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3 years ago

Please help :)

Physics

1 answer:

Orlov [11]3 years ago

8 0

Answer:

<em>3.5 A</em>

Explanation:

Current [A] =Electrical power [W] ÷ Potential difference [V]

KW TO W = ×1000

hence,

800W÷230V= 3.5A (1 d.p)

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A meteorite has a speed of 90.0 m/s when 850 km above the Earth. It is falling vertically (ignore air resistance) and strikes a

Schach [20]

Before it hits the sand bed, the meteorite is accelerating uniformly with $g=9.80\dfrac{\rm m}{\mathrm s^2}$ , so that its speed $v$ satisfies

$v^2-{v_0}^2=-2g\Delta y$

where $v_0=90.0\dfrac{\rm m}{\rm s}$ is its initial speed and $\Delta y=(0-850)\,\mathrm{km}=-850\,\mathrm{km}$ is its change in altitude. Notice that we're taking the meteorite's starting position in the atmosphere to be the origin, and the downward direction to be negative. Now,

$v^2-\left(-90.0\dfrac{\rm m}{\rm s}\right)^2=-2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(-850,000\,\mathrm m)\implies\boxed{v=4080\dfrac{\rm m}{\rm s}}$

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3 years ago

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30˚ angle. The block’s initial speed is 10 m/s. What vert

WARRIOR [948]

Answer:

The wood block reaches a height of 4.249 meters above its starting point.

Explanation:

The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.

$\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta$

$\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta$

$\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h$

$\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h$

$h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g}$ (1)

Where:

$h$ - Maximum height of the wood block, in meters.

$v$ - Initial speed of the block, in meters per second.

$\mu_{k}$ - Kinetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$m$ - Mass, in kilograms.

$s$ - Distance travelled by the wood block along the wooden ramp, in meters.

$\theta$ - Inclination of the wooden ramp, in sexagesimal degrees.

If we know that $v = 10\,\frac{m}{s}$ , $\mu_{k} = 0.20$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the height reached by the block above its starting point is:

$h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h = 4.249\,m$

The wood block reaches a height of 4.249 meters above its starting point.

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3 years ago

A world-class sprinter can reach a top speed (of about 11.8 m/s ) in the first 19.0 m of a race. what is the average acceleratio

Whitepunk [10]

We use the kinematic equation of motion, to calculate the average acceleration

$v^{2} =u^{2} +2a (h-h_{0} )$

or $a=\frac{v^{2} -v_{0}^2}{2(h-h_{0}) }$

Here, $v_{0}$ is initial speed , v is final speed h is covered distance $h_{0}$ is initial position.

Given $v= 11.8 m/s$ , $v_{0} =0$ , $h = 19.0 m$ and $h_{0} =0$ .

Substituting these values in above equation we get,

$a=\frac{(11.8)^2-0}{2(19.0-0)} \\\\\ a =3.66 m/s^2$

Thus, the average acceleration of the sprinter is $3.66 m/s^2$ .

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3 years ago

A ball is thrown straight upward and returns to the thrower's hand after 3 seconds in the air. A second ball is thrown at an ang

Katena32 [7]

Explanation:

Given that the first ball is thrown vertically and upward.

since it reaches the throwers hand after 3 seconds, then the time taken to reach maximum height is

= T/2

=3/2

=1.5 seconds

Also, the projection angle for the second ball is 30°

we know that at maximum height vertical velocity is 0

therefore the final velocity of the first ball is

Vy= 0

Also, since the first ball is thrown upward the horizontal velocity is 0

Vx=0

from the first law of motion

Vy=u+at

Vy=0, a= -g

Uy=gt

Uy=9.81*1.5

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since both reach the same height, they will have the same vertical velocity

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Vo=Uy/sin∅

Vo= 14.7/sin 30

Vo=29.4m/s

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4 years ago

The roller coaster car has a mass of 700 kg, including its passenger. If it is released from rest at the top of the hill A, dete

sweet [91]

Answer:

$h = 18.75 m$

Now when it will reach at point B then its normal force is just equal to ZERO

$N_B = 0$

$F_n = 1.72 \times 10^4$

Explanation:

Since we need to cross both the loops so least speed at the bottom must be

$v = \sqrt{5 R g}$

also by energy conservation this is gained by initial potential energy

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

so we will have

$\sqrt{2gh} = \sqrt{5Rg}$

now we have

$h = \frac{5R}{2}$

here we have

R = 7.5 m

so we have

$h = \frac{5(7.5)}{2}$

$h = 18.75 m$

Now when it will reach at point B then its normal force is just equal to ZERO

$N_B = 0$

now when it reach point C then the speed will be

$mgh - mg(2R_c) = \frac{1}{2]mv_c^2$

$v_c^2 = 2g(h - 2R_c)$

$v_c = 13.1 m/s$

now normal force at point C is given as

$F_n = \frac{mv_c^2}{R_c} - mg$

$F_n = \frac{700\times 13.1^2}{5} - (700 \times 9.8)$

$F_n = 1.72 \times 10^4$

7 0

4 years ago