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4 years ago

If your speech changes from 10 km/h to 6 km, you have a(n) ___ acceleration. A. negative B. Positive C. Neutral D. Oscillating

Physics

1 answer:

Vesnalui [34]4 years ago

6 0

I think your second figure should be 6km/hour if that is the case then this is the answer.

Answer

A. Negative. this is because your speed is reducing which means you are decelerating or rather going at a negative (acceleration).

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Two friends are walking on a hot beach on a hot summer day. one friend reports her feet are becoming hot she needs her sandals.

nata0808 [166]

Answer:

conduction

Explanation:

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3 years ago

two objects were lifted by machine one object had a massive 2 kg it was lifted at a speed of 2 m/s the other had a massive 4 kg

rosijanka [135]

Kinetic energy =(1/2) (mass) (speed²)

First object: KE = (1/2) (2 kg) (2m/s)² = 4 joules during the lift.

Second object: KE = (1/2) (4kg) (3 m/s)² = 18 joules during the lift.

The second object has more kinetic energy while it's being lifted
than the first object has while it's being lifted. Once they reach their
final heights and stop, neither object has any kinetic energy.

8 0

4 years ago

HEY CAN ANYONE PLS PLS ANSWER DIS I RLY NEED IT

jeka94

Place the object in an electronic balance and measure its mass.
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4 years ago

At divergent boundaries, hot mantle rock rises and occurs.

andrey2020 [161]

The answer is decompression melting

5 0

3 years ago

A rock band playing an outdoor concert produces sound at 120 dB 5.0 m away from their single working loudspeaker. What is the so

joja [24]

ANSWER:100dB

from the sound intensity level,the sound intensity is calculated as:

$\beta$ ₁= $\beta$ =(10dB)㏒₁₀(l₁/l₀)

inserting numbers:

120dB=(10dB)㏒₁₀[l₁/10⁻¹²W/m⁸] or 12=㏒₁₀[l₁/(10⁻¹²)W/m²

Getting the antilog of both sides and obtain 10¹²=l₁(10⁻¹²W/m²)which

can be used to solve for l₁ and get

l₁=(10⁻¹²W/m²)(10¹²)=1 W/m²

since the sound intensity is related to the power and that the power does not change,the sound intensity at any other point can be solved.Plugging-in

ᵃ = 4πr²,into P=l₁ₐ₁=l₂ₐ₂ and get:

l₂=l₁(r₁/r₂)² =(1W/m²)(5/35) =2.04×10⁻²W/m²

since we know the sound intensity at the sound point 2r,the sound intensity level at the point can be solved.We have:

$\beta$ ₂= $\beta$ =(10dB)㏒₁₀(l₂/l₀)=(10dB)㏒₁₀(2.04×10⁻²/1×10⁻²)

$\beta$ ₂=(10dB)㏒₁₀(2.04×10¹⁰)

$\beta$ =(10dB)[㏒₁₀(2.04)+㏒₁₀(10¹⁰)]=10dB[0.32+10]=103dB=100dB

3 0

4 years ago