All categories

2 years ago

A stone is projected horizontally from the

Physics

1 answer:

Nadya [2.5K]2 years ago

6 0

Answer:

80 m tall

Explanation:

*** EDITED *** (I left out the '1/2' in the equation....D'Oh!)

It was descending 4 seconds ...because it landed 20 m away (at 5 m/s ...so 4 seconds)

displacement =<u> 1/2</u> a t^2

= 1/2 * 10 ( 4)^2 = 80 m tall

You might be interested in

Two bumper cars collide into each other and each car jolts backwards. which one of Newton's laws is this?

LekaFEV [45]

I think it would be Newton’s second law

7 0

3 years ago

Which property of the wave makes it-(C)

AlekseyPX

Answer:

low amplitude hope it will help you

4 0

3 years ago

Read 2 more answers

A low orbit satellite at 100 km altitude had a camera that resolved a car location to within 0.3 meter. Find the minimum diamete

NeX [460]

Answer:

Minimum diameter of the camera lens is 22.4 cm

The focal length of the camera's lens is 300cm

Explanation:

y = Resolve distance = 0.3 m

h = Height of satellite = 100 km

λ = Wavelength = 550 nm

Angular resolution

$tan\theta\approx \theta =\frac{y}{h}\\\Rightarrow \theta=\frac{0.3}{100\times 10^3}=3\times 10^{-6}$

From Rayleigh criteria

$sin\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{sin\theta}\\\Rightarrow D=1.22\frac{550\times 10^{-9}}{sin3\times 10^{-6}}=0.2236\ m=22.4\ cm$

Minimum diameter of the camera lens is 22.4 cm

Relation between resolvable feature, focal length and angular resolution

$d=f\Delta \theta\\\Rightarrow f=\frac{d}{\Delta \theta}\\\Rightarrow f=\frac{9\times 10^{-6}}{3\times 10^{-6}}=3\ m=300\ cm$

The focal length of the camera's lens is 300cm

3 0

3 years ago

If an object was accelerating at 10 m/s2, and a mass of 1 kg, what was size of the force acting on the object?

ryzh [129]

Answer:

10 N

Explanation:

$f = ma$

= 10m/s^2 * 1 kg

=10N

7 0

3 years ago

During a tennis serve, a racket is given an angular acceleration of magnitude 150 rad/s^2. At the top of the serve, the racket h

QveST [7]

Answer:

270 m/s²

Explanation:

Given:

α = 150 rad/s²

ω = 12.0 rad/s

r = 1.30 m

Find:

The acceleration will have two components: a radial component and a tangential component.

The tangential component is:

at = αr

at = (150 rad/s²)(1.30 m)

at = 195 m/s²

The radial component is:

ar = v² / r

ar = ω² r

ar = (12.0 rad/s)² (1.30 m)

ar = 187.2 m/s²

So the magnitude of the total acceleration is:

a² = at² + ar²

a² = (195 m/s²)² + (187.2 m/s²)²

a = 270 m/s²

3 0

4 years ago