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3 years ago

Plz help

Physics

2 answers:

RSB [31]3 years ago

7 0

Answer:

Explanation:

K.E. = 1/2 mv2

3000= 1/2 *100 *v2

60= v2

v = 7.7459

Scrat [10]3 years ago

6 0

Answer:

3000 J/100 kg = 30 kg/J

hope this helps

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Find the GCF 12, 18, and 84

ahrayia [7]

To find the answer, plot down the factors for every number.

12: 1, 2 ,3 ,4, 6, 12

18: 1, 2, 3, 6, 9, 18

84: 1, 2, 3, 4, 6, 7, 12

If you noticed, the number that was common to the 3 numbers, were 1, 2, 3, and 6
And 6 is the bigger number
So 6 is your GCF

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3 years ago

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PLEASE HELP i’m giving 32 points!!

meriva

Answer: 1 How many grams of KCl will dissolve in 1 liter of H2O at 50 °C? 5. 58.0 g of K2Cr2O7 is added to 100 g H2O at. 0 °C. With constant stirring, to what temp-. 2 34 °C? 4. How many grams of KCl will dissolve in 1 liter of H2O at 50 °C? 5. 58.0 g of ... A saturated solution of KClO3 was made with 300 g of H2O at. 34 °C.

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3 years ago

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Suppose y = 2x + 10. Find y if x = -2

san4es73 [151]

Answer:

y = 6

Explanation:

Given parameters:

Given equation:

y = 2x + 10

And supposing; x = -2

Unknown:

y = ?

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We are going to solve this problem by substitution;

y = 2x + 10

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y = 2(-2) + 10 = -4 + 10 = 6

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3 years ago

A fish is hooked on a line that is wound around the spool of a fishing reel. The reel has a mass of 652 g and a diameter of 10 c

iren2701 [21]

The magnitude of the force exerted by the fish is 3,052.5 N.

The given parameters;

mass of the reel, m = 652 g = 0.652 kg

diameter of the reel, d = 10 cm

radius of the reel, r= 5 cm = 0.05 m

time of motion, t = 5 s

angular speed of the reel, ω = 306 rad/s

According to Newton's third law, the force exerted by the reel is equal in magnitude to the force fish exerted on the reel.

The centripetal force of the reel = force exerted by the fish

$F_c = m\omega^2 r\\\\F_c = 0.652 \times (306)^2 \times 0.05\\\\F_c = 3,052.5 \ N$

Thus, the magnitude of the force exerted by the fish is 3,052.5 N.

Learn more here: brainly.com/question/14905888

3 0

3 years ago

A student that jumps a vertical height of 50 cm during the hang time activity.

muminat

The hang time of the student is 0.64 seconds, and he must leave the ground with a speed of 3.13 m/s

Why?

To solve the problem, we must consider the vertical height reached by the student as max height.

We can use the following equations to solve the problem:

Initial speed calculations:

$v_{f}^{2}=v_{o}^{2}+2*a*d$

At max height, the speed tends to zero.

So, calculating, we have:

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Hang time calculations:

We must remember that the total hang time is equal to the time going up plus the time going down, and both of them are equal,so, calculating the time going down, we have have:

$y-yo=vo.t+\frac{1}{2}*9.81\frac{m}{s^{2}}*t^{2} \\\\0.5m-0=0*t+4.95*t^{2}\\\\t^{2}=\frac{0.5}{4.95}\\\\t=\sqrt{0.101}=0.318s$

Then, for the total hang time, we have:

$TotalHangTime=2*0.318seconds=0.64seconds$

Have a nice day!

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4 years ago