All categories

3 years ago

I need help ASAPP

Physics

1 answer:

Tresset [83]3 years ago

6 0

Answer: For the first part the answer is Radio Waves and the second part’s answer is gamma ray.

You might be interested in

A gun has a muzzle speed of 150 m/s. Find two angles of elevation that can be used to hit a target 800m away.

lorasvet [3.4K]

Answer:

Explanation:

u = 150 m/s

R = 800 m

The formula for the horizontal range is

$R=\frac{u^{2}Sin2\theta }{g}$

where, θ is angle of projection.

$800=\frac{150^{2}Sin2\theta }{9.8}$

Sin2θ = 0.348

2θ = 20.4°

θ = 10.2°

The range is same for the complementary angles.

So, the other angle is 90 - 10.2 = 79.8°

4 0

2 years ago

Leverrier predicted that an invisible planet was pulling the planet Uranus off its predicted course around the sun. Likewise, hu

Alexandra [31]

Human beings are pulled off course as a result of the invisible forces of the <u>unconscious.</u>

According to Leverrier, he stated that an invisible planet was pulling the planet Uranus off its predicted course around the sun.

In such a way, human beings are pulled off course by the invisible forces of their unconscious minds. The unconscious minds of people control the thoughts of people.

Read related link on:

brainly.com/question/25588203

3 0

2 years ago

A person in a kayak starts paddling, and it accelerates from 0 to 0.61 m/s in a distance of 0.39 m. If the combined mass of the

Iteru [2.4K]

Answer:

35.3 N

Explanation:

U = 0, V = 0.61 m/s, s = 0.39 m

Let a be the acceleration.

Use third equation of motion

V^2 = u^2 + 2 as

0.61 × 0.61 = 0 + 2 × a × 0.39

a = 0.477 m/s^2

Force = mass × acceleration

F = 74 × 0.477 = 35.3 N

6 0

2 years ago

Explain why the angle is important when lauching projectiles

yaroslaw [1]

Answer:

to have an accurate measure

Explanation:

6 0

3 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago