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lyudmila [28]
3 years ago
12

What is the eccentricity of an ellipse with a foci distance of 50,000,000 km and

Physics
1 answer:
inysia [295]3 years ago
5 0

Answer:

25,000,000 Km ;)

Explanation:

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Which element of a valid contract is established by getting the signatures of all parties?
sweet [91]

The element of a valid contract which is established by getting the signatures of all parties is <u>mutual agreement</u>

<h3>What is an element of a valid contract?</h3>

An element of a valid contract simply refers to that promise made between two or more parties that which allow the courts to make judgement.

Some elements of valid contract are:

  • Offer
  • Acceptance
  • Consideration
  • Intention to create legal relation
  • Certainty and capacity.

So therefore, the element of a valid contract which is established by getting the signatures of all parties is mutual agreement

Learn more about elements of a valid contract:

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8 0
1 year ago
A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a
makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°

The distance r is:

r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m

You replace the values of all parameters in the equation (1):

\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0
3 years ago
A mother and her 35.0 -kg child are riding an escalator to the third level of a shopping mall. If the child's gravitational pote
notka56 [123]

The increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.

<h3>What is gravitational potential energy?</h3>

The energy that an item has due to its location in a gravitational field is known as gravitational potential energy.

The potential energy increases by 3773 J

PE₂-PE₁=mg(h₂-h₁)

3773 J = 35.0 × 9.81 × (h₂-h₁)

(h₂-h₁) = 10.98

Case 2 ;

ΔPE =?

ΔPE=mg(h₂-h₁)

ΔPE=56.0 × 9.81 ×10.98

ΔPE=6031.97 J.

Hence, the increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.

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8 0
1 year ago
Which of the following statements below is correct?
Mamont248 [21]
C. is correct. when you make a pizza, you see the the meat or pepperoni or cheese heating up and sometimes melted. (thats physical). on the inside the crust in heated and the toppings are cooked (chemical)
5 0
3 years ago
When the mirror is rotated, the normal will turn as well, but will the incident Ray and reflected ray turn?
Inessa [10]

When a mirror is rotated . . .

-- The incident ray doesn't turn.  It's just the line from the source to the mirror. 
It would be there, in the same place, even if there was no mirror.

-- The normal turns.  It's the line perpendicular to the mirror, so it must turn
with the mirror.

-- Since the normal tuns and the incident ray doesn't, the angle between them
must change.  And since the angle of the reflected ray is equal to the angle of
the incident ray, the reflected ray must also turn.


6 0
3 years ago
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