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3 years ago

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Two ice skaters suddenly push off against one another starting from a stationary position. The 45-kg skater acquires a speed of

0.375 m/s relative to the ice. What speed does the 60-kg skater acquire relative to the ice?

1 answer:

timama [110]3 years ago

5 0

Answer:

Speed does the 60-kg skater acquire relative to the ice = 0.281 m/s

Explanation:

Momentum is conserved in this case.

Initial momentum = Final momentum.

Initial momentum = Initial momentum of skater 1 + Initial momentum of skater 2

Initial momentum = 45 x 0 + 60 x 0 = 0 kg m/s

Final momentum = 45 x 0.375 + 60 x u = 16.875 + 60 u

Here we need to find u.

We have

0 = 16.875 + 60 u

u = -0.281 m/s

Speed does the 60-kg skater acquire relative to the ice = 0.281 m/s

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Given required solution

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The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

A)

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

B)

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$

So we have:

$F_{2x}=F_2 cos \theta = (15)(cos 120)=-7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 120)=13.0 N$

So, the components of the resultant this time are:

$F_x = F_{1x}+F_{2x}=20-7.5 = 12.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{13.5^2+13.0^2}=18.7 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{13.5})=43.9^{\circ}$

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3 years ago

Which type of energy is stored in chemical compounds and released in chemical reactions?

astraxan [27]

Answer:

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Explanation:

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3 years ago

A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra

Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

$sin(\alpha) = \frac{Vyi}{Vi}$

$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

$Vy(t) = Vyi -g.t$

where g is the gravity acceleration

$Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t$

This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

$Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}$

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

$Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s$

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

$Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} } .(0.675s)^{2} \\Y(0.675s) =2.236 m$

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

$Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m$

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

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3 years ago