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2 years ago

What happens to ar object in free fall?

1 answer:

5 0

Energy from the gravitational potential store in converted to kinetic energy. Air friction acts against the object, dissipating some energy as heat or sound. The object will continuously accelerate until the acceleration is equal to the air friction acting against it. This is when it reaches terminal velocity

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The components of a 10.8-meters-per-second velocity at an angle of 34.° above the horizontal are

Kisachek [45]

Answer:

6.0 m/s vertical and 9.0 m/s horizontal

Explanation:

For the vertical component, we use the formula:

Sin(34°) = y / 10.8

Then we solve for y:

0.559 = y / 10.8

y = 6.0

And for the horizontal component, we use the formula:

Cos(34°) = x / 10.8

Then we solve for x:

0.829 = x / 10.8

y = 9.0

So the answer is " 6.0 m/s vertical and 9.0 m/s horizontal".

3 0

2 years ago

A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$

Differentiating with respect to t ( time ),

$0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ ( l = 26 feet = constant )

$\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}$

$\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

We have,

$y = 10, \frac{dx}{dt}= -3\text{ feet per min}$

$\frac{dy}{dt}=\frac{3x}{10}-----(X)$

(i) From equation (1),

$26^2 = x^2 + 10^2$

$676=x^2 + 100$

$576 = x^2$

$\implies x = 24\text{ feet}$

From equation (X),

$\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}$

(ii) From equation (X),

$\frac{dy}{dt}\propto x$

Thus, for different value of x the value of $\frac{dy}{dt}$ would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

$\frac{dy}{dt}=0\text{ feet per min}$

3 0

3 years ago

A wheel 1.0 m in radius rotates with an angular acceleration of 4.0rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 r

Oliga [24]

Answer:

(a) ωf= 42 rad/s

(b) θ = 220 rad

Explanation:

The uniformly accelerated circular movement, is a circular path movement in which the angular acceleration is constant.

There is tangential acceleration (at ) and is constant.

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t Formula (1)

θ= ω₀*t + (1/2)*α*t² Formula (2)

at = α*R Formula (3)

v= ω*R Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular velocity ( rad/s)

ωf: final angular velocity ( rad/s)

R : radius of the circular path (cm)

at : tangential acceleration (m/s²)

v : tangential speed (m/s)

Data

α = 4.0 rad/s² : wheel’s angular acceleration

t = 10 s

ω₀ = 2.0 rad/s : wheel’s initial angular velocity

R = 1.0 m : wheel’s radium

(a) Wheel’s angular velocity after 10 s

We replace data in the formula (1):

ωf= ω₀ + α*t

ωf= 2 + (4)*(10)

ωf= 42 rad/s

(b) Angle that rotates the wheel in the 10 s interval

We replace data in the formula (2):

θ= ω₀*t + (1/2)*α*t²

θ= (2)*(10) + (1/2)*(4)*(10)²

θ= 220 rad

θ= 220 rad

(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval

We replace data in the Formula (3)

at = α*R = (4)(1)

at = 4 m/s²

We replace data in the Formula (4)

v= ω*R = (42)*(1)

v = 42 m/s

6 0

3 years ago

Which of the following represents a chemical change to matter? A. digestion of food B. melting an ice cube C. shredding paper D.

guajiro [1.7K]

The answer is A. digestion of food. The chemical change to matter means that the molecule structure has been destroyed. For B, C and D. the matter are still ice, paper and ice. So they are physical change.

8 0

2 years ago

A child is riding a merry-go-round that is turning at 7.18 rpm. if the child is standing 4.65 m from the center of the merry-go-

Alisiya [41]

First we need to convert the angular speed from rpm to rad/s. Keeping in mind that
$1 rev= 2 \pi rad$
$1 min = 60 s$
the angular speed is
$\omega = 7.18 \frac{rev}{min} \cdot \frac{2 \pi}{60} = 0.75 rad/s$

And so now we can calculate the tangential speed of the child, which is the angular speed times the distance of the child from the center of the motion:
$v= \omega r = (0.75 rad/s)(4.65 m)=3.50 m/s$

3 0

3 years ago

What happens to ar object in free fall?​

What happens to ar object in free fall?