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4 years ago

The term pressure most nearly refers to which of the following?

2 answers:

5 0

The term pressure refers to the force per area created by the weight
of anything whose weight is distributed over an area, such as the
Earth's atmosphere, a lake, a gas inside a sealed jar, or a pointy
high heel.

jok3333 [9.3K]4 years ago

5 0

C. force per area created by the weight of water

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Now, let’s see what happens when the cannon is high above the ground. click on the wheel of the cannon, and drag it upward as fa

Setler79 [48]

The equation for range is:

R = v₀²sin(2θ)/g
To find the maximum R, differentiate the equation and equate to zero. The solution is as follows:

dR/dθ = (v₀²/g)(sin 2θ)
dR/dθ = (v₀²/g)(cos 2θ)(2) = 0
cos 2θ = 0
2θ = cos⁻¹ 0 = 90
θ = 90/2
<em>θ = 45°</em>

5 0

3 years ago

The gravitational force between two objects with masses, m1 and m2, separated by a distance "r" is given by F = (Gm1m2)/ r2, whe

diamong [38]

Answer:

it increases

Explanation:

4 0

3 years ago

A projectile is shot horizontally at 23.4 m/s from the roof of a building 55.0 m tall. (a) Determine the time necessary for the

jeka94

(a) 3.35 s

The time needed for the projectile to reach the ground depends only on the vertical motion of the projectile, which is a uniformly accelerated motion with constant acceleration

a = g = -9.8 m/s^2

towards the ground.

The initial height of the projectile is

h = 55.0 m

The vertical position of the projectile at time t is

$y = h + \frac{1}{2}at^2$

By requiring y = 0, we find the time t at which the projectile reaches the position y=0, which corresponds to the ground:

$0 = h + \frac{1}{2}at^2\\t=\sqrt{-\frac{2h}{a}}=\sqrt{-\frac{2(55.0 m)}{(-9.8 m/s^2)}}=3.35 s$

(b) 78.4 m

The distance travelled by the projectile from the base of the building to the point it lands depends only on the horizontal motion.

The horizontal motion is a uniform motion with constant velocity -

The horizontal velocity of the projectile is

$v_x = 23.4 m/s$

the time it takes the projectile to reach the ground is

t = 3.35 s

So, the horizontal distance covered by the projectile is

$d=v_x t = (23.4 m/s)(3.35 s)=78.4 m$

(c) 23.4 m/s, -32.8 m/s

The motion of the projectile consists of two independent motions:

- Along the horizontal direction, it is a uniform motion, so the horizontal velocity is always constant and it is equal to

$v_x = 23.4 m/s$

so this value is also the value of the horizontal velocity just before the projectile reaches the ground.

- Along the vertical direction, the motion is acceleration, so the vertical velocity is given by

$v_y = u_y +at$

where

$u_y = 0$ is the initial vertical velocity

Using

a = g = -9.8 m/s^2

and

t = 3.35 s

We find the vertical velocity of the projectile just before reaching the ground

$v_y = 0 + (-9.8 m/s^2)(3.35 s)=-32.8 m/s$

and the negative sign means it points downward.

3 0

3 years ago

Relate the changes in molecular motion of particles as hey change from solid to liquid to gas.

Drupady [299]

Answer:

as they go from solid to liquis to gas they heat up from thermal energy. that thermal energy then excites the molecules making them move faster causing them to change phases.

3 0

3 years ago

calcula el potencial electrico de un punto B que se encuentra a 75cm de una varga positica q=9×10-6c

ankoles [38]

Answer:

V = 72000 volts

Explanation:

calculates the electric potential of a point B that is 75cm from a positive charge q = 9 × 10-6c

Given that,

Electric charge, $q=9\times 10^{-6}\ C$

We need to find the electric potential of a point B that is 75 cm from this charge. The formula for the electric potential is given by :

$V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 6\times 10^{-6}}{0.75}\\\\V=72000\ V$

So, the electric potential is equal to 72000 Volts.

3 0

3 years ago