Explain each stage of communication process

The purpose of this lab is to explore the various ways to calculate projectile velocity using horizontal, vertical and angle inf

kolezko [41]

Answer: A projectile is any object in which the only force is gravity

Explanation: Equations on how to calculate projectile velocity is stated below:

The initial velocity Vo being a vector quantity, has two componentsVox and Voy

V0x = V0 cos(θ)

V0y = V0 sin(θ)

The acceleration A is a also a vector with two components Axand Ay given

Ax = 0 and Ay = - g = - 9.8 m/s2

Along the x axis the acceleration is equal to 0 and therefore the velocity Vx is constant

Vx = Vocos(θ)

Along the y axis, the acceleration is uniform and equal to - g and the velocity at time t is g

Vy = Vo sin(θ) - g t

Along the x axis the velocity Vx is constant and therefore the component x of the displacement is

x = Vocos(θ) t

Along the y axis, the motion is of uniform acceleration and the y component of the displacement is

y = Vo sin(θ) t - (1/2) g t2

3 0

3 years ago

A 45.2 kg softball player slides across dirt with Uk=0.340. What is her acceleration?

alisha [4.7K]

Answer:

the ball didnt hit my face so

Explanation:

8 0

3 years ago

A child and sled with a combined mass of 50.0 kg slide down a frictionless slope. if the sled starts from rest and has a speed o

Furkat [3]

<span> Using conservation of energy

Potential Energy (Before) = Kinetic Energy (After)

mgh = 0.5mv^2

divide both sides by m

gh = 0.5v^2

h = (0.5V^2)/g

h = (0.5*2.2^2)/9.81

h = 0.25m

</span>

4 0

2 years ago

Read 2 more answers

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward

Vladimir79 [104]

Answer:

The answer is " $1.01 \times 10^{-13}$ "

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

$KE=U=\frac{kqq'}{r}\\\\$

Calculating the closest distance:

$\to r=\frac{kqq'}{KE}\\\\$

$=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\$

$=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}$

5 0

2 years ago

In a chemical reaction, molecules of hydrogen gas (H2) react with molecules of oxygen gas (O2) in a sealed reaction chamber to p

11Alexandr11 [23.1K]

Answer:

Option (B) is correct.

Explanation:

Given that the molecules of hydrogen gas ( $H_2$ ) react with molecules of oxygen gas ( $O_2$ ) in a sealed reaction chamber to produce water ( $H_2O$ ).