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3 years ago

Pls help me this is being timed.

Physics

1 answer:

GrogVix [38]3 years ago

5 0

Answer:

Increase air resistance

Explanation:

Gravity forces the parachute down but air resistance pushing up on the flat surface of the parachute causes it to fall slower to the ground.

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Image caught on Screen is called

bixtya [17]

Answer:

<h2>Virtual image</h2>

Explanation:

<h3>Virtual image can be caught on a screen</h3>

hope this helps you.

will give the brainliest!

follow ~Hi1315~

4 0

3 years ago

Read 2 more answers

A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. T

Nina [5.8K]

Answer:

a)5.88J

b)-5.88J

c)0.78m

d)0.24m

Explanation:

a) W by the block on spring is given by

W= $\frac{1}{2}$ kx² = $\frac{1}{2}$ (530)(0.149)² = 5.88 J

b) Workdone by the spring = - Workdone by the block = -5.88J

c) Taking x = 0 at the contact point we have U top = U bottom

So, mg $h_o$ = $\frac{1}{2}$ kx² - mgx

And, $h_o$ = ( $\frac{1}{2}$ kx² - mgx )/(mg) = $[\frac{1}{2} (530)(0.149^2)-(0.645)(9.8)(0.149)$ ]/(0.645x9.8)

$h_o$ = 0.78m

d) Now, if the initial initial height of block is 3 $h_o$

$h_o$ = 3 x 0.78 = 2.34m

then, $\frac{1}{2}$ kx² - mgx - mg $h_o$ =0

$\frac{1}{2}$ (530)x² - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0

265x² - 6.321x - 14.8 = 0

a=265

b=-6.321

c=-14.8

By using quadratic eq. formula, we'll have the roots

x= 0.24 or x=-0.225

Considering only positive root:

x= 0.24m (maximum compression of the spring)

4 0

3 years ago

The two blocks in oscillate on a frictionless surface with a period of 1.5 s. The upper block just begins to slip when the ampli

Setler79 [48]

Answer:

0.72

Explanation:

$T$ = Time period of oscillation = 1.5 s

Angular frequency is given as

$w = \frac{2\pi }{T}\\w = \frac{2(3.14) }{1.5}\\w = 4.2 rad/s$

$A$ = Amplitude of oscillation = 40 cm = 0.40 m

$\mu$ = Coefficient of static friction = ?

$a$ = acceleration of the block

$m$ = mass of the block

Maximum acceleration of the block is given as

$a = Aw^{2}$

frictional force is given as

$f = \mu mg$

As per newton's second law

$f = ma \\\\\mu mg = ma \\\mu g = a\\\mu g = Aw^{2}\\\mu (9.8) = (0.40)(4.2)^{2}\\\mu = 0.72$

8 0

3 years ago

What is the center of gravity

Gala2k [10]

The center of gravity is a SIKKE bro you really thought I’d give the answer of something so simple

7 0

4 years ago

Read 2 more answers

How does the orbital speed of an asteroid in a circular solar orbit with a radius of 4.0 AU compare to a circular solar orbit wi

Murrr4er [49]

Answer:

The circular solar orbital speed at 4.0AU is 1/4( one fourth) that at 1.0AU

Explanation:

am = mvr= angular momentum

am4= 4mvt

am1= mvp1

Vt=1/4vp

Vp=4vt

am1= 4mvt

am1=am4

The circular solar orbital speed at 4.0AU is 1/4 (one fourth) that at 1.0AU

6 0

4 years ago