Is there a way to get sun lamps to work without replacing ballast?

(1 point) A projectile is fired from ground level with an initial speed of 600 m/sec and an angle of elevation of 30 degrees. Us

konstantin123 [22]

Answer:

(a) The range of the projectile is 31,813.18 m

(b) The maximum height of the projectile is 4,591.84 m

(c) The speed with which the projectile hits the ground is 670.82 m/s.

Explanation:

Given;

initial speed of the projectile, u = 600 m/s

angle of projection, θ = 30⁰

acceleration due to gravity, g = 9.8 m/s²

(a) The range of the projectile in meters;

$R = \frac{u^2sin \ 2\theta}{g} \\\\R = \frac{600^2 sin(2\times 30^0)}{9.8} \\\\R = \frac{600^2 sin (60^0)}{9.8} \\\\R = 31,813.18 \ m$

(b) The maximum height of the projectile in meters;

$H = \frac{u^2 sin^2\theta}{2g} \\\\H = \frac{600^2 (sin \ 30)^2}{2\times 9.8} \\\\H = \frac{600^2 (0.5)^2}{19.6} \\\\H = 4,591.84 \ m$

(c) The speed with which the projectile hits the ground is;

$v^2 = u^2 + 2gh\\\\v^2 = 600^2 + (2 \times 9.8)(4,591.84)\\\\v^2 = 360,000 + 90,000.064\\\\v = \sqrt{450,000.064} \\\\v = 670.82 \ m/s$

5 0

3 years ago

If you increase the frequency of a sound wave four times, what will happen to its speed?

ozzi

The speed will increase four (4) times when the frequency of sound wave is increased four (4) times.

<h3>How to calculate the frequency of a wave.</h3>

Mathematically, the frequency of a sound wave is given by this formula:

$F = \frac{V}{\lambda}$

<u>Where:</u>

F is the frequency of a sound wave.

V is the velocity of a sound wave.

$\lambda$ is the wavelength of a sound wave.

<u>Note:</u> The frequency of the sound wave is increased four (4) times.

F = 4F

Substituting the given parameters into the formula, we have;

$4F = \frac{V}{\lambda}\\\\V=4F\lambda$

Therefore, the speed will increase four (4) times when the frequency of sound wave is increased four (4) times.

Read more on sound wave here: brainly.com/question/19250621

4 0

3 years ago

•What power does a pump develop to lift 35 L of water per minute

guajiro [1.7K]

Answer:

$p = \frac{w}{t} \\ p = \frac{f \times d}{t} \\ p = \frac{m \times g \times d}{t} \\ p = \frac{35 \times 9.8 \times 110}{60.0} \\ p = 629w$

8 0

3 years ago

A 1000 kg train car traveling at 12 m/s on a level track has 400 kg of sand dumped, vertically, into it. What is the final speed

crimeas [40]

Answer:

The final speed of the train car is 8.57 m/s.

Explanation:

Given that,

Mass of train car = 1000 kg

Mass of sand = 400 kg

Speed of train car = 12 m/s

We need to calculate the final speed of the train car after the sand is dumped

Using conservation of momentum

$m_{1}u_{1}=(m_{1}+m_{2})v$ ...(I)

Put the value in the equation (I)

$1000\times12=(1000+400)v$

$v=\dfrac{1000\times12}{1400}$

$v=8.57\ m/s$

Hence, The final speed of the train car is 8.57 m/s.

5 0

3 years ago

What is malware short for?

yaroslaw [1]

The correct answer is C

8 0

3 years ago