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2 years ago

5

What is the distance from the crest to the equilibrium of a wave called? A. amplitude B. period C. frequency D. phase

2 answers:

Goryan [66]2 years ago

8 0

its the first one. A.

saw5 [17]2 years ago

7 0

Your answer is A. Amplitude

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Newton's First and Second Laws of Motion:Question 3Kepler-10b, the first confirmed terrestrial extrasolar planet, is about 564 l

OleMash [197]

Answer:

F = 85696.5 N = 85.69 KN

Explanation:

In this scenario, we apply Newton's Second Law:

$Unbalanced\ Force = ma\\Upthrust - Weight\ of\ Space\ Craft = ma\\F - W = ma\\F - mg = ma\\F = m(g + a)\\$

where,

F = Upthrust = ?

m = mass of space craft = 5000 kg

g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)

g = 15.0093 m/s²

a = acceleration required = 2.13 m/s²

Therefore,

$F = (5000\ kg)(15.0093\ m/s^{2} + 2.13\ m/s^{2})\\$

<u>F = 85696.5 N = 85.69 KN</u>

8 0

2 years ago

A solution is prepared by dissolving 49.3 g of KBr in enough water to form 473 mL of solution. Calculate the mass % of KBr in th

son4ous [18]

Answer:

9.31%

Explanation:

We are given that

Mass of KBr=49.3 g

Volume of solution=473 mL

Density of solution =1.12g/mL

We have to find the mass% of KBr.

Mass = $volume\times density$

Using the formula

Mass of solution= $1.12\times 473=529.76 g$

Mass % of KBr= $\frac{mass\;of\;KBr}{Total\;mass\;of\;solution}\times 100$

Mass % of KBr= $\frac{49.3}{529.76}\times 100$

Mass % of KBr=9.31%

Hence, the mass% of KBr=9.31%

7 0

2 years ago

A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

Explanation:

$m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg$

$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

3 0

3 years ago

How much heat is evolved when 1234 g of water condenses to a liquid at 100.°c?

lbvjy [14]

During the phase transition vapour --> liquid water, the temperature of the water does not change; the molecules of water release heat and the amounf of heat released is equal to
$Q=mL_e$
where
m is the mass of the water
$L_e$ is the latent heat of evaporation.

For water, the latent heat of evaporation is $L=2257 kJ/kg$ , while the mass of the water is
$m=1234 g=1.234 kg$

so, the amount of heat released in the process is
$Q=mL_e = (1.234 kg)(2257 kJ/kg)=2785 kJ$

6 0

3 years ago

An electric field of 3x10^16 n/c is needed to create a spark in the air. If the charged particles in the field are separated by

meriva

The answer of this problem is positive 3

6 0

3 years ago