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sattari [20]
3 years ago
6

When Lisa placed a balloon to her ear, she was able to hear what her sister was whispering from the other side of the room. When

she removed the balloon, she was no longer able to hear what her sister was saying. What is the best explanation for what Lisa discovered?
Physics
2 answers:
Aleks04 [339]3 years ago
5 0

Explanation :

Helium gas is present inside air balloon. Its density is less than air i.e. it is lighter than air.

The speed of sound in air is 343 m/s and around 972 m/s in He.

So, when Lisa placed a balloon to her ear, she was able to hear what her sister was whispering from the other side of the room because the sound waves are travelling with faster speed to Lisa.

When she removed the balloon, she was no longer able to hear what her sister was saying. It is because speed of sound in air is <em>343 m/s</em>.

kipiarov [429]3 years ago
4 0

Answer:

The answer is D

Explanation:

Sound does not travel around the air outside of the balloon as well as it does through the denser air pressure inside the balloon

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A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule
ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

p_i = 0

The final total momentum is instead:

p_f = m_a v_a + m_c v_c

where

m_a = 125 kg is the mass of the astronaut

v_a = 2.50 m/s is the velocity of the astronaut

m_c = 1900 kg is the mass of the capsule

v_c is the velocity of the capsule

Since the total momentum must be conserved, we have

p_i = p_f = 0

so

m_a v_a + m_c v_c=0

Solving the equation for v_c, we find

v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

F \Delta t = \Delta p

The change in momentum of the astronaut is

\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s

And the duration of the push is

\Delta t = 0.600 s

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J

3 0
3 years ago
Block on inclined plane experience a force due to gravity of 300N straight down. If the slope is inclined at 67.8°to the horizon
Tems11 [23]

Answer:

The component of the force due to gravity perpendicular and parallel to the slope is  113.4 N and 277.8 N respectively.

Explanation:

Force is any cause capable of modifying the state of motion or rest of a body or of producing a deformation in it. Any force can be decomposed into two vectors, so that the sum of both vectors matches the vector before decomposing. The decomposition of a force into its components can be done in any direction.

Taking into account the simple trigonometric relations, such as sine, cosine and tangent, the value of their components and the value of the angle of application, then the parallel and perpendicular components will be:

  • Fparallel = F*sinα =300 N*sin 67.8° =300 N*0.926⇒ Fparallel =277.8 N
  • Fperpendicular = F*cosα =  300 N*cos 67.8° = 300 N*0.378 ⇒ Fperpendicular= 113.4 N

<u><em>The component of the force due to gravity perpendicular and parallel to the slope is  113.4 N and 277.8 N respectively.</em></u>

6 0
3 years ago
If the bar magnet is flipped over and the south pole is brought near the hanging ball, the ball will be?
disa [49]

The ball may attracted to the magnet.

<h3>How can we understand that the hanging ball will be attracted to the magnet or not?</h3>
  • From the question, we understand that the ball is attracted by the north pole of the bar magnet, then the bar magnet flipped over and the south pole is brought near the hanging ball.
  • As we know, in this type of experiments of bar magnet most of the times the ball is made out of steel.
  • Steel is a magnetic material.
  • Magnetic materials gets attracted  to the magnet at both the North and South pole.
  • This can be compared to how neutral objects also gets attracted to the positively and negatively charged rods through the Polarization force.

So, If the bar magnet is flipped over and the south pole is brought near the hanging ball, The ball will be attracted to the magnet.

Learn more about the bar magnet:

brainly.com/question/27943723

#SPJ4

7 0
2 years ago
Oil at 150 C flows slowly through a long, thin-walled pipe of 30-mm inner diameter. The pipe is suspended in a room for which th
chubhunter [2.5K]

Answer:

<em>1.01 W/m</em>

Explanation:

diameter of the pipe d = 30 mm = 0.03 m

radius of the pipe r = d/2 = 0.015 m

external air temperature Ta = 20 °C

temperature of pipe wall Tw = 150 °C

convection coefficient at outer tube surface h = 11 W/m^2-K

From the above,<em> we assumed that the pipe wall and the oil are in thermal equilibrium</em>.

area of the pipe per unit length A = \pi r ^{2} = 7.069*10^{-4} m^2/m

convectional heat loss Q = Ah(Tw - Ta)

Q = 7.069 x 10^-4 x 11 x (150 - 20)

Q = 7.069 x 10^-4 x 11 x 130 = <em>1.01 W/m</em>

4 0
3 years ago
Water emerges straight down from a faucet with a 2.51-cm diameter at a speed of 3.04 m/s. (because of the construction of the fa
Mkey [24]
This is a question on conservation of energy. That is,
mgh + KE1 = KE2
mgh +1/2mv1^2 = 1/2mv2^2
gh + 1/2v1^2 = 1/2v2^2

Where, h = 0.2 m, v1 =3.04 m/s
Therefore,
v2 = Sqrt [2(gh+1/2v1^2)] = Sqrt [2(9.81*0.2 + 1/2*3.04^2)] = 7.26 m/s

Now, Volumetric flow rate, V/time, t = Surface area, A*velocity, v
Where,
V = Av = πD^2/4*3.04 = π*(2.51/100)^2*1/4*3.04 = 1.504*10^-3 m^3/s

At 0.2 m below,
V = 1.504*10^-3 m^3/s = A*7.26
A = (1.504*10^-3)/7.26 = 2.072*10^-4 m^2

But, A = πr^2
Then,
r = Sqrt (A/π) = Sqrt (2.072*10^-4/π) = 0.121*10^-3 m
Diameter = 2r = 0.0162 m = 1.62 cm
3 0
3 years ago
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