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olga2289 [7]
3 years ago
7

Would u rather/ be a polar bear or a grizzly bear

Physics
2 answers:
butalik [34]3 years ago
6 0
Grizzly bear bro they big and tuff and catch fish
nekit [7.7K]3 years ago
5 0

Answer:

Polar bear

Explanation:

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Suppose you increase your walking speed from 6 m/s to 14 m/s in a period of 1 s. What is your acceleration?
klasskru [66]
A = delta v over delta t delta v is calculated with final velocity less initial velocity so delta v is 14 - 6 that is 8m/s and delta t is calculated with final time less initial time as initial always is 0 then is 1 - 0 that is 1 then a = 8m/s over 1 that is 8 then the acceleration is 8m/s^2 (remember that is squared.)
6 0
3 years ago
Read 2 more answers
A body is moving with velocity 30 m pr sec Wt will b the distance covered by a body in 1 sec??
madam [21]

Answer:

30 metres

Explanation:

The velocity is 30m/s and the time is one second.

Distance =Velocity×time..

Distance=30m/s×1 second

=30×1

=30 metres

6 0
3 years ago
if a ball on a string swung in a circles and string suddenly breaks.the ball will move into what direction?
Akimi4 [234]

Answer: b

Explanation: did the quiz got it right

3 0
3 years ago
1) Consider a source particle of charge qS=9 C located at (−9,5) [distances in meters]. Find the electric field vector at the ta
xeze [42]

Answer:

1)  E = 2.25 i^+ 0.809j^)  10⁹ N / C , 2)   E = 2.39 10⁹ N / C , 3)    θ = 19.8º , 4)   F = 19.12 10⁹ N , 5)  E = (1.32 i^+ 3.56 j^) 109 N/C

Explanation:

1) The equation for the electric field is

       E = k q / r²

Where K is the Coulomb constant that is worth 8.99 10⁹ N m² /C², q is the load and r is the distance of the load to the test point

Since the electric field is a vector magnitude, we can find its component

X axis

     Ex = k q / x²

where the distance on the axis is

      x = √ (X₂-x₁)²

      x = √ (-15 + 9)² = 6 m

      Eₓ = 8.99 10⁹ 9/6²

      Eₓ = 2.25 10⁹ N /C

Y Axis

     y = √ (y₂-y₁)² = √ (15-5)² = 10 m

     E_{y} = 8.99 10⁹ 9/10²

      E_{y}  = 0.809 10⁹ N / C

     

     E = Eₓ i^+   E_{y}  j^

     E = 2.25 i^+ 0.809j^)  10⁹ N / C

2) the magnitude can be found using the Pythagorean triangle

        E = √ (Eₓ² +  E_{y}²)

        E = √ (2.25² + 0.809²) 10⁹

        E = 2.39 10⁹ N / C

3) to find the angle let's use trigonometry

       tan θ = E_{y} / Eₓ

       θ = tan⁻¹ E_{y} / Eₓ

       θ = tan⁻¹ (0.809 / 2.25)

       θ = 19.8º

Regarding the positive side of the x axis

4) a charge  q2 = 8C is placed, let's calculate the force

      F = q E

      F = 8 2.39 10⁹

      F = 19.12 10⁹ N

5) The total electric field at the origin, let's look for its components

     q₁ = 9C

     r₁ = -9 i ^ + 5 j ^

     q₂ = 8 C

     r₂ = -15 i ^ + 15 j ^

X axis

     Eₓ = E₁ₓ + E₂ₓ

     Eₓ = k q₁ / Dx₁² + k q₂ / Dx₂²

     Eₓ = 8.99 10⁹ (9 / (9-0)² + 8 / (15-0)²)

     Eₓ = 1.32 109 N / A

Y Axis

    E_{y} =  E_{1y} + E_{2y}

      E_{y} = k q₁ / Δy₁² + k q₂ / Δy₂²

      E_{y} = 8.99 109 (9/5² + 8/15²)

      E_{y} = 3.56 109 N / A

     E = (1.32 i^+ 3.56 j^) 109 N/C

7 0
4 years ago
A man pulls his dog (m=20kg) on a sled with a force of 100N at a 60° angle from the horizontal. What is the horizontal component
sertanlavr [38]

Answer:

the horizontal component of the force is 50 N

Explanation:

Given;

force applied by the man, F = 100 N

angle of inclination of the force, θ = 60⁰

mass of the dog, m = 20 kg

The horizontal component of the force is calculated as;

F_x = F\times cos(\theta)\\\\F_x = 100 \ N \times cos(60^0)\\\\F_x = 100\ N \times 0.5\\\\F_x = 50 \ N

Therefore, the horizontal component of the force is 50 N

4 0
4 years ago
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