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Masja [62]
2 years ago
6

What is your displacement if you travel 5 miles north to the store

Physics
1 answer:
kumpel [21]2 years ago
3 0

Answer:

5miles North

Explanation:

The displacement to the store is 5miles northward.

Displacement is the distance traveled in a specific direction. Displacement is a vector quantity.

A vector has both magnitude and direction.

The magnitude is the amount of that quantity

The direction is its orientation from a reference.

 Therefore, the displacement is 5miles north

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Help me please, i need it fast
s2008m [1.1K]

Answer:

2)  C.  (x - 3)² + (y + 2)² = 25

5)   x² +  y² - 8x - 16y + 54 = 0

6)   x² + y²  - 10x - 12y + 36  = 0

Explanation:

2)

center of circle = 3, -2

                            x1, y1

end point of circle = 7, 1

                                 x2, y2

 

the equation of a circle is Pythagorean theorem

x² + y² = r²    (where r is the radius of a circle)

distance between points  

(x2 - x1)² + (y2 - y1)² = r²

(7 - 3)² + (1 - (-2))² = r²

r² = 25

therefore the equation to the circle is

(x - 3)² + (y + 2)² = 25

=========================================

5)

write the general form of a circle with the center (4,8)

and containing the point (-1, 7)

distance between points  

(x2 - x1)² + (y2 - y1)² = r²

(-1 - 4)² + (7 - 8)² = r²

r² = 26

(x - 4)² + (y - 8)² = 26

(x - 4)(x - 4) +  (y - 8)(y - 8) = 26

x² - 8x + 16 + y² - 16y + 64 -26 = 0

x² +  y² - 8x - 16y + 54 = 0

=========================================

6)

find the general form of a circle with center (5,6)

and tangent to the y-axis.

           

center (5,6)

           h, k

radius = r²

r = 5

(x - h)² + (y - k)² = r²

(x - r)² + (y - k)² = r²

(x - 5)(x - 5) + (y - 6)(y - 6) = r²

x² - 10x + 25 + y² - 12y + 36 = 25

x² + y²  - 10x - 12y + 36  = 0

=========================================

8 0
2 years ago
A 70 kilogram hockey player skating east on an ice rink is hit by a 0.1 kilogram hockey puck moving toward the west. The puck ex
guajiro [1.7K]

Given:

70 Kilogram hockey

Hit by 0.1 kilogram

50 Newton

Description:

So in this case this bacially means that one object is experiencing a force then another object.  So the answer to the question will be 50N.

Answer: 50N (3rd Law)

Hope this helps.

4 0
3 years ago
A car is moving with an initial relocity of
MA_775_DIABLO [31]

Answer:

The final acceleration of the car, v = 70 m/s

Explanation:

Given,

The initial velocity of the car, u = 20 m/s

The acceleration of the car, a = 10 m/s²

The time period of travel, t = 5 s

Using the I equations of motion

                     v = u + at

                        = 20 + 10(5)

                        = 20 + 50

                        = 70 m/s

Hence, the final acceleration of the car, v = 70 m/s

4 0
2 years ago
An object is moving with the speed of light around the Earth. How much time will it take to complete one round trip along the eq
dimulka [17.4K]

Answer:

0.14 seconds

Explanation:

The speed of light in vacuum is approximately 3.0*10^8. The distance that would be covered by the object would be equivalent to the circumference of the cross-section of the earth on the equator.

Circumference = 2\pi*6400000 =4.02*10^7

Time = distance/speed = 4.2*10^7 / 3.0*10^8 =0.14s

8 0
3 years ago
Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
3 years ago
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