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3 years ago

If a students clothing catches fire the student should

Physics

1 answer:

Roman55 [17]3 years ago

4 0

Answer:

stop, drop and roll.

Explanation:

This is because rolling on the ground can help put out the fire by depriving it of oxygen.

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Your friend won the race that you were both running. Instead of being gracious, your friend made fun of you for losing, and now

nika2105 [10]

Negative peer influence

7 0

3 years ago

Read 2 more answers

An air-track cart is attached to a spring and completes one oscillation every 5.67 s in simple harmonic motion. at time t = 0.00

Masteriza [31]

The cart is moving by simple harmonic motion, and its position at time t is described by
$x(t) = A \cos (\omega t)$
where
A is the amplitude of the oscillation
$\omega$ is the angular frequency

The amplitude of the oscillation corresponds to the maximum displacement of the spring, which corresponds to the initial position where the spring was released:
A=0.250 m

The period of the motion is T=5.67 s, and the angular frequency is related to the period by
$\omega = \frac{2 \pi}{T}= \frac{2 \pi}{5.67 s} =1.11 rad/s$

Therefore now we can calculate the position of the system at the time t=29.6 s:
$x(29.6 s)=(0.250 m)\cos ((1.11 rad/s)(29.6 s))=+0.033 m$

3 0

3 years ago

Read 2 more answers

An airplane is seller rates down a runway at 3.10 M/S squared 441.5 seconds from rest until finally left off the ground ,what is

sineoko [7]

Answer:

302129m

Explanation:

Using s= ut + 1/2at²

But u= 0

S= 1/2(3.1)(441.5)²

=302129m

6 0

4 years ago

Which statement is

ELEN [110]

Answer:

Explanation: This is the correct answer

8 0

3 years ago

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What ocean depth would the volume of an aluminium sphere be reduced by 0.10%

yKpoI14uk [10]

Answer:

6400 m

Explanation:

You need to use the bulk modulus, K:

K = ρ dP/dρ

where ρ is density and P is pressure

Since ρ is changing by very little, we can say:

K ≈ ρ ΔP/Δρ

Therefore, solving for ΔP:

ΔP = K Δρ / ρ

We can calculate K from Young's modulus (E) and Poisson's ratio (ν):

K = E / (3 (1 - 2ν))

Substituting:

ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)

Before compression:

ρ = m / V

After compression:

ρ+Δρ = m / (V - 0.001 V)

ρ+Δρ = m / (0.999 V)

ρ+Δρ = ρ / 0.999

1 + (Δρ/ρ) = 1 / 0.999

Δρ/ρ = (1 / 0.999) - 1

Δρ/ρ = 0.001 / 0.999

Given:

E = 69 GPa = 69×10⁹ Pa

ν = 0.32

ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)

ΔP = 64.0×10⁶ Pa

If we assume seawater density is constant at 1027 kg/m³, then:

ρgh = P

(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa

h = 6350 m

Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.

6 0

4 years ago