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1 year ago

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A cyclist moving towards right with an acceleration of 4m/s² at t = 0 he has travelled 5 m moving towards the right at 15 m/s wh

at will be his final position

1 answer:

ira [324]1 year ago

4 0

Answer:

$x=2t^2+15t+5$

Explanation:

$x=\frac{1}{2}at^2+v_0t+x_0$

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The distance of the earth from the sun is 93 000 000 miles. if there are 3.15 × 107 s in one year, find the speed of the earth i

faltersainse [42]

The angular velocity of the orbit about the sun is:

w = 1 rev / year = 1 rev / 3.15 × 10^7 s

Now in 1 rev there is 360° or 2π rad, therefore:

w = 2π rad / 3.15 × 10^7 s

To convert in linear velocity, multiply the rad /s by the radius:

v = (2π rad / 3.15 × 10^7 s) * 93,000,000 miles

<span>v = 18.55 miles / s = 29.85 km / s</span>

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2 years ago

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The mass per unit length of a 14-gauge copper wire is 18.5 g/m. If the wire is placed running along the horizontal x-axis (east-

zmey [24]

Answer:

0.6295 A

Explanation:

I=mg/BL put values in this formula.

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1 year ago

Astronomers initially had difficulty identifying the emission lines in quasar spectra at optical wavelengths because

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No one expected violet & ultraviolet spectral lines to be shifted towards the red.

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2 years ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

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1 year ago

The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope. The sl

KonstantinChe [14]

Answer:

Acceleration, $a=1.2\ m/s^2$

Explanation:

Given that,

The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope, m = 100 kg

Force exerted by the doges on the rope attached to the front sled, F = 240 N

To find,

The acceleration of the sleds.

Solution,

Let a is the acceleration of the sleds. The product of mass and acceleration is called force. Its expression is given by :

F = ma

$a=\dfrac{F}{m}$

$a=\dfrac{240\ N}{2\times 100\ kg}$ (m = 2m)

$a=1.2\ m/s^2$

So, the acceleration of the sleds is $1.2\ m/s^2$ .

6 0

1 year ago