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-BARSIC- [3]
3 years ago
12

Cloudy nights can be warmer than clear nighys because clouds trap heat

Physics
1 answer:
Rasek [7]3 years ago
4 0
<span>Is it true that nighttime air temperatures on a cloudy night are lower than they would be on a clear night?</span>
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Water falls without splashing at a rate of 0.370 l/s from a height of 2.90 m into a 0.690-kg bucket on a scale. if the bucket is
raketka [301]
<span>R = rate of flow = 0.370 L/s H = height = 2.9 m T= time = 3.9 s V = velocity of water when it hits the bucket = sqrt(2gh) = sqrt(2 x 9.8 x 2.9) =7.539 m/s2 G value = 9.8 m/s2 Wb = weight of bucket = 0.690 kg x 9.8 m/s2 = 6.762 N Wa = weight of accumulated water after 3.9 s Fi = force of impact of water on the bucket S = reading on the scale = Wa + Wb + Fi mass of water accumulated after 3.9 s = R x T = 0.370 x 3.9 = 1.443 L = 1.443 kg Therefore, Wa = 1.443 x 9.8 = 14.1414 N Fi = rate of change of momentum at the impact point = R x V (because R = dm/dt) = 0.37 x 7.539 = 2.78943 N S = 14.1414 + 6.762 + 2.78943 = 23.692 N</span>
8 0
3 years ago
onsider the free body diagram. If the sum of the tension forces is equal to the force of gravity, which description BEST applies
cluponka [151]
The answer is C) A girl hangs by both hands, motionless, from a trapeze.
5 0
3 years ago
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
3 years ago
W=90 kg×4n/kg<br>Please halp me i need it ​
kiruha [24]

I'm guessing that this is a problem to find the weight of a 90kg mass on a planet where the acceleration of gravity is 4 m/s^2. (Much less gravity than Earth, a little more than Mars.)

Just do the multiplication, and you get

360 Newtons.

5 0
3 years ago
White light is an _____________ mixture of all the colors.<br> a) observable b) unequal c) equal
Musya8 [376]

Answer:

unequal

Explanation:

White light is a combination of all colors in the color spectrum.

6 0
3 years ago
Read 2 more answers
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