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2 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]2 years ago

8 0

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

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A student initially 10.0 m East of his school walks 17.5 m West. The magnitude of the student's displacement, relative to the sc

Fantom [35]

Answer:

1. 7.5 m

2. towards west side

explanation:

I hope it will help you

3 0

2 years ago

A 61.0-kg person jumps from rest off a 10.0-m-high tower straight down into the water. Neglect air resistance. She comes to rest

9966 [12]

Answer:

Explanation:

In this case, law of conservation of energy will be implemented. It states that "the energy of the system remains conserved until or unless some external force act on it. Energy of the system may went through the conversion process like kinetic energy into potential and potential into kinetic energy.But their total always remain the same in conserved systems."

Given data:

Height of tower = 10.0 m

Depth of the pool = 3.00 cm

Mass of person = 61.0 kg

Solution:

Initial energy = Final energy

$U_{i} = (K.E) + U_{f}$

As the person was at height initially so it has the potential energy only.

$mg(h_{1} +h_{2}) = K.E + mgh_{2}$

$K.E = mgh_{1}$

$K.E = (61.0)(9.8)(10)\\K.E = 5978 J$

Lets find out the magnitude of the force that the water is exerting on the diver.

W =ΔK.E

$F.h_{2} = 5978\\$

$F = \frac{5978}{3}$

F = 1992.67 N

7 0

3 years ago

A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target

stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

$\theta=35^{\circ}$

a.Let $v_0$ be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

$v_x=v_0cos\theta=v_0cos35$

$v_y=v_0sin\theta=v_0sin35$

$x=v_0cos\theta\times t=v_0cos35\times t$

$t=\frac{30}{v_0cos35}$

$h=v_yt-\frac{1}{2}gt^2$

Substitute the values

$1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2$

$1.8=30tan35-\frac{6574.6}{v^2_0}$

$\frac{6574.6}{v^2_0}=21-1.8=19.2$

$v^2_0=\frac{6574.6}{19.2}$

$v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s$

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

$t=\frac{30}{18.5cos35}$

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0

3 years ago

Without completing the calculations, determine what the new volume will be in the problem below. Also, explain how you were able

Talja [164]

Boyles law

Pressure and volume are inversely proportional as the new variable changes from the known.

Double the pressure equals 1/2 of original volume, assuming temperature remains the same.

So 40.0 mL is the new volume as it is compressed.

7 0

3 years ago

A player is positioned 35 m[40 degrees W of S] of the net. He shoot the puck 25 m [E] to a teammate. What second displacement do

Lubov Fominskaja [6]

Answer:

x=22.57 m

Explanation:

Given that

35 m in W of S

angle = 40 degrees

25 m in east

From the diagram

The angle

$\theta=90-40=50^o$

From the triangle OAB

$cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}$

$1340.57=35^2+25^2-x^2$

x=22.57 m

Therefore the answer of the above problem will be 22.57 m

4 0

2 years ago