Which of these is the most effective way for Leanna to cool down after an intense bike ride

g A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0

Margarita [4]

Answer:

Explanation:

If a particle move with time and expressed according to the formula:

f(t) = 0.01t⁴ − 0.03t³

a) Velocity is the change in motion of the particle with respect to time and it is expressed as;

$v(t) =\frac{d(f(t))}{dt}$

$v(t) = 4(0.01)t^{4-1} - 3(0.03)t^{3-1}\\v(t) = 0.04t^3 - 0.09t^2$

Hence the velocity of the particle at time t is $v(t) = 0.04t^3 - 0.09t^2$

b) To calculate the velocity after 1 second, we will substitute t = 1 into the function v(t) in (a) as shown:

$v(t) = 0.04t^3 - 0.09t^2\\v(1) = 0.04(1)^3 - 0.09(1)^2\\v(t) = 0.04 - 0.09\\v(t) = -0.05$

Hence the velocity after 1second is -0.05

c) The particle is at rest when when the time is zero.

Initially, the body is not moving and the time during this time is 0. Hence the particle is at rest when t = 0second

6 0

1 year ago

The cheetah is considered the fastest running animal in the world. Cheetahs can accelerate to a speed of 21.7 m/s in 2.50 s and

viktelen [127]

Answer:

1) 64.2 mi/h

2) 3.31 seconds

3) 47.5 m

4) 5.26 seconds

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 0 m/s

v = Final velocity = 21.7 m/s

s = Displacement

a = Acceleration

1) Top speed = 28.7 m/s

1 mile = 1609.344 m

$1\ m=\frac{1}{1609.344}\ miles$

1 hour = 60×60 seconds

$1\ s=\frac{1}{3600}\ hours$

$28.7\ m/s=\frac{\frac{28.7}{1609.344}}{\frac{1}{3600}}=64.2\ mi/h$

Top speed of the cheetah is 64.2 mi/h

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow t=\frac{21.7-0}{2.5}\\\Rightarrow a=8.68\ m/s^2$

Acceleration of the cheetah is 8.68 m/s²

2)

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{28.7-0}{8.68}\\\Rightarrow t=3.31\ s$

It takes a cheetah 3.31 seconds to reach its top speed.

3)

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{28.7^2-0^2}{2\times 8.68}\\\Rightarrow s=47.5\ m$

It travels 47.5 m in that time

4) When s = 120 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 120=0\times t+\frac{1}{2}\times 8.68\times t^2\\\Rightarrow t=\sqrt{\frac{120\times 2}{8.68}}\\\Rightarrow t=5.26\ s$

The time it takes the cheetah to reach a rabbit is 120 m is 5.26 seconds

8 0

2 years ago

Consider a wave along the length of a stretched slinky toy, where the distance between coils increases and decreases. What type

GrogVix [38]

Answer:

"Longitudinal wave" is the appropriate answer.

Explanation:

Generating waves whenever the form of communication being displaced in a similar direction as well as in the reverse way of the wave's designated points, could be determined as Longitudinal waves.
A wave running the length of something like a Slinky stuffed animal, which expands as well as reduces the spacing across spindles, produces a fine image or graphic.

3 0

1 year ago

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The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational

s344n2d4d5 [400]

Answer:

v = √2G $M_{earth}$ / R

Explanation:

For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)

Eo = K + U = ½ m1 v² - G m1 m2 / r1

Ef = - G m1 m2 / r2

When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf

Eo = Ef

½ m1v² - G m1 $M_{earth}$ / R = - G m1 $M_{earth}$ / R

v² = 2G $M_{earth}$ (1 / R - 1 / Rinf)

If we do Rinf = infinity 1 / Rinf = 0

v = √2G $M_{earth}$ / R

Ef = = - G m1 m2 / R

The mechanical energy is conserved

Em = -G m1 $M_{earth}$ / R

Em = - G m1 $M_{earth}$ / R

R = int ⇒ Em = 0

6 0

1 year ago

The launching velocity of a missile is 20.0 m/s, and it is shot at 53º above the horizontal. Which of the velocity components (n

Gekata [30.6K]

Neglecting air resistance, the horizontal component remains constant. The angle doesn't matter.

5 0

2 years ago

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