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3 years ago

(1.08×10-3)(9.3×10-4)

1 answer:

4 0

First you do the first parenthesis, (1.08 x 10 - 3) and you do it in the order of operations! (parenthesis, exponents, multiplication/division, add/subtract) to get 7.8. Then you take the second parenthesis (9.3 x 10 - 4) and do the same thing to get 89! You then times 7.8 by 89 to get 694.2! If it needs more elaboration just ask ^.^

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A 2.1 kg block is dropped from rest from a height of 5.5 m above the top of the spring. When the block is momentarily at rest, t

ella [17]

Answer:

The speed of the block is 8.2 m/s

Explanation:

Given;

mass of block, m = 2.1 kg

height above the top of the spring, h = 5.5 m

First, we determine the spring constant based on the principle of conservation of potential energy

¹/₂Kx² = mg(h +x)

¹/₂K(0.25)² = 2.1 x 9.8(5.5 +0.25)

0.03125K = 118.335

K = 118.335 / 0.03125

K = 3786.72 N/m

Total energy stored in the block at rest is only potential energy given as:

E = U = mgh

U = 2.1 x 9.8 x 5.5 = 113.19 J

Work done in compressing the spring to 15.0 cm:

W = ¹/₂Kx² = ¹/₂ (3786.72)(0.15)² = 42.6 J

This is equal to elastic potential energy stored in the spring,

Then, kinetic energy of the spring is given as:

K.E = E - W

K.E = 113.19 J - 42.6 J

K.E = 70.59 J

To determine the speed of the block due to this energy:

KE = ¹/₂mv²

70.59 = ¹/₂ x 2.1 x v²

70.59 = 1.05v²

v² = 70.59 / 1.05

v² = 67.229

v = √67.229

v = 8.2 m/s

8 0

3 years ago

Read 2 more answers

!!!!HURRY 30 POINTS!!!!

Nataly_w [17]

The anwser is C flooding

3 0

3 years ago

PLS HELP IM SOO CONFUSED!

Feliz [49]

False.

As temperature increases the more the electrons begin to vibrate more, as it decreases they vibrate less.

7 0

2 years ago

Read 2 more answers

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

4 0

2 years ago

By the term universe astronomers mean

OverLord2011 [107]

'Universe' means 'Everything'. That is, all matter, all space, all time.

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3 years ago