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IrinaVladis [17]

3 years ago

14

During a football game, the running back changes direction from running 4.5 m/s [N] to 4.5 m/s [S] over 8 s. What is the acceler

ation of the running back during that time?

1 answer:

Nimfa-mama [501]3 years ago

8 0

Explanation:

Take north to be positive and south to be negative.

a = (v − v₀) / t

a = (-4.5 m/s − 4.5 m/s) / 8 s

a = -1.125 m/s²

The acceleration is 1.125 m/s² south.

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The speed limit on an interstate highway is posted at 75mi/h. What is the speed in kilometers per hour? In feet per second?

Tatiana [17]

Answer:

See explanation

Explanation:

The conversion factor from miles per hour to kilometers per hour is 1.609344.

Hence;

To convert 75mi/h to kilometers per hour,

75mi/h * 1.609344 = 120.7 kilometers per hour

b)

To convert 75mi/h to feet per second

Since

1 mile per hour = 1.46667 foot per second

75mi/h = 75 * 1.46667 = 110 feet per second

5 0

3 years ago

A coin is resting on the bottom of an empty container. The container is then filled to the brim three times, each time with a di

Virty [35]

Answer:

Refractive index of liquid C > Refractive index of liquid B > Refractive index of liquid A

Explanation:

Let the depth of each section is h.

That means the real depth for each section is h.

Apparent depth is liquid A is 7 cm.

Apparent depth in liquid B is 6 cm.

Apparent depth in liquid C is 5 cm.

by the formula of the refractive index

n = real depth / apparent depth

where, n is the refractive index of the liquid.

For liquid A:

$n_{A}=\frac{h}{7}$ .... (1)

For liquid B:

$n_{B}=\frac{h}{6}$ ..... (2)

For liquid C:

$n_{C}=\frac{h}{5}$ ..... (3)

By comparing all the three equations

nc > nB > nA

Refractive index of liquid C > Refractive index of liquid B > Refractive index of liquid A

5 0

3 years ago

A small object with mass 1.30 kg is mounted on one end of arod

Julli [10]

Answer:

(a) $I_{system} = 1.014\ kg.m^{2}$

(b) $\tau = 0.0179\ N-m$

Solution:

As per the question:

Mass of the object, m = 1.30 kg

Length of the rod, L = 0.780 m

Angular speed, $\omega = 5010\ rev/min$

Now,

(a) To calculate the rotational inertia of the system about the axis of rotation:

Since, the rod is mass less, the moment of inertia of the rotating system and that of the object about the rotation axis will be equal:

$I_{system} = ML^{2} = 1.30\times (0.780)^{2} = 0.791\ kg.m^{2}$

(b) To calculate the applied torque required for the system to rotate at constant speed:

Drag Force, F = $2.30\times 10^{- 2}\ N$

$\tau = FLsin\theta 90 = 2.30\times 10^{- 2}\times 0.780\times 1 = 0.0179\ N-m$

3 0

3 years ago

Which of the following better describe planets most comparable to Earth?

wariber [46]

Answer:

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8 0

3 years ago

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An ice cube can slide around the inside of a vertical circular hoop of radius . It undergoes small-amplitude oscillations if dis

klasskru [66]

Answer:

$T=2\pi \times \sqrt {\frac {g}{\mu R}$

Explanation:

$\mu mw^{2}R=mg$

where m is the mass, g is acceleration due to gravity

The masses m are on both sides hence they cancel

$W=\sqrt {\frac {g}{\mu R}$

We know that T=2\pi W and substituting W with the above equation then

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4 0

4 years ago