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alekssr [168]
3 years ago
9

What is the final temperature when 50 g of water at 20°c is mixed with 75 g of water at 60°c

Physics
1 answer:
tatyana61 [14]3 years ago
4 0
44C i think. Do you have options for the answer?
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A vocalist with a bass voice can sing as low as 92 Hz.
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  • 3.26 x 10 to the power of 6

Explanation:

c = lambda × frequency

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Which of these pollutants is transferred from soil to water by organic pesticides and fertilizer runoff from farms?
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I would say your answer to this question would be D
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Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence
sineoko [7]

Answer:

From smallest ratio to the largest ratio:

Coasting Universe - Critical Universe - Recollapsing Universe(From left to right)

Explanation:

The coasting universe is one that expands at a constant rate given by the Hubble constant throughout all of cosmic time. It has a ratio of actual density to critical density that is less than 1

The critical universe is one that is at balance with no expansion .I.e. the actual density and the critical density are equal, which makes the ratio of actual density to critical density to be equal to 1

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3 years ago
A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider
shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

d=v_x t

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

y(t) = h + v_y t - \frac{1}{2}gt^2

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s

So now we can find the magnitude of the initial velocity:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s

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