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3 years ago

12

A stone is thrown vertically straight up with a velocity of 2800cm/s . By neglecting the air resistance , find the maximum heigh

t achieved

1 answer:

yarga [219]3 years ago

6 0

We know that t=2.8571428571428568
and therefore 28*2.8571428571428568 - 9.8*2.8571428571428568^2/2
is around 40 m

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In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80m/

polet [3.4K]

V o - initial velocity
v = velocity at the maximum height,
v² = v o² - 2 g h
v = 0
0 = v o² - 2 g h
v o² = 2 g h = 2 · 9.80 · 0.460
v o² = 9.052
v o = √9.052 = 3.004197 m/s ≈ 3 m/s

8 0

3 years ago

Why pressure above the surface is greater than in the air

nexus9112 [7]

Answer:

At higher elevations, there are fewer air molecules above a given surface than a similar surface at lower levels. ... Since most of the atmosphere's molecules are held close to the earth's surface by the force of gravity, air pressure decreases rapidly at first, then more slowly at higher levels.

Explanation:

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3 years ago

A gray kangaroo can bound across level ground with each jump carrying it 9.1 m from the takeoff point. Typically the kangaroo le

Alona [7]

Answer:

u = 10.63 m/s

h = 1.10 m

Explanation:

For Take-off speed ..

by using the standard range equation we have

$R = u² sin2θ/g$

R = 9.1 m

θ = 26º,

Initial velocity = u

solving for u

$u² = \frac{Rg}{sin2\theta}$

$u^2 = \frac{9.1 x 9.80}{sin26}$

$u^2 = 113.17$

u = 10.63 m/s

for Max height

using the standard h(max) equation ..

$v^2 = (v_osin\theta)^2 -2gh$

$h =\frac{(v_o^2sin\theta)^2}{2g}$

$h = \frac{(113.17)(sin26)^2}{(2 x 9.80)}}$

h = 1.10 m

7 0

3 years ago

To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d

Nimfa-mama [501]

Complete Question

The compete question is shown on the first uploaded question

Answer:

The speed is $v = 350 \ m/s$

Explanation:

From the question we are told that

The distance of separation is d = 4.00 m

The distance of the listener to the center between the speakers is I = 5.00 m

The change in the distance of the speaker is by $k = 60 cm = 0.6 \ m$

The frequency of both speakers is $f = 700 \ Hz$

Generally the distance of the listener to the first speaker is mathematically represented as

$L_1 = \sqrt{l^2 + [\frac{d}{2} ]^2}$

$L_1 = \sqrt{5^2 + [\frac{4}{2} ]^2}$

$L_1 = 5.39 \ m$

Generally the distance of the listener to second speaker at its new position is

$L_2 = \sqrt{l^2 + [\frac{d}{2} ]^2 + k}$

$L_2 = \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}$

$L_2 = 5.64 \ m$

Generally the path difference between the speakers is mathematically represented as

$pD = L_2 - L_1 = \frac{n * \lambda}{2}$

Here $\lambda$ is the wavelength which is mathematically represented as

$\lambda = \frac{v}{f}$

=> $L_2 - L_1 = \frac{n * \frac{v}{f}}{2}$

=> $L_2 - L_1 = \frac{n * v}{2f}$

=> $L_2 - L_1 = \frac{n * v}{2f}$

Here n is the order of the maxima with value of n = 1 this because we are considering two adjacent waves

=> $5.64 - 5.39 = \frac{1 * v}{2*700}$

=> $v = 350 \ m/s$

7 0

3 years ago

You drop a rock off a bridge. The rock's height, h (in feet above the water), after t seconds is modeled by h = – 16 t2 + 541. W

marin [14]

<span>h ( t) = h(1 sec) = -16t^2 + 541

so h (2 sec) = -16*(2)^2 + 541 = -64 + 541 = <span>477 ft

Therefore, </span></span>the height of the rock after 2 seconds is 477 feet.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

5 0

3 years ago