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3 years ago

1 point

Physics

2 answers:

Sphinxa [80]3 years ago

7 0

Answer:

the answer is doppler effect

Explanation:

the doppler effect is the change in frequency of a wave in realation to an observer who is moving relative to the wave source. it is named after the austrian physicist christan doppler,who described the phenomenon in 1842. :) happy to help

Anna11 [10]3 years ago

5 0

Answer:

doppler effect

Explanation:

took the quiz

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Sphinxa [80]

Answer:

1/2 of your plate

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3 years ago

Read 2 more answers

Gayle runs at a speed of 3.85 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After s

enot [183]

Answer:

Final velocity at the bottom of hill is 15.56 m/s.

Explanation:

The given problem can be divided into four parts:

1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)

From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:

$p_i = p_f$

$m_1u_1 + m_2v_2 = (m_1 + m_2)v$

$v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)}$

$v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s$

2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.

The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}$

Here, initial velocity is the final velocity from the first stage. Therefore:

$v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s$

3. Use conservation of momentum to find the combined speed of Gayle and her brother.

Given:

Initial velocity of Gayle and sled is, $u_1(i)=10.5$ m/s

Initial velocity of her brother is, $u_2(i)=0$ m/s

Mass of Gayle and sled is, $m_1=55.0$ kg

Mass of her brother is, $m_2=30.0$ kg

Final combined velocity is given as:

$v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}$

$v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79$ m/s

4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.

Using conservation of energy, the final velocity at the bottom of the hill is:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s$

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3 years ago

What heat transfer method usually occurs through movement of hot smoke and fire gases?

Vladimir [108]

<span>What heat transfer method usually occurs through movement of hot smoke and fire gases?
</span>
Convention

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3 years ago

Describe a structural adaption that a snake has. What challenge does it overcome? PLZ HELP ASAP

PolarNik [594]

It has a flexible vertibrae which allows it to move faster and fit into places . Plus it also allows it to get its prey easier

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3 years ago

An electric drill

Lera25 [3.4K]

Answer:

Incomplete question. Complete question is: An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is twice the magnitude of the tangential acceleration. Determine the angle through which the drill rotates by this point.

The answer is : Δ θ = 1 rad

Explanation:

Ok, so the condition involves the centripetal acceleration and the tangential acceleration, so let’s start by writing expressions for each:

Ac= centripetal acceleration At= tangential acceleration

Ac = V² / r At = r α

Because we have to determine the angle ultimately, therefore we should convert the linear velocity into angular velocity in the expression for centripetal acceleration

V = r ω

Ac = (r ω)² / r = r² ω² / r

Ac = r ω²

now that we have expressions for the centripetal and tangential acceleration, we can write an equation that expresses the condition given: The magnitude of the centripetal acceleration is twice the magnitude of the tangential acceleration.

Ac = 2 At

That is,

r ω² = 2 r α

it is equivalent to;

ω² = 2 α

now we have the relation between angular speed and angular acceleration, but we also need to determine the angular displacement as well. Therefore choose a kinematics equation that doesn’t involve time because time is not mentioned in the question. Thus,

ω² – ω°² = 2 α Δ θ

such that ω° = 0

and ω² = 2 α

therefore;

2 α - 0 = 2 α Δ θ

2 α = 2 α Δ θ

So the angle will be : Δ θ = 1 rad

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3 years ago