You observe that a mass suspended by a spring takes 0.25 s to make a full oscillation. What is the frequency of this oscillation

Katarina [22]

Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, $\omega = 25.13 rad/s$

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

f = $\frac{1}{T}$

f = $\frac{1}{0.25}$

Frequency of oscillation, f = 4 Hz

The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

Therefore, time period, T = 0.25 s

Angular frequency of oscillation is given by:

$\omega = 2\pi \times f$

$\omega = 2\pi \times 4$

$\omega = 25.13 rad/s$

5 0

3 years ago

Trevor places a mass on a spring. When he releases it, the mass and spring move in simple harmonic motion.

777dan777 [17]

The correct answer is D He could pull the mass down farther.

Amplitude, is the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. So to increase the amplitude the spring should be pulled down further which can increase the amplitude.

8 0

3 years ago

Read 2 more answers

In a nuclear fusion reaction, the mass of the products is _____ the mass of the reactants.

JulijaS [17]

Answer:

more than

Explanation:

In a nuclear fusion reaction, the mass of the products is more than the mass of the reactants.

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3 years ago