Drag the right word to the word that means the opposite

The mass of the particles that a river can transport is proportional to the sixth power of the speed of the river. A certain riv

kow [346]

Answer:

1.122 m/s

Explanation:

So usually a river with a speed of 1 meters per second can transport particle that weighs:

$1^6 = 1 kg$

If the particle is twice as massive as usual, then its weights would be 1 * 2 = 2kg

This means the river must be flowing at a speed of

$2^{\frac{1}{6}} = 1.122 m/s$

5 0

3 years ago

A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a

Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C = 50 - x cm

We know that ,electric field is given as

$E=K\dfrac{q}{r^2}$

$K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\$

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

3 0

3 years ago

Ten high-technology batteries are tested for 200 hours each. One failed at 20 hours; another failed at 140 hours; all others com

Bas_tet [7]

Answer:

Failure rate = 20%

MTBF = 880 hours

Explanation:

given data

batteries = 10

tested = 200 hours

one failed = 20 hours

another fail at = 140 hours

solution

we know that Mean Time between Failures is express as = (Total up time) ÷ (number of breakdowns) ....................1

so here Total up time will be

Total up time = 200 × 10

Total up time = 2000

and here

Number of breakdown = 1 at 20 hour and another at 140 hour = 2

so it will be = (Total up time) ÷ (number of breakdowns) .......2

= $\frac{2000}{2}$ = 1000

so here gap between occurrences is

gap between occurrences= 140 - 20

gap between occurrences = 120 hour

and

MTBF will be

MTBF = 1000 - 120

MTBF = 880 hours

and

Failure rate (FR) will be

Failure rate (FR) = 1 ÷ MTBF ................3

Failure rate (FR) = R÷T ......................4

as here R is the number of failures and T is total time

so Failure rate (FR) = 20%

4 0

3 years ago

Problem #2: An apple is thrown upward with an initial velocity of +24.0 m/s. a. Sketch the apple's trip and label what you know.

bogdanovich [222]

Answer:

The answer is below

Explanation:

a) The initial velocity (u) = 24 m/s

We can solve this problem using the formula:

v² = u² - 2gh

where v = final velocity, g= acceleration due to gravity = 9.8 m/s², h = height.

At maximum height, the final velocity = 0 m/s

v² = u² - 2gh

0² = 24² - 2(9.8)h

2(9.8)h = 24²

2(9.8)h = 576

19.6h = 576

h = 29.4 m

b) The time taken to reach the maximum height is given as:

v = u - gt

0 = 24 - 9.8t

9.8t = 24

t = 2.45 s

The total time needed for the apple to return to its original position = 2t = 2 * 2.45 = 4.9 s

4 0

2 years ago

Use the definition of scalar product, a overscript right-arrow endscripts times b overscript right-arrow endscripts = ab cos θ,

makkiz [27]

Answer: $\theta=cos^{-1}0.991=7.69^o$

The following vectors have been given: $\vec{a}=3.0\widehat{i}+3.0\widehat{j}+3.0\widehat{k}\\ \vec{b}=5.0\widehat{i}+7.0\widehat{j}+6.0\widehat{k}$

The angle between these two vectors can be found by:

$cos\theta=\frac{\vec{a}.\vec{b}}{||\vec{a}|| ||\vec{b}||}\\
||\vec{a}=\sqrt{a_x^2+a_y^2+a_z^2}$

$\vec{a}.\vec{b}=a_xb_x+a_yb_y+a_zb_z\\ \vec{a}.\vec{b}=3\times5+3\times7+3\times6=15+21+18=54$

$||\vec{a}||=\sqrt{3^2+3^2+3^2}=\sqrt{27}\\ ||\vec{b}||=\sqrt{5^2+7^2+6^2}=\sqrt{110}$

$cos\theta=\frac{54}{\sqrt{27}\times\sqrt{110}}\\=0.991\\ \Rightarrow \theta=cos^{-1}0.991=7.69^o$

7 0

3 years ago