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3 years ago

a cheetah starting from rest accelerates at a rate of 2.5m/s^2 to reach a speed of 45 m/s.how fare does the cheetah travel

Physics

1 answer:

Helen [10]3 years ago

7 0

Answer:

405 m

Explanation:

Given:

v₀ = 0 m/s

a = 2.5 m/s²

v = 45 m/s

Find: Δx

v² = v₀² + 2aΔx

(45 m/s)² = (0 m/s)² + 2 (2.5 m/s²) Δx

Δx = 405 m

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During a tornado in 2008 the Peachtree Plaza Westin Hotel in downtown Atlanta suffered damage. Suppose a piece of glass dropped

Darina [25.2K]

Answer:

Time = t = 6.62 s

Explanation:

Given data:

Height = h = 215 m

Initial velocity = $v_{i}$ = 0 m/s

gravitational acceleration = g = 9.8 m/s²

Time = t = ?

According to second equation of motion

$h = v_{i}t + \frac{1}{2}gt^{2}$

As initial velocity is zero, So the first term of right hand side of above equation equal to zero.

$h = \frac{1}{2}gt^{2}$

t² = $\frac{2h}{g}$

t = $\sqrt{\frac{2h}{g} }$

t = $\sqrt{\frac{(2)(215) }{9.8} }$

t = 6.62 s

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3 years ago

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3 years ago

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mafiozo [28]

Answer:

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2 years ago

A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric

jeka94

Answer:

a) F₃₁ = 63.0 μN

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

$F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}} = 63.0 \mu N (1)$

Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

$F_{32} = F_{3} - F_{31} = 49.0\mu N - 63.0\mu N = -14.0 \mu N (2)$

Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

$q_{2} = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC (3)$

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3 years ago

A 1,000 kg car is travelling at 6.5 m/s to the North. A 3,500 kg truck is travelling South at the same velocity. What is the tot

Molodets [167]

Answer:

16250 kgm/s due south

Explanation:

Applying,

M = mv................. Equation 1

Where M = momentum, m = mass, v = velocity.

From the car,

Given: m = 1000 kg, v = 6.5 m/s

Substitute these values into equation 1

M = 1000(6.5)

M = 6500 kgm/s

For the truck,

Given: m = 3500 kg, v = 6.5 m/s

Substitute these values into equation 1

M' = 3500(6.5)

M' = 22750 kgm/s.

Assuming South to be negative direction,

From the question,

Total momentum of the two vehicles = (6500-22750)

Total momentum of the two vehicles = -16250 kgm/s

Hence the total momentum of the two vehicles is 16250 kgm/s due south

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2 years ago