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OLEGan [10]
3 years ago
5

Sylvia and Jadon now want to work a problem. Imagine a puck of mass 0.5 kg moving as in the simulation. Suppose that the tension

in the string is 1.0 N, and that the radius of its circular path is 0.7 m. What will Jadon and Sylvia find for the tangential speed of the puck? m/s
Physics
1 answer:
Pavlova-9 [17]3 years ago
5 0

Answer:

1.2 m/s

Explanation:

The puck in this problem is moving with uniform circular motion, so the net force acting on it (the tension in the string) must be equal to the centripetal force.

So we can write:

T=m\frac{v^2}{r}

where:

T is the tension in the string

m is the mass of the puck

v is its tangential speed

r is the radius of the circular path

For the puck in this problem, we have:

m = 0.5 kg

T = 1.0 N

r = 0.7 m

Substituting and solving for v, we find the tangential speed:

v=\sqrt{\frac{Tr}{m}}=\sqrt{\frac{(1.0)(0.7)}{0.5}}=1.2 m/s

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Satellite A has an orbital radius 3.00 times greater than that of satellite B. Satellite B's orbital period around Earth is 120
igomit [66]

Answer:

To find the circumference (orbit) of an object, you use Pi x Diameter. 

As you have the circumference of B, you divide it by Pi to get the Diameter. 

So 120 divided by 3.141592654 = 38.2 minutes for the Diameter. 

As' radius and Diameter will be 3x greater than B. 

38.2 x 3 = 114.6 

To get back to the orbital period, times 114.6 by Pi, and you will get 360 minutes

HOPE THIS HELPS AND PLS MARK AS BRAINLIEST

THNXX :)

7 0
3 years ago
In a fireworks display, a rocket is launched from the ground with a speed of 18.0 m/s and a direction of 51.0° above the horizon
Masja [62]

Answer:38.66 m

Explanation:

Given

launch angle \theta =51^{\circ}

launch velocity u=18 m/s

center of mass continue to travel its original Path so it center of mass will be at a distance of

R=\frac{u^2\sin 2\theta }{g}

R=\frac{18^2\sin 102}{9.8}

R=32.33 m

Center of mass will be at x=32.33 m

(b)if one of the piece will be at x=26  m then other will be at

x_{com}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

x_{com}=\frac{\frac{m}{2}\cdot 26+\frac{m}{2}\cdot x_0}{\frac{m}{2}+\frac{m}{2}}

32.33=\frac{26+x_0}{2}

x_0=38.66 m

8 0
3 years ago
What is the nucleation mechanisms in which nuclei of the solid phase form in the interior of the liquid as atoms cluster togethe
pshichka [43]
I  think its q because i suck on toes 


8 0
3 years ago
A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body di
Pachacha [2.7K]

Answer:

m=146.277kg which is rounded to 146kg

Explanation:

Remember that F=ma

But F represents not 250N, but 250cos(35)N since the force is being pulled above the horizontal.

So 250cos(35)=204.7880111 approximately, and since a=1.4m/s^2, we have 204.7880111=m(1.4m/s^2). Then we divide both sides by the acceleration to get the mass. So m=146.2771508kg which the nearest number is 146kg

Mass is always in kg, unless stated otherwise.

4 0
2 years ago
Read 2 more answers
What can you infer about the density of two objects if they object remains suspended in the liquid?
ZanzabumX [31]

Answer:

Equal Densities

Explanation:

if the density of the object was greater than that of the liquid, it would sink to the bottom. if the density od the object was lesser than the liquid, it would float :)

5 0
3 years ago
Read 2 more answers
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