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Amanda [17]
3 years ago
12

Find the pressure exerted by a 3000 N crate that has an area of 2m squared

Physics
1 answer:
ipn [44]3 years ago
8 0
Pressure = Force/ Area = 3000/2 = 1500 pascal.
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What are two advantages of using renewable resources?
Pavel [41]
One advantage is that whatever resource it is, it will never run out and you wont have to worry about not having it. A second is that there is going to be enough for everyone to use however much they want without there having to be a limit on how much you use.
3 0
3 years ago
A piece of plastic with a mass of 15 g is
satela [25.4K]

Answer:

The density of plastic is equal to 0.6 g/mL.

Explanation:

Given that,

The mass of piece of plastic, m = 15 g

It is placed in a graduated cylinder. The water  level in the graduated cylinder rises from  30 mL to 55 mL when the plastic is added.

We need to find the density of plastic.

Rise in volume = 55 mL - 30 mL

= 25 mL

The density of an object is given by :

d=\dfrac{m}{V}\\\\d=\dfrac{15\ g}{25\ mL}\\\\d=0.6\ g/mL

So, the density of plastic is equal to 0.6 g/mL.

5 0
3 years ago
A wall clock has a second hand 22.0 cmcm long. For related problem-solving tips and strategies, you may want to view a Video Tut
olga55 [171]

Answer:

Explanation:

The tip of the second hand moves on a circular path having radius equal to .22 m . Redial acceleration is given by the expression

ω²R where ω is angular velocity and R is radius of the circular path .

angular velocity of second hand = 2π / T where T is time period of circular motion . For second hand it is 60 s.

ω = 2π / T

= 2π / 60

= .1047

angular acceleration =  .1047² x .22

= 2.41 x 10⁻³ rad / s² .

8 0
3 years ago
Scientists discover two planets orbiting a distant star. The average distance from the star to Planet A is 8 AU, and it takes 71
Mama L [17]

Answer: 11.2 AU

Explanation:

Applying Kepler's 3rd law, we can find out the average distance of planet B to the star.

This Law states that for planets orbiting a same star, there exists a fixed relationship between the average distance to  the star, and the period of his orbit  around it, as follows:

K = T² / d³

So , in this case, we can write:

(da)³ / Ta² = (db)³ / (Tb)²

Solving for db:

db = ∛8³.(1170)² / 710² = ∛1390.4 = 11.2 AU

5 0
3 years ago
A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The
lina2011 [118]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s  

Explanation:

1) We can find the speed of the object by conservation of energy:

E_{i} = E_{f}

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}                              

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s      

 

3) When the object is at the equilibrium position, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s

I hope it helps you!                                                                                        

8 0
3 years ago
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