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AlladinOne [14]
3 years ago
9

Picture attached please help! 2 different questions!

Physics
2 answers:
Maksim231197 [3]3 years ago
7 0

Wave A would have higher amplitude

Hope this helps :D

saw5 [17]3 years ago
5 0

Wave A has higher amplitude because it is stretched out and the energy of the sound wave is loud

P.S. hope this is helpful

Explanation:

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Explanation:

epero que te curva

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A Ping-Pong ball has a mass of 2.3 g and a terminal speed of 9.4 m/s. The drag force is of the form bv^2 What is the value of b?
julia-pushkina [17]
Drag Force = bv^2 = ma; a = g = 9.81 m/s^2

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2 years ago
WILL GIVE BRAINLIEST! QUICK PLEASE HELP!!!
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Answer:Weight is the result of gravity . The gravitational field strength of Earth is 10 N/kg (ten newtons per kilogram). This means an object with a mass of 1 kg would be attracted towards the centre of Earth by a force of 10 N. We feel forces like this as weight.

Explanation:

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3 years ago
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Why did the “yielders” conform in Asch’s experiment?
svet-max [94.6K]

Answer:

Asch's experiment showed that about 75% of people were "yielders" who conformed and 25% were "independent" who didn't conform. Asch concludes that people ignored reality and gave an incorrect answer in order to follow the rest of the group.

4 0
2 years ago
Block B (of mass m) is initially at rest. Block A (of mass 3m) travels toward B with an initial speed v0 (vee nought) and collid
NeTakaya

Answer:

The maximum height reached by the two blocks is approximately 0.1147959 × v₀²

Explanation:

The mass of block B = m

The mass of block A = 3·m

The initial velocity of block B, v₂ = 0 m/s

The initial velocity of block A, v₁ = v₀

The amount of friction between the blocks and the surface = Negligible friction

By the Law of conservation of linear momentum, we have;

Total initial momentum = Total final momentum

3·m·v₁ + m·v₂ = (3·m + m)·v₃ = 4·m·v₃

Plugging in the values for the velocities gives;

3·m × v₀ + m × 0 = (3·m + m)·v₃ = 4·m·v₃

∴ 3·m × v₀ =  4·m·v₃

\therefore v_3 = \dfrac{3}{4} \times v_0 = 0.75 \times v_0

The kinetic energy, K.E. of the combined blocks after the collision is given as follows;

K.E. = 1/2 × mass × v²

\therefore K.E. = \dfrac{1}{2} \times 4\cdot m \times \left (\dfrac{3}{4} \cdot v_0 \right )^2 = \dfrac{9}{8} \cdot m\cdot v_0^2

The potential energy, P.E., gained by the two blocks at maximum height = The kinetic energy, K.E., of the two blocks before moving vertically upwards

The potential energy, P.E. = m·g·h

Where;

m = The mass of the object at the given height

g = The acceleration due to gravity

h = The height at which the object of mass, 'm', is located

Therefore, for h = The maximum height reached by the two blocks, we have;

P.E. = K.E.

m \cdot g \cdot h =  \dfrac{9}{8} \cdot m\cdot v_0^2

h = \dfrac{\dfrac{9}{8} \cdot m\cdot v_0^2}{m \cdot g }  = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ g }  =  \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ 9.8} = \dfrac{42}{392} \cdot  v_0^2 \approx 0.1147959   \cdot  v_0^2

The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².

7 0
2 years ago
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