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Marysya12 [62]
3 years ago
14

Boat A and Boat B have the same mass. Boat A's velocity is three times greater than that of Boat B. Compared to

Physics
2 answers:
Zarrin [17]3 years ago
7 0

Answer:

nine times as much.

Explanation:

K.E of A = 9 times K.E of B

Nikitich [7]3 years ago
3 0

Answer:

D) nine times as much

Explanation:

its correct

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Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about 500 cm
mario62 [17]

Answer:

A) if each astronaut breathes about 500 cm³, the total volume of air breathed in a year is 14716.8m³.

B) The Diameter of this spherical space station should be 30.4m

Explanation:

The breathing frequency (according to Rochester encyclopedia) is about 12-16 breath per minute. if we take the mean value (14 breath per minute), we can estimate the total breaths of a person along a year:

f_b=14\frac{br}{min} \cdot \frac{60min}{1hr} \frac{24hr}{1day}\frac{365day}{1year}=29433600\frac{br}{year}

If we multiply this for the number of people in the station and the volume each breath needs, we obtain the volume breathed in a year.

The volume of a sphere is:

V_{sph}=\frac{4\pi}{3}r^3

So the diameter is:

D=2r=2\sqrt[3]{\frac{3V_{sph}}{4\pi}} =30.4m

6 0
3 years ago
A mass of .1539 kg moves down a 5 meter ramp in 2 seconds. What
enot [183]

Answer:

Velocity=2.5m/s

KE=4809.375J

Explanation:

Velocity=5m/2s=2.5m/s

KE=½×1539kg×(2.5m/s)²

KE=4809.375J

4 0
3 years ago
PLEASE HELP ME WITH THIS ONE QUESTION
Ipatiy [6.2K]

k = \dfrac{ (\dfrac{h}{ \lambda}  )^{2} }{2m}

k = (6.626×10-¹⁹/590 × 10-⁹ )^{2} /2 × 1.673 × 10-²⁷

k = (1.12 × 10-³⁰)^2/3.346×10-²⁷

k = 1.25 × 10-⁶⁰ /3.346×10-²⁷

k = 0

ldk why, my answer is coming this :(

4 0
2 years ago
A mass of (200 g) of hot water at (75.0°C) is mixed with cold water of mass M at (5.0°C). The final temperature of the mixture i
SIZIF [17.4K]

The mass of the cold water, given the data from the question is 500 g

<h3>Data obtained from the question</h3>
  • Mass of warm water (Mᵥᵥ) = 200 g
  • Temperature warm water (Tᵥᵥ) = 75 °C
  • Temperature of cold water (T꜀) = 5 °C
  • Equilibrium temperature (Tₑ) = 25 °C
  • Specific heat capacity of the water = 4.184 J/gºC
  • Mass of cold water (M꜀) =?

<h3>How to determine the mass of the cold water </h3>

Heat loss = Heat gain

MᵥᵥC(Tᵥᵥ – Tₑ) = M꜀C(Tₑ – T꜀)

200 × 4.184 (75 – 25) = M꜀ × 4.184(25 – 5)

41840 = M꜀ × 83.68

Divide both side 83.68

M꜀ = 41840 / 83.68

M꜀ = 500 g

Learn more about heat transfer:

brainly.com/question/6363778

#SPJ1

4 0
2 years ago
A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi
Vlad [161]

Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block m = 6.60 kg

Initial speed of block v _{i} = 1.56 \frac{m}{s}

Final speed of block v_{f} = 0 \frac{m}{s}

Coefficient of kinetic friction \mu _{k} = 0.62

Ramp inclined at angle \theta = 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

  \mu _{k} mgd \cos 28.4 =  \Delta K - \Delta U

Where \Delta U = mg d\sin 28.4

\mu _{k} mgd \cos 28.4 =  \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4

\mu _{k} mgd \cos 28.4 +mgd\sin 28.4  =  \frac{1}{2}mv_{i} ^{2}

d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}

 d = 0.122 m

Therefore, the distance travel by block before coming to rest is 0.122 m

7 0
3 years ago
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