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4 years ago

A skier reaches the end of a 0.9 km track in 1.5 hours, at what speed does it travel?

Physics

2 answers:

Wewaii [24]4 years ago

7 0

Speed = Distance/Time

S = D/T

D = 0.9Km, T = 1.5hours,

S = 0.9/1.5

S = 0.6 m/s

hope it helps!

mojhsa [17]4 years ago

4 0

Answer: Speed =

distance = 0.9 * 1000 = 900 metres, because speed is measured in m/s so the km has to be converted to m

time = 1.5 * 60 = 90 seconds , because speed is measured in m/s so the time in minutes(m) has to be converted to seconds(s)

Speed =

= 10m/s

Read more on Brainly.com - brainly.com/question/11367821#readmore

Explanation: I don't know what this is but this is what I found and I hope it's helpful

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Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid

Korolek [52]

Answer:

$z= 2.5 \ m$

$z = (1 \ m , 4 \ m )$

Explanation:

From the question we are told that

Their distance apart is $d = 5.00 \ m$

The wavelength of each source wave $\lambda = 6.0 \ m$

Let the distance from source A where the construct interference occurred be z

Generally the path difference for constructive interference is

$z - (d-z) = m \lambda$

Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0

$z - (5-z) = 0$

=> $2 z - 5 = 0$

=> $z= 2.5 \ m$

Generally the path difference for destructive interference is

$|z-(d-z)| = (2m + 1)\frac{\lambda}{2}$

=> $|2z - d |= (0 + 1)\frac{\lambda}{2}$

=> $|2z - d| =\frac{\lambda}{2}$

substituting values

$|2z - 5| =\frac{6}{2}$

=> $z = \frac{5 \pm 3}{2}$

$z = \frac{5 + 3}{2}$

$z = 4\ m$

and

$z = \frac{ 5 -3 }{2}$

=> $z = 1 \ m$

=> $z = (1 \ m , 4 \ m )$

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3 years ago

A sprinter in a 100-m race accelerates uniformly for the first 71 m and then runs with constant velocity. The sprinter’s time fo

adoni [48]

Answer:

The acceleration of the sprinter is 1.4 m/s²

Explanation:

Hi there!

The equation of position of the sprinter is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the sprinter at a time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

Since the origin of the frame of reference is located at the starting point and the sprinter starts from rest, then, x0 and v0 are equal to zero:

x = 1/2 · a · t²

At t = 9.9 s, x = 71 m

71 m = 1/2 · a · (9.9 s)²

2 · 71 m / (9.9 s)² = a

a = 1.4 m/s²

The acceleration of the sprinter is 1.4 m/s²

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4 years ago

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Two students push on a 5-kg cart from opposite sides

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Answer:

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3 years ago

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A worker wants to load a 12 kg crate into a truck by sliding the crate up a straight ramp which is 2.5 m long and which makes an

olga2289 [7]

Answer:

a) The magnitude of the friction force is 55.851 newtons, b) The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

Explanation:

a) This situation can be modelled by the Principle of Energy Conservation and the Work-Energy Theorem, where friction represents the only non-conservative force exerting on the crate in motion. Let consider the bottom of the straight ramp as the zero point. The energy equation for the crate is:

$U_{g,1}+K_{1} = U_{g,2}+K_{2}+ W_{fr}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energy, measured in joules.

$W_{fr}$ - Work losses due to friction, measured in joules.

By applying the defintions of translational kinetic and gravitational potential energies and work, this expression is now expanded:

$m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta$

Where:

$m$ - Mass of the crate, measured in kilograms.

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

$y_{1}$ , $y_{2}$ - Initial and final height of the crate, measured in meters.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the crate, measured in meters per second.

$\mu_{k}$ - Kinetic coefficient of friction, dimensionless.

$\theta$ - Ramp inclination, measured in sexagesimal degrees.

The equation is now simplified and the coefficient of friction is consequently cleared:

$y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot \cos \theta$

$\mu_{k} = \frac{1}{\cos \theta} \cdot \left[y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) \right]$

The final height of the crate is:

$y_{2} = (1.6\,m)\cdot \sin 30^{\circ}$

$y_{2} = 0.8\,m$

If $\theta = 30^{\circ}$ , $y_{1} = 0\,m$ , $y_{2} = 0.8\,m$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{1} = 5\,\frac{m}{s}$ and $v_{2} = 0\,\frac{m}{s}$ , the coefficient of friction is:

$\mu_{k} = \frac{1}{\cos 30^{\circ}}\cdot \left\{0\,m-0.8\,m+\frac{1}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}\cdot \left[\left(5\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] \right\}$

$\mu_{k} \approx 0.548$

Then, the magnitude of the friction force is:

$f =\mu_{k}\cdot m\cdot g \cdot \cos \theta$

If $\mu_{k} \approx 0.548$ , $m = 12\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\theta = 30^{\circ}$ , the magnitude of the force of friction is:

$f = (0.548)\cdot (12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}$

$f = 55.851\,N$

The magnitude of the force of friction is 55.851 newtons.

b) The energy equation of the situation is:

$m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta$

$y_{1}+\frac{1}{2\cdot g}\cdot v_{1}^{2} =y_{2} + \frac{1}{2\cdot g}\cdot v_{2}^{2} + \mu_{k}\cdot \cos \theta$

Now, the final speed is cleared:

$y_{1}-y_{2}+ \frac{1}{2\cdot g}\cdot v_{1}^{2} -\mu_{k}\cdot \cos \theta= \frac{1}{2\cdot g}\cdot v_{2}^{2}$

$2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta) + v_{1}^{2} = v_{2}^{2}$

$v_{2} = \sqrt{2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta)+v_{1}^{2}}$

Given that $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 0.8\,m$ , $y_{2} = 0\,m$ , $\mu_{k} \approx 0.548$ , $\theta = 30^{\circ}$ and $v_{1} = 0\,\frac{m}{s}$ , the speed of the crate at the bottom of the ramp is:

$v_{2}=\sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0.8\,m-0\,m-(0.548)\cdot \cos 30^{\circ}]+\left(0\,\frac{m}{s} \right)^{2}}$

$v_{2}\approx 2.526\,\frac{m}{s}$

The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

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3 years ago