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lozanna [386]
3 years ago
14

Please explain the process and equation to get the answer

Physics
1 answer:
Juliette [100K]3 years ago
8 0

The shell's horizontal position x and vertical position y are given by

x=v_0\cos55.0^\circ\,t

y=v_0\sin55.0^\circ\,t-\dfrac g2t^2

where v_0 is the given speed of 1.70 x 10^4 m/s, and g is the acceleration due to gravity (taken here to be 9.80 m/s^2).

To find the horizontal range, you can use the range formula,

x_{\rm max}=\dfrac{{v_0}^2\sin2\theta}g

with \theta being the angle at which the shell is fired.

Alternatively, we can work backwards and deal with part (b) first:

(b) The time spent in the air is the time it takes for the shell to reach the ground. To find that, you solve for t in y=0:

v_0\sin55.0^\circ\,t-\dfrac g2t^2=0\implies t\approx284\,\rm s

(a) After this time, the shell will have traveled horizontally

x=v_0\cos55.0^\circ(284\,\rm s)\approx2.77\times10^5\,\rm m

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Strong nuclear force Gravity.

Explanation:

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Gravity is considered the weak force out of the four fundamental forces because it has a very small strength of 10^40 than electromagnetic force.

7 0
3 years ago
The mean distance of an asteroid from the Sun is 2.98 times that of Earth from the Sun. From Kepler's law of periods, calculate
lutik1710 [3]

Answer:

The asteroid requires 5.14 years to make one revolution around the Sun.

Explanation:

Kepler's third law establishes that the square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit:

T^{2} = a^{3} (1)

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

In the other hand, the distance between the Earth and the Sun has a value of 1.50x10^{8} Km. That value can be known as well as an astronomical unit (1AU).

But 1 year is equivalent to 1 AU according with Kepler's third law, since 1 year is the orbital period of the Earth.

For the special case of the asteroid the distance will be:

a = 2.98(1.50x10^{8}Km)

a = 4.47x10^{8}Km

That distance will be expressed in terms of astronomical units:

4.47x10^{8}Km.\frac{1AU}{1.50x10^{8}Km} ⇒ 2.98AU

Finally, from equation 1 the period T can be isolated:

T = \sqrt{a^{3}}

T = \sqrt{(2.98)^{3}}  

T = \sqrt{26.463592}

T = 5.14AU

Then, the period can be expressed in years:

5.14AU.\frac{1yr}{1AU} ⇒ 5.14 yr

T = 5.14 yr

Hence, the asteroid requires 5.14 years to make one revolution around the Sun.

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A corvette starts from rest and travels 69.0 meters in 50 s. What is its acceleration?
devlian [24]

Answer:

0.0552 m/s²

Explanation:

Given:

Δx = 69.0 m

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