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miss Akunina [59]
3 years ago
15

Which artist of the northern European Renaissance, shown in this self-portrait, was also a block printer and engraver?

Physics
2 answers:
kicyunya [14]3 years ago
6 0
It's very hard to see the self-portrait, so I can't identify him.
zloy xaker [14]3 years ago
4 0

I just took the test with my son, the answere is Albrecht Durer. I hope this helps, you can check it on google :)

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Your average speed on the first half of a car trip is 69.0 km/h. How fast do you have to drive on the second half of the trip to
vova2212 [387]

Answer:

13 km/h

Explanation:

Average speed = distance/time

Let the total distance and total time taken for the whole trip be d km and t hours respectively

Average speed for the whole trip = 82 km/h

d = 82t

The distance covered in the first half = d1/2

Time taken = t/2

Average speed = 69 km/h

69 = d1/2 ÷ t/2

d1 = 69t

The distance covered in the second half = d2/2

Time taken = t/2

Let the average sly for the see half be A

A = d2/2 ÷ t/2

d2 = At

d = d1 + d2

82t = 69t + At

At = 82t - 69t

At = 13t

A = 13t/t = 13 km/h

3 0
3 years ago
1. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott
Kobotan [32]
 0.4 x 18 = 7.2 kg m/s

The momentum of the bottle after being hit is 0.2 x 25 = 5 kg m/s

7.2 - 5 = 2.2 kg m/s is the motmentum of the ball now 

 the velocity  is 2.2/0.4 = 5.5 m/s
7 0
3 years ago
Suppose that the dipole moment associated with an iron atom of an iron bar is 2.8 × 10-23 J/T. Assume that all the atoms in the
masya89 [10]

To solve this exercise it is necessary to apply the equations related to the magnetic moment, that is, the amount of force that an image can exert on the electric currents and the torque that a magnetic field exerts on them.

The diple moment associated with an iron bar is given by,

\mu = \alpha *N

Where,

\alpha = Dipole momento associated with an Atom

N = Number of atoms

\alpha y previously given in the problem and its value is 2.8*10^{-23}J/T

L = 5.8cm = 5.8*10^{-2}m

A = 1.5cm^2 = 1.5*10^{-4}m^2

The number of the atoms N, can be calculated as,

N = \frac{\rho AL}{M_{mass}}*A_n

Where

\rho = Density

M_{mass} = Molar Mass

A = Area

L = Length

A_n =Avogadro number

N = \frac{(7.9g/cm^3)(1.5cm)(5.8cm^2)}{55.9g/mol}(6.022*10^{23}atoms/mol)

N = 7.4041*10^{23}atoms

Then applying the equation about the dipole moment associated with an iron bar we have,

\mu = \alpha *N

\mu = (2.8*10^{-23})*(7.4041*10^{23})

\mu = 20.72Am^2

PART B) With the dipole moment we can now calculate the Torque in the system, which is

\tau = \mu B sin(90)

\tau = (20.72)(2.2)

\tau = 45.584N.m

<em>Note: The angle generated is perpendicular, so it takes 90 ° for the calculation made.</em>

3 0
3 years ago
The voltage supplied to a circuit is 17 V and the current running through is 10 A. What is the power generated?
Lelu [443]

Answer:

170 W

Explanation:

Applying

P = VI.................... Equation 1

Where P = Power generated in watt, V = Voltage supplied to the circuit, I = Current running through the circuit.

From the question,

Given: V = 17 V, I = 10 A

Substitute these values into equation 1

P = (17×10)

P = 170 Watt.

Hence the power generated is 170 W.

The right option is A. 170 W

3 0
2 years ago
______ is the best way to determine your own correct intensity level.
scZoUnD [109]
<span>Self-monitoring would be the best way to </span><span>determine your own correct intensity level. I hope this helps! <3
</span>
5 0
3 years ago
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