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Mumz [18]
3 years ago
8

Objects: A balloon full of helium in the air. Air- 1.27 kg/ml or Helium Balloon- 0.33 kg/ml

Physics
1 answer:
dem82 [27]3 years ago
6 0

Answer:

Air has greater density.

The object will float.

Explanation:

Density of a body is defined as the mass of a body per unit volume of it.

Here, air has a density of 1.27 kg per ml and Helium balloon has a density of 0.33 kg per ml. This means that if volume is 1 ml, then the mass of air in that volume will be 1.27 kg and mass of Helium balloon in that same volume will be 0.33 kg.

Therefore, the mass of Helium balloon is lighter than that of air in a given volume. Hence, air is heavier and has a greater density.

A lighter density object always floats over a greater density body.

Hence, Helium balloon will float over air as Helium balloon is less dense than that of air.

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Describe how can two or more velocities be combined
eduard
Two or more velocities add by vector addition
4 0
3 years ago
Confirm if this is correct or not. If it isn't correct, please correct it.
kow [346]

Answer:

d = 421.83 m

Explanation:

It is given that,

Height, h = 396.9 m

Horizontal speed, v = 46.87 m/s

We need to find the distance traveled by the ball horizontally. Let t is the time taken by the ball. Using second equation of motion for vertical direction. So,

396.9=0\times t+\dfrac{1}{2}\times 9.8 t^2\\\\t=9\ s

Now d is the distance covered by the cannonball. So,

d=vt\\\\d=46.87\times 9\\\\d=421.83\ m

Hence, this is the required solution.

3 0
2 years ago
A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 6.3 m/s2. For what value of the coe
Levart [38]

Answer:

Coefficient of static friction will be equal to 0.642  

Explanation:

We have given acceleration a=6.3m/sec^2

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

We have to find the coefficient of static friction between truck and a cabinet will

We know that acceleration is equal to a=\mu g, here \mu is coefficient of static friction and g is acceleration due to gravity

So \mu =\frac{a}{g}=\frac{6.3}{9.8}=0.642

So coefficient of static friction will be equal to 0.642

3 0
3 years ago
Which of these would most likely be a parts of a lab procedure?
vladimir2022 [97]
C . Record the time to complete a chemical reaction
6 0
3 years ago
Read 2 more answers
A contestant in a winter games event pulls a 36.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in
Novay_Z [31]

(a) The minimum force F he must exert to get the block moving is 38.9 N.

(b) The acceleration of the block is 0.79 m/s².

<h3>Minimum force to be applied </h3>

The minimum force F he must exert to get the block moving is calculated as follows;

Fcosθ = μ(s)Fₙ

Fcosθ = μ(s)mg

where;

  • μ(s) is coefficient of static friction
  • m is mass of the block
  • g is acceleration due to gravity

F = [0.1(36)(9.8)] / [(cos(25)]

F = 38.9 N

<h3>Acceleration of the block</h3>

F(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32

a = F(net)/m

a = 28.32/36

a = 0.79 m/s²

Thus, the minimum force F he must exert to get the block moving is 38.9 N.

The acceleration of the block is 0.79 m/s².

Learn more about minimum force here: brainly.com/question/14353320

#SPJ1

4 0
2 years ago
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