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SIZIF [17.4K]
3 years ago
8

True or false? A consequence that decrease the likelihood that a behavior will occur is known as reinforcement

Physics
1 answer:
dlinn [17]3 years ago
6 0

Answer:

I would choose false but that might be the wrong answer

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A. How does increasing the voltage of a power supply in an electromagnet affect the strength of the magnetic field?
seropon [69]

PART A)

If we increase the voltage supply in an electromagnet then it will increase the current that is flowing in it

So here due to increase in current there will be increase in the magnetic field due to that electromagnet

PART B)

Here in electric generator the current is produced by rotating a coil between two strong magnets.

So here mechanical energy of rotation of coil is converted into electromagnetic energy.

PART C)

Step up transformer convert the lower voltage input into higher voltage output

here number of turns of coil in output side or secondary number of coils is more than the number of coils in primary side or input side

PART D)

Force on a moving charge is given by

F = q(\vec v \times \vec B)

here we know that

q = 0.000600 C

v = 2.8 \times 10^5 m/s

B = 4.21 T

now from above equation we have

F = 0.0006(2.8\times 10^5)(4.21)

F = 707.3 N

direction of force is given by right hand thumb rule

using that rule we got force downwards

7 0
3 years ago
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If Jupiter were 10 times more massive, it would generate nuclear fusion in its core and be a star instead of a planet. If Jupite
OverLord2011 [107]

Answer:

False

Explanation:

Let me put it in numbers. Jupiter has mass of 1.898x10^27Kg, make it 10 times as much and it becomes 1.898x10^28 Kg.

TRAPPIST-1, smallest star ever found has Mass of 1.77x10^29Kg, that is around 93 times larger than the mass of Jupiter.

Clearly 10 times as much is not enough to generate fusion reaction and become a star, in theory Jupiter must be at least 85 times larger than it's current mass to be able to generate fusion reaction and become a star.

So i think we can conclude that if Jupiter were to have 10 times of its own mass, it would not become a star, therefore the given statement is False.

5 0
3 years ago
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All of the following are examples of hypnotism methods except...
MAVERICK [17]
C) Hypnosis Certification this is only showing your certification and not a method of hypnosis
5 0
3 years ago
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Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are
Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

d_{12} = \sqrt{3^{2}+3^{2}  }  = \sqrt{18} m : distance from q₁ to q₂

(d₁₂)² = 18 m²

d_{13} =\sqrt{1^{2}+3^{2}  } = \sqrt{10} m  : distance from q₁ to q₃

(d₁₃)² = 10 m²

d_{14} =\sqrt{3^{2}+2^{2}  } = \sqrt{13} m  : distance from q₁ to q₄

(d₁₄)² = 13 m²

K=  8.98755 × 10⁹ N *m²/C²

q₁=  7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45°  ,  β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N  

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }

F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }

Fn₁= 23.95*10⁻⁹ N

8 0
3 years ago
This force involves the attraction between objects with mass. (2 points) i need it asap
aev [14]

Answer:

Gravitational Mass

6 0
3 years ago
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