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3 years ago

1. Mendeleev arranged the known chemical elements in a table according to

Physics

2 answers:

marin [14]3 years ago

8 0

Answer:

Explanation:

Mendleev arranged the known chemical elements in the periodic table according to increasing atomic no.

V125BC [204]3 years ago

3 0

Answer:

the answer is A

Explanation:

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A rock tied to the end of a string moves in a circle at a constant speed of 2.5 m/s and experiences an acceleration of 4.0 m/s.

artcher [175]

The radius of the circular motion at the given speed is 1.56 m.

The given parameters;

speed of the rock, v = 2.5 m/s
acceleration of the rock, a = 4 m/s²

The radius of the circular motion is calculated by using centripetal acceleration formula as follows;

$a_c = \frac{v^2}{r} \\\\r = \frac{v^2}{a_c} \\\\r = \frac{2.5^2}{4} \\\\r = 1.56 \ m$

Thus, the radius of the circular motion at the given speed is 1.56 m.

Learn more about centripetal acceleration here: brainly.com/question/79801

8 0

2 years ago

24 POINTS!!!!!!!!!!!!

Karo-lina-s [1.5K]

Answer: Option (c) is the correct answer.

Explanation:

When the child is tossed up into the air then she gains kinetic energy as the child has moved from its initial position.

It is given that mass is 20 kg, velocity is $5 m/s^{2}$ , and height is 2 m.

Calculate the kinetic energy of child as follows.

kinetic energy = $\frac{1}{2}mv^{2}$

= $\frac{1}{2}20 kg \times (5 m/s)^{2}$

= $10 kg \times 25 m^{2}/s^{2}$

= $250 kg m^{2}/s^{2}$

Also, when child falls off the ground then she will have gravitational potential energy.

Calculate gravitational potential energy of child as follows.

Potential energy = m × g × h

= $20 kg \times 9.8 m/s^{2} \times 2 m$

= $392 kg m^{2}/s^{2}$

3 0

3 years ago

Read 2 more answers

In part one of this experiment, a 0.20 kg mass hangs vertically from a spring and an elongation below the support point of the s

statuscvo [17]

To solve this problem it is necessary to apply the concepts related to Hooke's Law as well as Newton's second law.

By definition we know that Newton's second law is defined as

$F = ma$

m = mass

a = Acceleration

By Hooke's law force is described as

$F = k\Delta x$

Here,

k = Gravitational constant

x = Displacement

To develop this problem it is necessary to consider the two cases that give us concerning the elongation of the body.

The force to keep in balance must be preserved, so the force by the weight stipulated in Newton's second law and the force by Hooke's elongation are equal, so

$k\Delta x = mg$

So for state 1 we have that with 0.2kg there is an elongation of 9.5cm

$k (9.5-l)=0.2*g$

$k (9.5-l)=0.2*9.8$

For state 2 we have that with 1Kg there is an elongation of 12cm

$k (12-l)= 1*g$

$k (12-l)= 1*9.8$

We have two equations with two unknowns therefore solving for both,

$k = 3.136N/cm$

$l = 8.877cm$

In this way converting the units,

$k = 3.136N/cm(\frac{100cm}{1m})$

$k = 313.6N/m$

Therefore the spring constant is 313.6N/m

3 0

3 years ago

A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When

natita [175]

Answer:

a) $\mu_{k} = 0.704$ , b) $R = 0.312\,m$

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s$

$\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}$

$\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}$

$\mu_{k} = 0.704$

b) The speed of the block is determined by using the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = x\cdot \sqrt{\frac{k}{m} }$

$v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }$

$v \approx 9.829\,\frac{m}{s}$

The radius of the circular loop is:

$\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}$

$\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}$

$R = 0.312\,m$

5 0

4 years ago

A Chinook salmon can jump out of water with a speed of 6.70 m/s6.70 m/s . How far horizontally dd can a Chinook salmon travel th

aksik [14]

Answer:

R = 3.88 m

Explanation:

As the Chinook salmon leaves the water till it gets back into the water it is performing a projectile motion with the following parameters:

V₀ = Launch Speed = 6.7 m/s

θ = Launch Angle = 29°

R= Range of Projectile= Horizontal Distance Covered by Chinook salmon= ?

The value of the range of a projectile is given by the following formula:

R = (V₀² Sin 2θ)/g

R = [(6.7 m/s)² Sin {(2)(29°)}/(9.8 m/s²)]

R = [(6.7 m/s)² Sin (58°)/(9.8 m/s²)]

R = 3.88 m

8 0

4 years ago