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3 years ago

4. Tectonic plates are located on the and . 6. Inner core is made of _ and __.

Physics

2 answers:

o-na [289]3 years ago

7 0

4. Tectonic plates are located on the earth's crust and upper most mantle(lithosphere).

6.Innercore is made of iron and nickel.

DanielleElmas [232]3 years ago

6 0

Answer:

Tectonic plates are located in the lithosphere.
Inner core is made of iron and nickel.

Hope it helps!

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Un cuerpo de 20 kg es lanzado verticalmente hacia arriba con una velocidad de 30 m/s. ¿Cuánto tiempo tardará el cuerpo en alcanz

nata0808 [166]

Answer:

tu hermana con salsa de concha

4 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

bothersome feature of many physical measurements is the presence of a background signal (commonly called "noise"). In Part 2.2.4

serious [3.7K]

Answer:

Incomplete question: "A signal of 20.7 mV is measured at a distance of 29 mm and 15.8 mV is measured at 32.5 mm. Correct the data for background and normalize the data to the maximum value. What is the normalized corrected value at 32.5 mm?"

The normalized corrected value at 32.5 mm is 0.1638

Explanation:

The corrected light measurement at 29 mm is equal to:

20.7 - 5.1 = 15.6 mV

The corrected light measurement at 32.5 mm is equal to:

15.6 - 5.1 = 10.5 mV

To normalize the data to its maximum value means that the maximum value must be calculated and the data must be scaled using that value, as in this case the maximum value is 15.6 mm, then the normalized corrected value at 32.5 mm is equal to:

10.5 * 15.6 = 163.8 = 0.1638

6 0

4 years ago

A daredevil is shot out of a cannon at 31.4 ◦ to the horizontal with an initial speed of 25.9 m/s. A net is positioned a horizon

Tanzania [10]

Answer:

The net should be placed at a height of 8.45 m

Explanation:

In order to calculate the height you have to apply the equations of <em>projectil motion.</em>

For x-axis, the position at a time t is given by:

Xf=Xo +(VoCosα)t (I)

Where Xf is the final position, Xo is the initial position, Vo is the initial speed and t is the time.

For the y-axis, the position at a time t is given by:

Yf=Yo + (VoSinα)t - 0.5g $t^{2}$ (II)

Where Yf is the final position, Yo is the initial position, Vo is the initial speed, t is the time and g is the acceleration of gravity.

For both equations, ∝ is the angle formed by the cannon and the horizontal

The height where the net should be placed in order to catch the daredevil is given by replacing the time when the daredevil will be in the horizontal position of 39.7 m in the equation of position for the y-axis.

So, you have to calculate the time using (I)

Let the system of reference be placed at x=o and y=o

Xf= 0 + 25.9Cos(31,4◦)t

Xf=39.7

39.7=22.1t

Dividing both sides by 22.1 to find the value of t:

t=1.79 seconds

Replacing the time in equation (II) to find the height:

Yf=0 +25.9Sin(31,4◦)(1.79) - 0.5(9.8) $1.79^{2}$

Yf= 8.45 m

So the height is 8.45 m

4 0

4 years ago

Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. A particle with charge q i

sergejj [24]

Answer: in this question, the only charge in the cavity is Q. Inside the conducting spherical shell, the electric field is zero.

While outside the shell, the electric field is given by: k(q + Q)/r²

Where;

K= is a constant which is given as, 8.99 x 10^9 N m² / C².

Q= source charge which creates the electric field

q= is the test charge which is used to measure the strength of the electric field at a given location.

r= is the radius

Explanation: Inside the conducting spherical shell, the electric field is zero since the Electric field vanishes everywhere inside the volume of a good conductor.

8 0

3 years ago

4. Tectonic plates are located on the __ and __. 6. Inner core is made of ___ and ____.

4. Tectonic plates are located on the and . 6. Inner core is made of _ and __.