Ask question

Ask question

All categories

3 years ago

5

How should i fix an overfilled ballon? its my sisters 3rd birthday and i have overfilled all the ballons and there is no other o

ption left than to only fix those ballons.please help.i am so nervous right now.

1 answer:

iVinArrow [24]3 years ago

8 0

Try to untie the knot and let a little air out and tie it back.....

You might be interested in

A cricket ball has mass 0.155 kg. If the velocity of a bowled ball has a magnitude of 35.0 m/s and the batted ball's velocity is

FinnZ [79.3K]

Answer:

Magnitude of change in momentum = 4.65 kg.m/s

Magnitude of impulse = 4.65 kg.m/s

Magnitude of the average force applied by the bat = 1550 N

Explanation:

Mass of the cricket ball, m = 0.155 kg

Initial velocity of the ball, u = 35.0 m/s

final velocity of the ball after hitting the bat, v = 65.0 m/s

Time of contact, t = 2.00 ms = 2.00 × 10⁻³ s

Now,

Magnitude of change in momentum = Final momentum - Initial momentum

or

Magnitude of change in momentum = ( m × v ) - ( m × u )

or

Magnitude of change in momentum = ( 0.155 × 65 ) - ( 0.155 × 35 )

or

Magnitude of change in momentum = 10.075 - 5.425 = 4.65 kg.m/s

Now, Magnitude of impulse = change in momentum

thus,

Magnitude of impulse = 4.65 kg.m/s

Now,

magnitude of the average force applied by the bat = $\frac{\textup{Impulse}}{\textup{Time}}$

or

magnitude of the average force applied by the bat = $\frac{\textup{4.65}}{\textup{3}\times\textup{10}^{-3}}$

or

Magnitude of the average force applied by the bat = 1550 N

6 0

4 years ago

At time t=0 , in your frame of reference Z, you measure the back of the spaceship to be at x=0 and the front of the ship to be a

Liula [17]

An equation relating the length that you measure l to the ship's proper length l0 is

l =l0/y. This is further explained below.

<h3>What is an equation relating the length that you measure l to the ship's proper length l0?</h3>

Generally, Any object's length in a moving frame will look shortened or contracted when seen in that direction. The Lorentz transformation may be used to determine the amount of contraction.

In conclusion, To use the Lorentz Lorentz transformation, the length Lo-x2 - may be determined if it is measured in the moving reference frame. Hence the Resultant l = l0/y.

Read more about Lorentz transformation

brainly.com/question/16284701

#SPJ1

7 0

3 years ago

If gravity did not affect the pain of a horizontally thrown ball the ball would

Drupady [299]

If gravity had no effect on a ball after you threw it ... and there also
were no air to slow it down ... then the ball would continue traveling
in a straight line, in whatever direction you threw it.

That's the heart and soul of Newton's laws of motion ... any object
keeps moving at the same speed, and in a straight line in the same
direction, until a force acts on it to change its speed or direction.\

If you threw the ball horizontally, then it would keep moving in the
same direction you threw it. But don't forget: The Earth is not flat.
The Earth is a sphere. So, as the ball kept going farther and farther
in the same straight line, the Earth would curve away from it, and it
would look like the ball is getting farther and farther from the ground.

4 0

3 years ago

Read 2 more answers

PLEASE SOLVE QUICKLY!!!<br> Solve for A, B, and C from graph

hram777 [196]

A = 59.35cm

B = 196.56g

C = 74.65g

<u>Explanation:</u>

We know,

$x = \frac{L}{\frac{W}{F} +1}$

and L = x+y

1.

Total length, L = 100cm

Weight of Beam, W = 71.8g

Center of mass, x = 49.2cm

Added weight, F = 240g

Position weight placed from fulcrum, y = ?

$L-y = \frac{L}{\frac{W}{F}+1 } \\100 - y = \frac{100}{\frac{71.8}{49.2}+1 } \\100 - y = \frac{100}{1.46+1}\\\\100 - y = \frac{100}{2.46} \\100-y = 40.65\\\\y = 59.35cm$

Therefore, position weight placed from fulcrum is 59.35cm

2.

Total length, L = 100cm

Center of mass, x = 47.8 cm

Added weight, F = 180g

Position weight placed from fulcrum, y = 12.4cm

Weight of Beam, W = ?

$47.8 = \frac{100}{\frac{W}{180}+1 }\\\47.8 = \frac{100}{\frac{W+180}{180} } \\\\47.8 = \frac{100 X 180}{W+180}\\ \\47.8W + 47.8 X 180 = 18000\\47.8W = 18000 - 8604\\W = \frac{9396}{47.8}\\ W = 196.56g$

Therefore, weight of the beam is 196.56g

3.

Total length, L = 100cm

Center of mass, x = 50.8 cm

Position weight placed from fulcrum, y = 9.8cm

Weight of Beam, W = 72.3g

Added weight, F = ?

$50.8 = \frac{100}{\frac{72.3}{F}+1 }\\\ 50.8 = \frac{100}{\frac{72.3+F}{F} } \\\\50.8 = \frac{100 X F}{72.3+F}\\ \\50.8 X 72.3 + 50.8 X F = 100F\\\\3672.84 = 100F-50.8F\\3672.84 = 49.2F\\F = 74.65g$

Therefore, Added weight F is 74.65g

A = 59.35cm

B = 196.56g

C = 74.65g

4 0

3 years ago

An object is thrown vertically up and attains an upward velocity of 9.6 m/s when it reaches one fourth of its maximum height abo

ki77a [65]

Answer:

11.09 m/s

Explanation:

Given that an object is thrown vertically up and attains an upward velocity of 9.6 m/s when it reaches one fourth of its maximum height above its launch point.

The parameters given are:

Initial velocity U = ?

Final velocity V = 9.6 m/s

Acceleration due to gravity g = 9.8m/s^2

Let first assume that the object is thrown from rest with the velocity U, at maximum height V = 0

Using third equation of motion

V^2 = U^2 - 2gH

0 = U^2 - 2 × 9.8H

U^2 = 19.6H ........ (1)

Using the formula again for one fourth of its maximum height

9.6^2 = U^2 - 2 × 9.8 × H/4

92.16 = U^2 - 19.6/4H

92.16 = U^2 - 4.9H

U^2 = 92.16 + 4.9H ...... (2)

Substitute U^2 in equation (1) into equation (2)

19.6H = 92.16 + 4.9H

Collect the like terms

19.6H - 4.9H = 92.16

14.7H = 92.16

H = 92.16/14.7

H = 6.269

Substitute H into equation 2

U^2 = 92.16 + 4.9( 6.269)

U^2 = 92.16 + 30.72

U^2 = 122.88

U = 11.09 m/s

Therefore, the initial velocity of the object is 11.09 m/s

3 0

4 years ago