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2 years ago

According to the author, Picasso’s artistic achievements were in large part the result of his:

Physics

1 answer:

Harrizon [31]2 years ago

4 0

Answer:

Picasso’s artistic achievements were in large part the result of his contribution to help bring the Nazi' devastating casual bombing and Spanish civil war in Guernica to the world's attention through his paintings.

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Diffraction spreading for a flashlight is insignificant compared with other limitations in its optics, such as spherical aberrat

lyudmila [28]

Answer:

$\theta_{min} = 1.21 \times 10^{-5}\ rad$

Explanation:

given,

diameter of the beam (d)= 5.85 cm

= 0.0585 m

average wavelength of the(λ) = 580 n m

angle of of spreading = ?

according to the Rayleigh Criterion the minimum angular spreading, for a circular aperture, is

$\theta_{min} = 1.22\ \dfrac{\lambda}{d}$

$\theta_{min} = 1.22\ \dfrac{580 \times 10^{-9}}{0.0585}$

$\theta_{min} = 1.22\times 9.145 \times 10^{-6}$

$\theta_{min} = 1.21 \times 10^{-5}\ rad$

the minimum angle of spreading is $\theta_{min} = 1.21 \times 10^{-5}\ rad$

5 0

2 years ago

You are at a stop light in your car, stuck behind a red light. Just before the light is supposed to change, a fire engine comes

nikdorinn [45]

Answer:

The frequency of the sound you will hear is 713.85 Hz

Explanation:

Given;

speed of your car, $v_s$ = 85.0 km/h

frequency of the siren, f = 665 Hz

Speed of sound in air, v = 345 m/s

The frequency of the sound you hear, can be calculated as;

$f' = f(\frac{v}{v-v_s})$

Convert the speed of the car to m/s

$85 \ km/h =\frac{85 \ km}{h} (\frac{1000\ m}{1 \ km})(\frac{1 \ h}{3600 \ s} ) = 23.61 \ m/s$

$f' = f(\frac{v}{v-v_s} )\\\\f' = 665(\frac{345}{345-23.61} )\\\\f' = 665 (1.07346)\\\\f' = 713.85 \ Hz$

Therefore, the frequency of the sound you will hear is 713.85 Hz

7 0

2 years ago

A square loop of wire surrounds a solenoid. The side of the square is 0.1 m, while the radius of the solenoid is 0.025 m. The sq

eduard

Answer:

Number of turns per metre, n= 500/0.3= (5000/3)m^-1

Cross sectional areaof the square loop of wire, A= (0.1^2)m^2= 0.01m^2dB/dt= μn(dI/dt)= (4.00π x10^-7)(5000/3)(0.7)= 1.46608x10^-3T/s

The induced emf in the square loop of wire, ε= the rate of change of magnetic flux of the square loop of wire(dΦ/dt)= A(dB/dt)= (0.01)(1.46608x10^-3)= 0.0146608x10^-5VA

current flows in the square loop of wire since a potential difference(induced emf in this case) exists. Its magnitude,

I= ε/R where R is the resistance of the square loop of wire.

I= (0.0146608x10^-5)/30= 4.89x10^-7A

6 0

2 years ago

Read 2 more answers

The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the ob

givi [52]

Answer:

Charge, $Q=1.56\times 10^{-8}\ C$