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aleksklad [387]

2 years ago

5

A toy car that is 0.12 m long is used to model the actions of an actual car that is 6 m long. which ratio shows the relationship

between the sizes of the model and the actual car?
a. 50:1
b. 1:50
c. 5:1
d. 1:5

1 answer:

GREYUIT [131]2 years ago

7 0

Answer:

its b) 1:50 just did it on apex

Explanation:

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Why is it important to assess flexibility both before and during a workout regimen? a. to assess flexibility progress b. to keep

lidiya [134]

It important to assess flexibility both before and during a workout regimen because to assess flexibility progress.

<h3>What is a flexibility exercise?</h3>

Flexibility exercises, as the name implies, help to increase the stretch of the muscles, improving the range of motion. More flexibility is also important to help prevent possible post-workout soreness.

In this case we have that the reason to continue doing stretching before and during the exercises is about to assess flexibility progress.

See more about flexibility at brainly.com/question/15395713

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5 0

2 years ago

Read 2 more answers

Outside my window a squirrel is scurrying up and down a tree. Its position function is given by s(t) = t 3 − 12t 2 + 36t for the

marin [14]

Answer:

Explanation:

Given

Position of squirrel is given by

$s(t)=t^3-12t^2+36t$

Velocity is given by

$v(t)=\frac{ds(t)}{dt}=\frac{d(t^3-12t^2+36t)}{dt}$

$v(t)=3t^2-12\times 2t+36$

$v(t)=3t^2-24t+36$

(b) acceleration is given by

$a(t)=\frac{da(t)}{dt}=\frac{d(3t^2-24t+36)}{dt}$

$a(t)=6t-24$

(c)at $s(3)=3^3-12(3)^2+36(3)$

$s(3)=27\ m$

at $s(4)=4^3-12(4)^2+36(4)$

$s(4)=16\ m$

at $t=3\ s$ Position is $27\ m$ and at $t=4\ s$ position is $16\ m$

therefore squirrel is moving down

6 0

3 years ago

A certain spring exerts a nonlinear force given by F(x) = −60x − 18x2 , where x is in meters and F is in newtons. A 0.90-kg bloc

tia_tia [17]

Answer:

a) W = 6.75 J and b) v = 3.87 m / s

Explanation:

a) In the problem the force is nonlinear and they ask us for work, so we must use it's definition

W = ∫ F. dx

Bold indicates vectors. In a spring the force is applied in the direction of movement, whereby the scalar product is reduced to the ordinary product

W = ∫ F dx

We replace and integrate

W = ∫ (-60 x - 18 x²) dx

W = -60 x²/2 -18 x³/3

Let's evaluate between the integration limits, lower W = 0 for x = -0.50 m, to the upper limit W = W for x = 0 m

W = -30 [0- (-0.50) 2] -6 [0 - (- 0.50) 3]

W = 7.5 - 0.75

W = 6.75 J

b) Work is equal to the variation of kinetic energy

W = ΔK

W = ΔK = ½ m v² -0

v =√ 2W/m

v = √(2 6.75/ 0.90)

v = 3.87 m / s

8 0

3 years ago

Fan object moves in uniform circular motion in a circle of radius R=200 meters, and the objectes 5.00 seconds to

Leviafan [203]

Answer:

The centripetal acceleration of the object is $31550.72\ m/s^2$ .

Explanation:

We have,

Radius of a circular path is 200 m

It takes 5 seconds to complete 10 revolutions. The angular velocity of the object is given by the rate of change of angular displacement per unit time :

$\omega=\dfrac{\Delta \theta}{\Delta t}\\\\\omega=\dfrac{2\pi \times 10}{5}\\\\\omega=12.56\ rad/s$

The centripetal acceleration of the object is given by :

$a=\omega^2 r\\\\a=(12.56)^2 \times 200\\\\a=31550.72\ m/s^2$

So, the centripetal acceleration of the object is $31550.72\ m/s^2$ .

6 0

3 years ago

You have landed on an unknown planet, newtonia, and want to know what objects will weigh there. you find that when a certain too

AleksandrR [38]

Refer to the diagram shown.

Because the surface is frictionless, the resistive for, R, is zero.

Let m = the mass of the object.
Let a = acceleration due to the applied force.
Therefore
12.7 N = (m kg)*(a m/s²)
a = 12.7/m m/s²

The object travels 16.1 m in 2.5 s, starting from rest. Therefore
16.1 N = (1/2)*(12.7/m m/s²)*(2.5 s)² = 39.6875/m N
m = 16.1/39.6875 = 0.4057 kg

For freefall, let g = acceleration due to gravity.
The time to fall from 10.3 m is 2.88 s, therefore
10.3 m = (1/2)*(g m/s²)*(2.88 s)² = 4.1472g m
g = 10.3/4.1472 = 2.484 m/s²

Answer:
The gravitational acceleration on the planet is 2.5 m/s² (nearest tenth)

3 0

3 years ago