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2 years ago

A jet of water squirts out horizontally from

Physics

2 answers:

Paul [167]2 years ago

7 0

Answer:

the answer is 6.06

t=0.501

v=x/t

0.5pv²=pgh

h=v²/2g= 1.09²/2(9.8)=0.0606 m=6.06 cm

Bond [772]2 years ago

5 0

The correct answer to your question is 3.62

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A hot iron ball of mass 200 g is cooled to a temperature of 22°C. 6.9 kJ of heat is lost to the surroundings during the process.

qaws [65]

Heat lost or gained, H = mc(θ₂ - θ₁)
Where m = mass, c = Specific heat capacity, θ₂= final temperature, θ₁ = initial temperature

m = 200g, c = 0.444 J/g°C, θ₁ = 22 °C (Since it was cooled).

H = 6.9 kj = 6.9 *1000J = 6900 J

6900 = 200*0.444* (θ₂ - 22)

6900/(200*0.444) = θ₂ - 22

77.70 = θ₂ - 22

θ₂ - 22 = 77.7

θ₂ = 77.7 + 22 = 99.7

So initial temperature before cooling ≈ 100°C . Option C.

5 0

3 years ago

Read 2 more answers

Chemistry Sem 2

3241004551 [841]

Answer:

Solids - Bricks , wood , Pottery, Bucket

Liquid - Water, soap, Sanitizers.

Gases - Aerosol in Deodorants, Chlorofluorocarbons in Fire extinguishers , Butane in lighters.

4 0

2 years ago

The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1

telo118 [61]

<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life $h$ of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: $h_{B-13}=5(10)^{-10}s=0.0000000005s$

beryllium-15: $h_{B-15}=2(10)^{-7}s=0.0000002s$

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

$h_{B-15}=X.h_{B-13}$

Where $X$ is the amount we want to find:

$X=\frac{h_{B-15}}{h_{B-13}}$

$X=\frac{2(10)^{-7}s}{5(10)^{-10}s}$

Finally:

$X=400$

Therefore:

The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.

8 0

3 years ago

A speed boat increases its speed uniformly from vi = 20.0 m/s to vf = 30.0 m/s in a distance of 2.00 x 10^2m. (a) Draw a coordin

pychu [463]

a) See graph in attachment

b) The suvat equation to use is $v_f^2 - v_i^2 = 2as$

c) The acceleration is $a=\frac{v_f^2-v_i^2}{2s}$

d) The acceleration is $1.25 m/s^2$

e) The time needed is 8 s

Explanation:

For this part, find in attachment the diagram representing this situation.

Since we are not given any particular direction for the motion, we choose the x-direction as the direction of motion of the boat.

Then we have the following:

- The initial position of the boat is $x_i = 0$ , the origin

- The final position of the boat is $x_f = 200 m$

- The initial velocity of the boat is $v_i = 20.0 m/s$

- The final velocity of the boat is $v_f = 30.0 m/s$

Note that the arrow representing the final velocity is longer than that of the initial velocity, since the final velocity is larger.

The motion of the speed boat is a uniformly accelerated motion (motion at constant acceleration), therefore we can use one of the suvat equations. In this particular problem, we know the following quantities:

$v_i = 20.0 m/s$ , the initial velocity

$v_f = 30.0 m/s$ , the final velocity

$s = x_f - x_i = 200 m$ , the displacement of the boat