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2 years ago

The following table lists the speed of sound in various materials. Use this table to answer the question.

Physics

2 answers:

Lisa [10]2 years ago

7 0

Answer: it will travel slowest in air.

aksik [14]2 years ago

4 0

Answer:

Sound will travel the slowest in Air

Explanation:

The higher the m/s is, the faster the sound travels

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Distant galaxies appear to be much larger than those nearby. true or false

jonny [76]

Answer:True

Explanation: Distant galaxies appear to be larger than those close by because of cosmic expansion. light from nearer galaxies travel farther and makes the other to be billions of years away. The farther it takes to get to another galaxy, the bigger it appears

3 0

3 years ago

A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 26.0m tall building. The rock do

Pachacha [2.7K]

Answer:

A) v = 28.3 m/s

B) t = 4.64 s

Explanation:

Assuming no other forces acting on the rock, since the accelerarion due to gravity close to the surface to the Earth can be taken as constant, we can use one of the kinematic equations in order to get first the maximum height (over the roof level) that the ball reaches:

$v_{f}^{2} - v_{o}^{2} = 2* g* \Delta h (1)$

Taking into account that at this point, the speed of the rock is just zero, this means vf=0 in (1), so replacing by the givens and solving for Δh, we get:

$\Delta h = \frac{-v_{o} ^{2}}{2*g} = \frac{-(17.0m/s)^{2} }{2*(-9.8m/s2)} = 14.8 m (2)$

So, we can use now the same equation, taking into account that the initial speed is zero (when it starts falling from the maximum height) and that the total vertical displacement is the distance between the roof level and the ground (26.0 m) plus the maximum height that we have just found in (2) , 14.8m:
Δh = 26.0 m + 14. 8 m = 40.8 m (3)
Replacing now in (1), we can solve for vf, as follows:

$v_{f} =\sqrt{2*g*\Delta h} = \sqrt{2*9.8m/s2*40.8m} = 28.3 m/s (4)$

In order to find the total elapsed from when the rock is thrown until it hits the street, we can divide this time in two parts:
1) Time elapsed from the the rock is thrown, until it reaches to its maximum height, when vf =0
2) Time elapsed from this point until it hits the street, with vo=0.
For the first part, we can simply use the definition of acceleration (g in this case), making vf =0, as follows:

$v_{f} = v_{o} + a*\Delta t = v_{o} - g*\Delta t = 0 (5)$

Replacing by the givens in (5) and solving for Δt, we get:

$\Delta t = \frac{v_{o}}{g} = \frac{17.0m/s}{9.8m/s2} = 1.74 s (6)$

For the second part, since we know the total vertical displacement from (3), and that vo = 0 since it starts to fall, we can use the kinematic equation for displacement, as follows:

$\Delta h = \frac{1}{2} * g * t^{2} (7)$

Replacing by the givens and solving for t in (7), we get:

$t_{fall} =\sqrt{\frac{2*\Delta h}{g}} = \sqrt{\frac{2*40.8m}{9.8m/s2} } = 2.9 s (8)$

So, total time is just the sum of (6) and (8):
t = 2.9 s + 1.74 s = 4.64 s

5 0

2 years ago

Lien uses a spring scale to pull a block toward the right across the lab table. The scale reads 8 N. Which force should Lien con

Alja [10]

Answer

The force on the left across the lab table.

Explanation

The Newton's third law of motion states that; <em>action and reaction are equal but a ct in opposite direction. </em>

When the block of is pulled on the right with a force of X Newtons then there is a force -X Newtons applying and equal force on the left. For every action there must be a reaction with is equal and applying in the opposite direction.

So, if the block is pulled on the right by a force of 8 N there is another equal force applying on the left.

5 0

3 years ago

Read 2 more answers

Sound level B in decibels is defined as

Lelechka [254]

Answer:

The approximate combined sound intensity is $I_{T}=1.1\times10^{-4}W/m^{2}$

Explanation:

The decibel scale intensity for busy traffic is 80 dB. so intensity will be

$10log(\frac{I_{1}}{I_{0}} )=80$ , therefore $I_{1}=1\times10^{8}I_{0}=1\times10^{8} * 1\times10^{-12}W/m^{2}=1\times10^{-4}W/m^{2}$

In the same way for the loud conversation having a decibel intensity of 70 dB.

$10log(\frac{I_{2}}{I_{0}} )=70$ , therefore $I_{2}=1\times10^{7}I_{0}=1\times10^{7} * 1\times10^{-12}W/m^{2}=1\times10^{-5}W/m^{2}$

Finally we add both of them $I_{T}=I_{1}+I_{2}=1\times10^{-4}W/m^{2}+1\times10^{-5}W/m^{2}=1.1\times10^{-4}W/m^{2}$ , is the approximate combined sound intensity.

3 0

2 years ago

DEFINE HEAT CAPACITY

Rudiy27

The number of heat units needed to raise the temperature of a body by one degree.

4 0

2 years ago

Read 2 more answers