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3 years ago

A cheetah can go from a state of resting to running at 20 m/s in just two seconds what is the cheetah's average acceleration?

Physics

2 answers:

Sunny_sXe [5.5K]3 years ago

8 0

It's C i believe. To solve it we just take 20 and divide it by 2. Which gives us the average of 10 m/s

-Steel jelly

Oxana [17]3 years ago

4 0

The answer would be C. 10 m/s

Hope this helps

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An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the

garik1379 [7]

(a) 6.04 rev/s

The speed of the ball is given by:

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the ball from the centre of the circle

In situation 1), we have

$\omega=8.13 rev/s \cdot 2\pi = 51.0 rad/s$

r = 0.600 m

So the speed of the ball is

$v=(51.0 rad/s)(0.600 m)=30.6 m/s$

In situation 2), we have

$\omega=6.04 rev/s \cdot 2\pi = 37.9 rad/s$

r = 0.900 m

So the speed of the ball is

$v=(37.9 rad/s)(0.900 m)=34.1 m/s$

So, the ball has greater speed when rotating at 6.04 rev/s.

(b) $1561 m/s^2$

The centripetal acceleration of the ball is given by

$a=\frac{v^2}{r}$

where

v is the speed

r is the distance of the ball from the centre of the trajectory

For situation 1),

v = 30.6 m/s

r = 0.600 m

So the centripetal acceleration is

$a=\frac{(30.6 m/s)^2}{0.600 m}=1561 m/s^2$

(c) $1292 m/s^2$

For situation 2 we have

v = 34.1 m/s

r = 0.900 m

So the centripetal acceleration is

$a=\frac{v^2}{r}=\frac{(34.1 m/s)^2}{0.900 m}=1292 m/s^2$

5 0

3 years ago

A 2.4 mm -diameter copper wire carries a 37 A current (uniform across its cross section). Determine the magnetic field at the su

cluponka [151]

Answer:

Explanation:

We shall apply Ampere's circuital law to find out magnetic field . It is given as follows.

∫B.dl = μ₀ I , B is magnetic field , I is current , μ₀ is permeability .

Radius of the wire r = 1.2 x 10⁻³ m

magnetic field B will be circular in shape around the wire. If B is uniform

∫B.dl = B x 2πr

B x 2πr = μ₀ I

B = μ₀ I / 2πr

= 4π x 10⁻⁷ x 37 /2πx1.2 x 10⁻³

= 10⁻⁷ x 2x37 / 1.2 x 10⁻³

= 61.67 x 10⁻⁴ T

= 62 x 10⁻⁴ T

7 0

3 years ago

Two particles oscillate in simple harmonic motion along a common straight-line segment of length 1.0 m. Each particle has a peri

igor_vitrenko [27]

Answer:

a) the particles are 0.217 m apart

b) the particles are moving in the same direction.

Explanation:

a) The amplitude of the oscillations is A/2 and the period of each particle is

T = 1.5 s however, they differ by a phase of π/6 rad. Let the phase of the first particle be zero so that the phase of the second particle is π/6. So we can write the coordinates of each of the particles as,

x₁ = A/2 cos(ωt)

x₂ = A/2 cos(ωt + π/6)

we can write the angular frequency ω, as

ω = 2π / T

so,

x₁ = A/2 cos(2π / T)

x₂ = A/2 cos(2π / T + π/6)

Thus, the coordinates of the particles at t = 0.45 s are,

x₁ = A/2 cos((2π × 0.45) / 1.5)) = -0.155 A

x₂ = A/2 cos((2π × 0.45) / 1.5) + π/6) = -0.372 A

Their separation at that time is, therefore,

Δx = x₁ - x₂

= -0.155 A + 0.372 A

= 0.217 A

since A = 1 m

Thus,

Δx = 0.217 m

b) In order to find their directions, we must take the derivatives at t = 0.45 s.

Therefore,

v₁ = dx₁ / dt

= (-πA / T) sin(2πt / T)

= -(π(1) / 1.5) sin(2π(0.45) / 1.5)

= -1.99

and,

v₂ = dx₂ / dt

= (-πA / T) sin((2πt / T) + π/6)

= -(π(1) / 1.5) sin((2π(0.45) / 1.5) + π/6)

= -1.40

Since both v₁ and v₂ are negative, this shows that the particles are moving in the same direction.

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Whose geocentric model of the solar system was accepted for 1400 years

-BARSIC- [3]

Answer:

Plato, Aristotle developed it further and used for 1400 years till Copernicus.

Explanation:

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natita [175]

Answer:

2. erect and magnified

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3 years ago

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