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olasank [31]
2 years ago
8

Who was the scientist who gave us the laws of motion?.

Physics
2 answers:
kicyunya [14]2 years ago
8 0
Isaac Newton was the one who gave his three laws of motion
timurjin [86]2 years ago
3 0

Answer:

Isaac Newton

Explanation:

Newton's laws of motion, three statements describing the relations between the forces acting on a body and the motion of the body.

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How does the resistance change as you add bulbs to a series circuit?
Ronch [10]
The resistance is increased when more and more bulbs are added to the circuit.
7 0
2 years ago
Read 2 more answers
Light enters water from air at an angle of 25° with the normal, Θ1. If water has an index of refraction of 1.33, determine Θ2.
vladimir1956 [14]
By Snell's law:

η = sini / sinr.        i = 25,  η = 1.33

1.33 = sin25° / sinr

sinr = sin25° / 1.33 = 0.4226/1.33 = 0.3177    Use a calculator.

r = sin⁻¹(0.3177)

r ≈ 18.52°

Option A.

God's grace.
6 0
3 years ago
What pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c?
Ronch [10]

The pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

Therefore, option A is correct option.

Given,

Mass m = 14g

Volume= 3.5L

Temperature T= 75+273 = 348 K

Molar mass of CO = 28g/mol

Universal gas constant R= 0.082057L

Number of moles in 14 g of CO is

n= mass/ molar mass

= 14/28

= 0.5 mol

As we know that

PV= nRT

P × 3.5 = 0.5 × 0.082057 × 348

P × 3.5 = 14.277

P = 14.277/3.5

P = 4.0794 atm

P = 4.1 atm.

Thus we concluded that the pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

learn more about pressure:

brainly.com/question/22613963

#SPJ4

5 0
10 months ago
Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge
Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃=  2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So,  the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs.  F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

\frac{k*q_{1}*q_3 }{(d_{13})^{2}  } = \frac{k*q_{2}*q_3 }{(d_{23})^{2}  }

We divide by (k * q3) on both sides of the equation

\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }

q_{1} = \frac{q_{2}*(d_{13})^{2}   }{(d_{23} )^{2}  }

q_{1} = \frac{5*(2)^{2} }{(4 )^{2}  }

q₁ = + 1.25 nC

3 0
3 years ago
How fr do you think you would go on a car while sneezing for 2.5 seconds ?
arsen [322]

depends on the speed limit

4 0
3 years ago
Read 2 more answers
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