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Luda [366]
3 years ago
5

Anyone get this can you please help me, can you also please do it step by step

Physics
1 answer:
Leya [2.2K]3 years ago
6 0

Answer:

7.69 hours

Explanation:

3 hr/ 1000mi x 130mi = 7.69 hours or 7.7 hours

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A force of 20N is applied to a 10kg mass on a level frictionless surface. What is the acceleration of the mass
34kurt

Answer:

2m/s^2

Explanation:

f = ma

20 = 10 * a

2m/s^2 = a

8 0
3 years ago
What was the Law of Triads? A. Elements were arranged into three rows. B. Elements were arranged in groups of three. C. Elements
shtirl [24]

Answer:

its B

Explanation: i searched it up...

8 0
4 years ago
A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly
Mnenie [13.5K]

Answer:

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The acceleration of the tortoise is 0.28 m/s²

Explanation:

The equations that describe the position and velocity of the hare and the tortoise are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

To find the acceleration of the hare once it begins to slow down, we have to find how much time the hare traveled during the deceleration and what was its initial speed.

First, the hare moves with constant acceleration for 4.7 s. Then, its velocity at  t = 4.7 s will be:

v = v0 + a · t    (v0 = 0 because the hare starts form rest)

v = a · t = 0.9 m/s² · 4.7 s = <u>4.2 m/s</u>

<u />

The distance traveled by the hare while accelerating can be calculated using the equation of the position:

x = x0 + v0 · t + 1/2 · a · t²      (x0 = 0 and v0 = 0)

x = 1/2 · a · t² = 1/2 · 0.9 m/s² · (4.7)² = <u>9.9 m</u>

<u />

Then, the hare runs at a constant speed of 4.2 m/s for 11.7 s. The distance traveled at constant speed will be:

x =  v · t

x = 4.2 m/s · 11.7 s = <u>49.1 m</u>

<u />

Then, the distance traveled by the hare while slowing down was:

Distance traveled while slowing down = 72 m - 49.1 m - 9.9 m = 13 m

Let´s find how much time it took the hare to come to stop, so we can calculate the acceleration. We know that when the position is 13 m, the velocity is 0.

v = v0 + a · t

0 = 4.2 m/s + a · t

-4.2 m/s / t = a

Replacing in the equation of the position:

x = v0 · t + 1/2 · a · t²      (considering x0 as the point at which the hare started to slow down)

13 m = 4.2 m/s · t - 1/2 · 4.2 m/s / t · t²

13 m = 4.2 m/s · t - 2.1 m/s · t

13 m = 2.1 m/s · t

t = 13 m / 2.1 m/s

t = 6.2 s

Then, the acceleration of the hare while slowing down will be:

-v0/t = a

-4.2 m/s / 6.2 s = a

a = -0.68 m/s²

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The hare traveled 72 m in (6.2 s + 11.7 s + 4.7 s) 22.6 s. The tortoise reaches the final position of the hare at the same time, so, using the equation of the position we can calculate the acceleration of the tortoise:

x = x0 + v0 · t + 1/2 · a · t²     (x0 = 0 and v0 = 0)

x = 1/2 · a · t²

72 m = 1/2 · a · (22.6 s)²

144 m / (22.6 s)² = a

a = 0.28 m/s²

The acceleration of the tortoise is 0.28 m/s²

6 0
3 years ago
Consider the following four objects: a hoop, a flat disk, a solid sphere, and a hollow sphere. Each of the objects has mass M an
Mumz [18]

Answer:

The hoop

Explanation:

We need to define the moment of inertia of the different objects, that is,

DISK:

I_{disk} = \frac{1}{2} mR^2

HOOP:

I_{hoop} = mR^2

SOLID SPHERE:

I_{ss} = \frac{2}{5}mR^2

HOLLOW SPHERE

I_{hs} = \frac{2}{3}mR^2

If we have the same acceleration for a Torque applied, then

mR^2>\frac{2}{3}mR^2>\frac{1}{2} mR^2>\frac{2}{5}mR^2

I_{hoop}>I_{hs} >I_{disk}>I_{ss}

The greatest momement of inertia is for the hoop, therefore will require the largest torque to give the same acceleration

4 0
3 years ago
A tennis ball is dropped from 1.16 m above the
sweet-ann [11.9K]

Answer:

Vf = 4.77 m/s

Explanation:

During the downward motion we can easily find the final velocity or the velocity with which the ball hits the ground, by using third equation of motion. The third equation of motion is given as follows:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity = 9.8 m/s²

h = height = 1.16 m

Vf = Final Velocity of Ball = ?

Vi = Initial Velocity of Ball = 0 m/s (Since, ball was initially at rest)

Therefore, using these values in the equation, we get:

(2)(9.8 m/s²)(1.16 m) = Vf² - (0 m/s)²

Vf = √(22.736 m²/s²)

<u>Vf = 4.77 m/s</u>

6 0
3 years ago
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