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tangare [24]
3 years ago
7

Who is the founding father of modern psychology?

Physics
2 answers:
zmey [24]3 years ago
7 0

Answer:

Wilhelm Wundt

Zepler [3.9K]3 years ago
4 0

Answer:

Sigmund Freud

Explanation:

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Before the experiment, the total momentum of the system is 2.5 kg m/s to the right and the kinetic energy is 5J. After the exper
finlep [7]

Answer:

Option (b) is correct.

Explanation:

Elastic collision is defined as a collision where the kinetic energy of the system remains same. Both linear momentum and kinetic energy are conserved in case of an elastic collision.

Inelastic collision is defined as a collision where kinetic energy of the system is not conserved whereas the linear momentum is conserved. This loss of kinetic energy may due to the conversion to thermal energy or sound energy or may be due to the deformation of the materials colliding with each other.

As given in the problem, before the collision, total momentum of the system is 2.5~Kg~m~s^{-1} and the kinetic energy is 5~J. After the collision, the total momentum of the system is  2.5~Kg~m~s^{-1}, but the kinetic energy is reduced to 4~J. So some amount of kinetic energy is lost during the collision.

Therefor the situation describes an inelastic collision (and it could NOT be elastic).

5 0
3 years ago
What is the resistance of a 3.5 m copper wire (Rho= 1.7x10-8 Ohm·m) that 1 point
VikaD [51]

Answer:

(D)

Explanation:

Given :

l=3.5 m

A=5.26*10^{-6} m^{2}

p=1.7*10^{-8}  ohm.m

Resistance can be calculated as :

R=p\frac{l}{A} \\R=1.7*10^{-8} \frac{3.5}{5.26*10^{-6} }

R=\frac{5.95*10^{-2} }{5.26} \\R=1.13*10^{-2}

Resistance of the wire will be 1.1×10^{-2} ohms

Option D is correct

4 0
2 years ago
How is electrolysis used in order to prevent materials from corrosion or rusting?​
KengaRu [80]

Answer:

We use electrolysis to prevent a material from rusting,

The metal forms a coating around the material and hence prevents any contact between the material and the environment

This process also gives us the physical strength of the material and the aesthetic properties of the coated metal

the metal commonly used to coat the object is Zinc and the process is called galvanisation

6 0
3 years ago
Amplitude modulation is used in _____.
spayn [35]

Amplitude modulation is a modulation technique used in electronic communication, most commonly for transmitting information via a radio carrier wave. In amplitude modulation, the signal strength of the carrier wave is varied in proportion to that of the message signal being transmitted. The message signal is, for example, a function of the sound to be reproduced by a loudspeaker, or the light intensity of pixels of a television screen. This technique contrasts with frequency modulation, in which the frequency of the carrier signal is varied, and phase modulation, in which its phase is varied.

AM was the earliest modulation method used to transmit voice by radio. It was developed during the first quarter of the 20th century beginning with Landell de Moura and Reginald Fessenden's radiotelephone experiments. It remains in use today in many forms of communication; for example it is used in portable two-way radios, VHF aircraft radio, citizens band radio, and in computer modems in the form of QAM. AM is often used to refer to mediumwave AM radio broadcasting.

6 0
3 years ago
Read 2 more answers
An object with mass 100 kg moved in outer space. When it was at location <8, -30, -4> its speed was 5.5 m/s. A single cons
Alenkasestr [34]

Answer:

v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s

Explanation:

We can solve this problem using the kinematic relations, we have a three-dimensional movement, but we can work as three one-dimensional movements where the only parameter in common is time (a scalar).

X axis.

They indicate the initial position x = 8 m, its initial velocity v₀ = 5.5 m / s, the force Fx₁ = 220 N x₁ = 14 m, now the force changes to Fx₂ = 100 N up to the point xf = 17 m. The final speed is wondered.

