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Paraphin [41]
4 years ago
9

The cheetah can maintain its maximum speed for only 7.5 s. What is the minimum distance the gazelle must be ahead of the cheetah

to have a chance of escape? (After 7.5 s the speed of cheetah is less than that of the gazelle.) Answer in units of m.
Physics
1 answer:
amm18124 years ago
6 0

Basically the cheetah is running 31.5km/h faster than the gazelle. So to determone how long it will take to cover 9mm at that speed, you have to a lot of work. If you skip all of that work, the answer is 1.60m seconds

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Two coils that are separated by a distance equal to their radius and that carry equal currents such that their axial fields add
jok3333 [9.3K]

Answer:

When x = 2.8 cm, B_{x1} = 0.0265 T

When x = 5.5 cm, B_{x2} = 0.0209 T

when x = 7.3 cm, B_{x3} = 0.0169 T

When x = 11.0 cm, B_{x4} = 0.0103 T

Explanation:

According to Biot-Savart law,

B_{x} = \frac{N \mu_{o}IR^{2}  }{2(x^{2} +R^{2}  )^{3/2} }\\.......................(1)

R = 11.0 cm = 0.11 m

I = 17.0 A

N = 300 turns

\mu_{o}  = 4\pi  * 10^{-7} N/A^{2}

When x₁ = 2.8 cm = 0.028 m

B_{x1} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.028^{2} +0.11^{2}  )^{3/2} }\\B_{x1} = 0.0265 T

When x₂ = 5.5cm = 0.055 m

B_{x2} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.055^{2} +0.11^{2}  )^{3/2} }\\B_{x2} = 0.0209 T

When x₃ = 7.3 cm = 0.073 m

B_{x3} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.073^{2} +0.11^{2}  )^{3/2} }\\B_{x3} = 0.0169 T

When X₄ = 11.0 cm = 0.11 m

B_{x4} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.11^{2} +0.11^{2}  )^{3/2} }\\B_{x4} = 0.0103 T

4 0
4 years ago
Help me someone ?...
Burka [1]
Gradpoint? The answer is B
8 0
3 years ago
A particle moves in a circular path of radius 0.10 m with a constant angular speed of 5 rev/s. The acceleration of the particle
Darya [45]

Answer:

2.5 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s²

For circular motion, the expression for acceleration is given as,

a = ω²r ................ Equation 1

Where a = acceleration of the particle, ω = angular speed of the particle, r = radius of the circular path.

Given: ω = 5 rev/s = 31.42 rad/s, r = 0.10 m.

Substitute into equation 1

a = 5²(0.10)

a = 25(0.10)

a = 2.5 m/s²

Hence the acceleration of the particle = 2.5 m/s²

Hence, none of the option  is correct

7 0
3 years ago
Copper wire is 1. 9 mm in diameter and carries a current of 20 a. What is the electric field stregnth inside this wire?
Tju [1.3M]

The electric field strength will be 0.6252 V/m. It is the strength at which the field is created by charges.

<h3>What is electric file strength?</h3>

The electric field strength is defined as the ratio of electric force and charge.

The electric field strength is found as;

\rm E = \frac{I \rho }{A} \\\\ \rm E = \frac{20  \times 1.68 \times 10^{-8} }{ (0.6385 \times 10^{-6}} \\\\ E= 0.5262 \  V/m

Hence, the electric field strength will be 0.6252 V/m.

To learn more about the electric field strength, refer to the link;

brainly.com/question/4264413

#SPJ4

3 0
2 years ago
A man can walk three blocks east and four blocks north in 20 min. He can drive 12 times that distance in the same amount of time
skad [1K]

Answer:  25.2 km

<u>Explanation:</u>

3 blocks east and four blocks east = 7 blocks

1 block = 0.3 km  --> 7 blocks = 2.1 km

12 times 2.1 km = 25.2 km

4 0
3 years ago
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