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3 years ago

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Question 15: At the edge of a lake you throw a stone with a velocity of 29 m/s at an angle of 45°. The stone is in the air for 4

.23 s. How far did you throw the stone? (What is the range?)
I think I have this question's right answer, but I would like to check.

1 answer:

Marianna [84]3 years ago

8 0

Get your numbers gathered up and solve the problem in the ordered step

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The length of the student desk is measured using a

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3 significant figures are in 1.02m

Explanation:

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Does the size of a paper airplane affect how far it flies

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No the only thing that affects it is how it is built

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You have just landed on Planet X. You take out a ball of mass 101 g , release it from rest from a height of 16.0 m and measure t

IRINA_888 [86]

Answer:

0.3817 N

Explanation:

Remark

One thing is certain: the ball has a mass of 101 grams wherever it is in the universe. That is not true of the force. The force on the moon is a whole lot less than it is on earth, and maybe planet x as well.

Givens

m = 101 g

vi = 0 That's what at rest means.

t = 2.91 s

d = 16 m

F= ?

Formulas

d = vi*t + 1/2*a * t^2

Force = m * a

Solution

16 = 0 + 1/2 a * 2.91^2

16 = 4.234 a Divide by 4.234

16/4.234 = a

a = 3.779

F = m * a

a = 3.779

m = 101 g = 1 kg / 1000 grams

m = 0.101 kg

F = 0.101 * 3.779

F = 0.3817N

8 0

3 years ago

Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is

dimaraw [331]

Answer:

15.106 N

Explanation:

From the given information,

The weight of the bucket can be calculated as:

$W_b = m_bg = \\ \\ W_b = (0.730 \ kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N$

The mass of the water accumulated in the bucket after 3.20s is:

$m_w= (0.20 \ L/s) ( 3.20)s$

$m _w=0.64 \ kg$

To determine the weight of the water accumulated in the bucket, we have:

$W_w = m_w g$

$W_w = ( 0.64 \ kg )(9.80\ m \ /s^2)$

$W_w = 6.272 \ N$

For the speed of the water before hitting the bucket; we have:

$v = \sqrt{2gh}$

$v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}$

v = 8.4 m/s

Now, the force required to stop the water later when it already hit the bucket is:

$F = v ( \dfrac {dm}{dt} )$

$F = (8.4 \ m/s)( 0.200 \ L/s)$

F = 1.68 N

Finally, the reading scale is:

$F_{scale$ = 7.154 N + 6.272 N + 1.68 N

= 15.106 N

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3 years ago

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So velocity of a balloon? Use V = 2KE over v. A hot air balloon with a mass of 100 kg moves across the sky with 3200 J of kineti

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