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3 years ago

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Which formula can be used to calculate the horizontal displacement not of a horizontally launched projectile

1 answer:

svetlana [45]3 years ago

8 0

<span>If you are looking to get an object up the highest, shoot it straight up. If you want to go for a specific horizontal displacement, use the range equation. R = v2sin(twice the launch angle)/ g. g is the gravitaional constant, 9.8 meters per second. Use degrees for the angle. v is the launch velocity. R is the horizontal displacement. This formula only works if your start altitude and end altitude are the same, i.e. you must shoot over a level field. it depends on the gravitational force of attraction of earth and air resistance. if we are neglecting air resistance, the max.horizontal distance is according to this formulae, V0/2 * sin (2theta) where V0 is the initial velocity theta is the angle with x axis and the projection. There are a number of ways that you could find a horizontally displaced object. You could for example just look.</span>

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3 years ago

A boat takes off from a dock at 2.5 m/s and speeds up at 4.2 m/s squared for six seconds how far has the most traveled

GaryK [48]

The boat's position $x$ relative to its starting point $x_0=0$ is determined by

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$x=\left(2.5\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(4.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2$

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3 years ago

How to represent milligram in kilogram by standard formula?

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Answer:

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Explanation:

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3 years ago

(a) How much work is required to lift a 35-kg object from the ground 3.0 m into the air? (b) How much gravitational potential en

V125BC [204]

Answer:

(a) work required to lift the object is 1029 J

(b) the gravitational potential energy gained by this object is 1029 J

Explanation:

Given;

mass of the object, m = 35 kg

height through which the object was lifted, h = 3 m

(a) work required to lift the object

W = F x d

W = (mg) x h

W = 35 x 9.8 x 3

W = 1029 J

(b) the gravitational potential energy gained by this object is calculated as;

ΔP.E = Pf - Pi

where;

Pi is the initial gravitational potential energy, at initial height (hi = 0)

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ΔP.E = 1029 J

7 0

3 years ago

Find the final temperature of 375 grams of tea (c = 4.184 J/g°C) if its initial temperature is 95°C just before it is placed in

Minchanka [31]

Answer:

the final temperature of the tea is 7.39⁰C.

Explanation:

Given;

mass of the tea, m = 375 g

specific heat capacity of the tea, C = 4.184 JJ/g°C

initial temperature of the tea, t₁ = 95°C

the final temperature of the tea, t₂ = ?

Energy lost by the refrigerator, Q = 137,460 J

The energy lost by the refrigerator is given by the following formula;

-Q = mc(t₂ - t₁)

-137,460 =375 x 4.184(t₂ - 95°C)

-137,460 = 1569(t₂ - 95°C)

$t_2-95 = \frac{-137,460}{1569} \\\\t_2-95 = -87.61\\\\t_2 = -87.61 + 95\\\\t_2 = 7.39 \ ^0 C$

Therefore, the final temperature of the tea is 7.39⁰C.

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3 years ago