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cricket20 [7]
3 years ago
6

Help plz anyone ?????

Physics
1 answer:
Alexus [3.1K]3 years ago
3 0

Answer:

Most likely, it will be harder to get strong magnets to change phase because they have more density.

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How do physicists distinguish one type of electromagnetic wave from another?
GenaCL600 [577]
A wave is characterized by the cyclic occurrences of crests and troughs. Wavelengthis defined as the distance between two consecutive troughs or two crests and the Frequency is defined as the number of cycles that pass through a point per second
7 0
3 years ago
A brick in the shape of a cube with sides 10 cm has a density of 2500 kg/m^3. What is its weight? a.) 250 N
Arisa [49]

Answer:

c.) 25 N

Explanation:

 We find the volume of the brick, knowing that the volume of a cube is given by the formula:

l=0,1 metros \\V=l^{3}\\V=(0,1\: \: m)^{3}=0,001\: \: m^{3}

being l the side of the cube, which in this case is 10 cm or 0,1 m. Now we find the mass of the object, knowing the density and the Volume of the cube:

m=V*d\\m=(0,001 \:\:m^{3})(2500 \: \: \frac{Kg}{m^{3}})=2,5 \:\: Kg

We find the weight by multiplying the mass of the object with the gravity constant.

W=m*g=(2.5 \:Kg)*(9,81 \:m/s^{2} )=24,5 N\approx25\: N

8 0
3 years ago
plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 85.0 nC . The plates are in vacuu
tensa zangetsu [6.8K]

Answer:

12500 V

Explanation:

The electric field in the gap of a parallel-plate capacitor is uniform, so the following relationship between electric field strength, potential difference and distance can be used:

\Delta V = E d

where

\Delta V is the potential difference between the plates

E is the electric field strength

d is the distance between the plates

For the capacitor in this problem, we have

E=5.00\cdot 10^6 V/m

d = 2.50 mm = 2.50\cdot 10^{-3} m

Substituting, we find

\Delta V = (5.00\cdot 10^6)(2.50\cdot 10^{-3})=12500 V

3 0
3 years ago
What is the mechanical advantage of the screw shown below? O A. 14.1 O B. 2 O C. 12.6 O D. 8.2.​
Vilka [71]

Answer: C. 12.6

Explanation: 2*pi*1.8= 11.304

11.304/0.9= 12.56

3 0
2 years ago
A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude
Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0
3 years ago
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