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Goryan [66]
3 years ago
13

Why are magnetic fields evidence of sea floor spreading

Physics
1 answer:
xenn [34]3 years ago
8 0
Eruptions of molten material, magnetic stripes in the rock of the ocean floor, and the ages of the rocks themselves.
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what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe
ArbitrLikvidat [17]

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2

Hence, this is the required solution.

8 0
3 years ago
Read 2 more answers
Which of the following statements best summarizes the main points of the passage?
Deffense [45]

Answer:the summery is ...i really don't know the picture is blurry and I cant see can you make it clear?

Explanation:

6 0
2 years ago
Which statement is true of cooling down after physical activity? A cool-down is a period of semistrenuous activity after physica
Xelga [282]
The statement that is true of cooling down after physical activity is that you should cool down for about 5 to 10 minutes after being physically active.
8 0
3 years ago
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A child of mass m is standing at the edge of a carousel. Both the carousel and the child are initially stationary. The carousel
suter [353]

Answer:

the angular velocity of the carousel after the child has started running =

\frac{2F}{mR} \delta t

Explanation:

Given that

the mass of the child = m

The radius of the disc = R

moment of inertia I = \frac{1}{2} mR^2

change in time = \delta \ t

By using the torque around the inertia ; we have:

T = I×∝

where

R×F = I × ∝

R×F = \frac{1}{2} mR^2∝

F = \frac{1}{2} mR∝

∝ = \frac{2F}{mR}           ( expression for angular  angular acceleration)

The first equation of motion of rotating wheel can be expressed as :

\omega = \omega_0  + \alpha  \delta t

where ;

∝ = \frac{2F}{mR}    

Then;

\omega = 0+ \frac{2F}{mR} \delta t

\omega =  \frac{2F}{mR} \delta t

 

∴ the angular velocity of the carousel after the child has started running =

\frac{2F}{mR} \delta t

7 0
3 years ago
A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn
Lelechka [254]

Answer:

a. 16 s b. -1.866 kJ

Explanation:

a. Since the initial rotational speed ω₀= 3313 rev/min = 3313/60 × 2π rad/s = 346.94 rad/s. Its rotational speed becomes ω₁ = 0.75ω₀ in time t = 4 s.

We find it rotational acceleration using α = (ω₁ - ω₀)/t = (0.75ω₀ - ω₀)/t = ω₀(0.75 - 1)/t = -0.25ω₀/t = (-0.25 × 346.94 rad/s)/4 s = -21.68 rad/s².

Since the turntable stops at ω = 0, the time it takes to stop is gotten from

ω = ω₀ + αt and t = (ω - ω₀)/α = (0 - 346.94 rad/s)/-21.68 rad/s² = (-346.94/-21.68) s = 16 s.

So it takes the turntable 16 s to stop.

b. The workdone by the turntable to stop W equals its rotational kinetic energy change.

So, W = 1/2Iω² - 1/2Iω₀² = 1/2 × 0.031 kgm² × 0² - 1/2 × 0.031 kgm² × (346.94 rad/s)² = 0 - 1865.7 J = -1865.7 J = -1.8657 kJ ≅ -1.866 kJ

3 0
3 years ago
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