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3 years ago

What causes the earths plates to move

Physics

2 answers:

Norma-Jean [14]3 years ago

8 0

The intense heat in the earth's core that causes molten rock in the mantle layer to move.

Kaylis [27]3 years ago

3 0

Answer: The intense heat

Explanation:

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How much of Earth's surface is dry, ice-free land? *

AleksandrR [38]

Answer:

Less than a quarter of Earth's surfaces are dry, ice free lands

8 0

2 years ago

two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco

Reika [66]

Answer:

$t=750s$

Explanation:

The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

$\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt$

$x_f$ is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

$x_0_1=6km\\x_0_2=0$

We have $x_f_1=x_f_2$ . Thus, solving for t:

$x_0_1+v_1t=x_0_2+v_2t\\x_0_1=t(v_2-v_1)\\t=\frac{x_0_1}{v_2-v_1}\\t=\frac{6*10^3m}{28\frac{m}{s}-20\frac{m}{s}}\\t=750s$

8 0

2 years ago

Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad

love history [14]

Answer:

a) The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ , b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

$\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where:

$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

$\alpha$ - Angular acceleration, measured in radians per square second.

$r$ - Radius of rotation (Radius of a tire), measured in meters.

Given that $\alpha = 15\,\frac{rad}{s^{2}}$ and $r = 0.31\,m$ . The linear acceleration experimented by the car is:

$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

$\theta = 46.875\,rad$

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0

3 years ago

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

2 years ago

Read 2 more answers

Which of the following objects is accelerating?

jenyasd209 [6]

Answer:

Explanation:

Constant speed (without change in direction) is not accelerating. If you are slowing down, speeding up, or changing direction, you are accelerating

4 0

3 years ago