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Harlamova29_29 [7]
2 years ago
12

Write the equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and

Gravitation.
Physics
1 answer:
yKpoI14uk [10]2 years ago
3 0

Answer:

The newton’s second law is F=ma

The Gravitational force is F=\dfrac{Gm_{1}m_{2}}{r^2}

Explanation:

Given that,

The equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and Gravitation.

We know that,

Velocity :

The velocity is equal to the rate of position of the object.

v=\dfrac{dx}{dt}....(I)

Acceleration :

The acceleration is equal to the rate of velocity of the object.

a=\dfrac{dv}{dt}....(II)

Newton’s second Laws

The force is equal to the change in momentum.

In mathematically,

F=\dfrac{d(p)}{dt}

Put the value of p

F=\dfrac{d(mv)}{dt}

F=m\dfrac{dv}{dt}

Put the value from equation (II)

F=ma

This is newton’s second laws.

Gravitational force :

The force is equal to the product of mass of objects and divided by square of distance.

In mathematically,

F=\dfrac{Gm_{1}m_{2}}{r^2}

Where, m₁₂ = mass of first object

m= mass of second object

r = distance between both objects

Hence, The newton’s second law is F=ma

The Gravitational force is F=\dfrac{Gm_{1}m_{2}}{r^2}

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sattari [20]

Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

We know Fc = \frac{m v^{2} }{r}

==> 1.1 × 10^{19} = (9.4 × 10^{21} × v^{2} ) ÷ 1.5 × 10^{8}

==> 1.1 × 10^{19} = v^{2} × 6.26×10^{13}

==> v^{2} =  1.1 × 10^{19} ÷ 6.26×10^{13}

==> v^{2} = 0.17571885 × 10^{6}

==> v= 0.419188323 × 10^{3} m/sec

==> v= 419.188322834 m/s

Putting value of r and v from above in ;

T= 2πr ÷ v

==> T= 2×3.14×1.5×10^{8} ÷ 0.419188323 × 10^{3}

==> T = 22.472× 100000 = 2247200 sec

but

86400 sec = 1 day

==> 2247200 sec= 2247200 ÷ 86400 = 26 days

3 0
2 years ago
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A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra
skelet666 [1.2K]

Answer:

So the acceleration of the child will be 8.05m/sec^2

Explanation:

We have given angular speed of the child \omega =1.25rad/sec

Radius r = 4.65 m

Angular acceleration \alpha =0.745rad/sec^2

We know that linear velocity is given by v=\omega r=1.25\times 4.65=5.815m/sec

We know that radial acceleration is given by a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2

Tangential acceleration is given by

a_t=\alpha r=0.745\times 4.65=3.464m/sec^

So total acceleration will be a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2

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The diagram shows a heater above a thermometer. The thermometer bulb is in the position shown. Which row shows how the heat ener
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Answer:

The diagram shows a heater above a thermometer. The thermometer bulb is in the position shown. How the heat

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Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy
BaLLatris [955]

Answer:

E) are almost circular, with low eccentricities.

Explanation:

Kepler's laws establish that:

All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).

A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

T^{2} = a^{3}

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)

In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).

Therefore, option A and B can not be true.

In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.  

4 0
2 years ago
A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnet
Mumz [18]
<h3>Question:</h3>

A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.

<h3>Answer:</h3>

1.6nT [in the negative z direction]

<h2>Explanation:</h2>

The magnetic field, B, due to a distance of finite value b, is given by;

B = (μ₀IL) / (4πb\sqrt{b^2 + L^2})                -----------(i)

Where;

I = current on the wire

L = length of the wire

μ₀ = magnetic constant = 4π × 10⁻⁷ H/m

From the question,

I = 20A

L = 2.0cm = 0.02m

b = 5.0m

Substitute the necessary values into equation (i)

B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0 \sqrt{5.0^2 + 0.02^2})

B = (10⁻⁷ x 20 x 0.02) / (5.0 \sqrt{5.0^2 + 0.02^2})

B = (10⁻⁷ x 20 x 0.02) / (5.0 \sqrt{25.0004})

B = (10⁻⁷ x 20 x 0.02) / (25.0)

B = 1.6 x 10⁻⁹T

B = 1.6nT

Therefore, the magnetic field at the point x = 5.0m  on the x-axis is 1.6nT.

PS: Since the current is directed in the positive y direction, from the right hand rule, the magnetic field is directed in the negative z-direction.

5 0
2 years ago
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