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3 years ago

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

Read 2 more answers

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