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Anika [276]
3 years ago
9

A projectile is fired with an initial velocity of 450 feet per second at an angle of 70° with the horizontal.

Physics
1 answer:
pav-90 [236]3 years ago
5 0
<h2>After 26.28 seconds projectile returns 26.28 seconds.</h2>

Explanation:

Initial velocity = 450 ft/s = 137.16 m/s

Angle, θ = 70°

Consider the vertical motion of projectile,

When the projectile return to the ground we have

           Displacement, s = 0 m

           Acceleration, a = -9.81 m/s²

            Initial velocity, u = 137.16 x sin70 = 128.89 m/s

Substituting in s = ut + 0.5 at²

                s = ut + 0.5 at²

                0 = 128.89 x t + 0.5 x (-9.81) x t²

                t² - 26.28 t = 0

                t ( t- 26.28) = 0

               t = 0 s or t = 26.28 s

After 26.28 seconds projectile returns 26.28 seconds.

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The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
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Answer:

  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

3 0
3 years ago
A Meteorite Strikes On October 9, 1992, a 27-pound meteorite struck a car in Peekskill, NY, leaving a dent 22 cm deep in the tru
vagabundo [1.1K]

Answer:

Acceleration of the meteorite, a=-38409.09\ m/s^2

Explanation:

It is given that,

A Meteorite after striking struck a car, v = 0

Initial speed of the Meteorite, u = 130 m/s

Distance covered by Meteorite, s = 22 cm = 0.22 m

We need to find the magnitude of its deceleration. It can be calculated using the third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{-(130)^2}{2\times 0.22}

a=-38409.09\ m/s^2

So, the deceleration of the Meteorite is -38409.09\ m/s^2. Hence, this is the required solution.

7 0
2 years ago
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