As this movement is in three dimensions we must find the projection of the initial velocity in each axis, for this we can use trigonometry

the angle fi is with respect to the in z and the angle theta with respect to the x axis.

               sin φ = z / r

                Cos φ = r_p / r

               z = r sin φ

               r_p = r cos φ

the modulus of the vector r can be found with the Pythagorean theorem

               r² = (x-x₀) ² + (y-y₀) ² + (z-z₀) ²

               r² = (14-8) 2 + (-21 + 30) 2+ (-7 +4) 2

               r = √126

               r = 11.23 m

Let's find the angle with respect to the z axis (φfi)

                φ = sin⁻¹ z / r

                φ = sin⁻¹ ( \frac{-7+4}{11.23} )

                φ = 15.5º

Let's find the projection of the position vector (r_p)

                r_p = r cos φ

                r_p = 11.23 cos 15.5

                r_p = 10.82 m

This vector is in the xy plane, so we can use trigonometry to find the angle with respect to the x axis.

                 cos θ = x / r_p

                 θ = cos⁻¹ x / r_p

                 θ = cos⁻¹ ( \frac{14-8}{10.82})  

                 θ = 56.3º

taking the angles we can decompose the initial velocity.

               sin φ = v_z / v₀

               cos φ = v_p / v₀

               v_z = v₀ sin φ

               v_z = 5.5 sin 15.5 = 1.47 m / z

               v_p = vo cos φ

               v_p = 5.5 cos 15.5 = 5.30 m / s

                 

               cos θ = vₓ / v_p

                sin θ = v_y / v_p

                vₓ = v_p cos θ

                v_y = v_p sin θ

                vₓ = 5.30 cos 56.3 = 2.94 m / s

                v_y = 5.30 sin 56.3 = 4.41 m / s

 

                 

we already have the components of the initial velocity

                v₀ = (2.94 i ^ + 4.41 j ^ + 1.47 k ^) m / s

let's find the acceleration on this axis (ax1) using Newton's second law

                Fₓx = m aₓ₁

                aₓ₁ = Fₓ / m

                aₓ₁ = 220/100

                aₓ₁ = 2.20 m / s²

Let's look for the velocity at the end of this interval (vx1)

Let's be careful if the initial velocity and they relate it has the same sense it must be added, but if the velocity and acceleration have the opposite direction it must be subtracted.

                 vₓ₁² = v₀ₓ² + 2 aₓ₁ (x₁-x₀)

                 

let's calculate

                 vₓ₁² = 2.94² + 2 2.20 (14-8)

                 vₓ₁ = √35.04

                 vₓ₁ = 5.92 m / s

to the second interval

they relate it to xf

                   aₓ₂ = Fₓ₂ / m

                   aₓ₂ = 100/100

                   aₓ₂ = 1 m / s²

final speed

                    v_{xf}²  = vₓ₁² + 2 aₓ₂ (x_f- x₁)

                    v_{xf}² = 5.92² + 2 1 (17-14)

                    v_{xf} =√41.05

                    v_{xf} = 6.41 m / s

We carry out the same calculation for each of the other axes.

Axis y

acceleration (a_{y1})

                      a_{y1} = F_y / m

                      a_{y1} = 460/100

                      a_[y1} = 4.60 m / s²

the velocity at the end of the interval (v_{y1})

                      v_{y1}² = v_{oy}² + 2 a_{y1{ (y₁ -y₀)

                      v_{y1}2 = 4.41² + 2 4.60 (-21 + 30)

                      v_{y1} = √102.25

                       v_{y1} = 10.11 m / s

second interval

acceleration (a_{y2})

                      a_{y2} = F_{y2} / m

                      a_{y2} = 260/100

                      a_{y2} = 2.60 m / s2

final speed

                     v_{yf}² = v_{y1}² + 2 a_{y2} (y₂ -y₁)

                     v_{yf}² = 10.11² + 2 2.60 (-27 + 21)

                      v_{yf} = √ 71.01

                      v_{yf} = 8.43 m / s

here there is an inconsistency in the problem if the body is at y₁ = -27m and passes the position y_f = -21 m with the relationship it must be contrary to the velocity

z axis

 

first interval, relate (a_{z1})

                      a_{z1} = F_{z1} / m

                      a_{z1} = -200/100

                      a_{z1} = -2 m / s

the negative sign indicates that the acceleration is the negative direction of the z axis

the speed at the end of the interval

                    v_{z1}² = v_{zo)² + 2 a_{z1} (z₁-z₀)

                    v_{z1}² = 1.47² + 2 (-2) (-7 + 4)

                    v_{z1} = √14.16

                    v_{z1} = -3.76 m / s

second interval, acceleration (a_{z2})

                    a_{z2} = F_{z2} / m

                    a_{z2} = 210/100

                    a_{z2} = 2.10 m / s2

final speed

                    v_{fz}² = v_{z1}² - 2 a_{z2} | z_f-z₁)

                    v_{fz}² = 3.14² - 2 2.10 (-3 + 7)

                     v_{fz} = √6.94

                     v_{fz} = 2.63 m / s

speed is     v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s

5 0
3 years ago
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