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3 years ago

6

"A toy airplane flies clockwise at a constant speed in a horizontal circle of radius 8.0 meters. The magnitude of the accelerati

on of the airplane is 25 meters per second squared. The diagram shows the path of the airplane as it travels around the circle.
Calculate the speed of the airplane. [Show all work, including the equation and substitution with units.]"

2 answers:

Citrus2011 [14]3 years ago

6 0

For the answer to the question above, so at the instant, the acceleration of the airplane is southward, the direction of the velocity is also southward. The direction should be the same because it is both a vector quantity and it does not make sense if the direction and acceleration have different direction.

ryzh [129]3 years ago

3 0

Answer:

speed of the airplane: 14.14 m/s

Explanation:

radius, r = 8.0 m

centripetal acceleration, a = 25 m/(s^2)

velocity, v = ?

Centripetal acceleration can be computed as follows:

a = v^2/r

Solving for v and replacing with data:

v = √a*r

v = √25 m/(s^2)*8.0 m

v = 14.14 m/s

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Two 1-kg objects, C and D, increase in temperature by the same amount, but the

Ede4ka [16]

The object D is made up of material Lead. The correct option is D.

<h3>What is specific heat?</h3>

The specific heat is the amount of heat required to change the temperature by 1°C. It is denoted by C.

Two 1-kg objects, C and D, increase in temperature by the same amount, but the thermal energy transfer of object C is greater than the thermal energy transfer of object D. The object C has a specific heat of 235 J/kg-K.

Q = m C ΔT

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The energy transfer is proportional to specific heat.

Specific heat of D must be less. The possible material with specific heat less than the given value is for Lead material.

Thus, the correct option is D.

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2 years ago

A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are

Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of $x= 5t^3+1$

Coordinate of position of $y=3t^2+2$

Time = 2.00 s

We need to calculate the acceleration

$a = \dfrac{d^2x}{dt^2}$

For x coordinates

$x=5t^3+1$

On differentiate w.r.to t

$\dfrac{dx}{dt}=15t^2+0$

On differentiate again w.r.to t

$\dfrac{d^2x}{dt^2}=30t$

The acceleration in x axis at 2 sec

$a = 60i$

For y coordinates

$y=3t^2+2$

On differentiate w.r.to t

$\dfrac{dy}{dt}=6t+0$

On differentiate again w.r.to t

$\dfrac{d^2y}{dt^2}=6$

The acceleration in y axis at 2 sec

$a = 6j$

The acceleration is

$a=60i+6j$

We need to calculate the net force

$F = ma$

$F = 3.00\times(60i+6j)$

$F=180i+18j$

The magnitude of the force

$|F|=\sqrt{(180)^2+(18)^2}$

$|F|=180.89\ N$

Hence, The net force acting on this object is 180.89 N.

3 0

3 years ago

What is the formula you use to determine the gravitational potential energy of a object

kozerog [31]

Gravitational potential energy, relative to some level =

   (mass of the object)
times
       (height above the reference level)
times
   (acceleration due to gravity) .

4 0

3 years ago

A person is riding a bicycle, and its wheels have an angular velocity of 10.7 rad/s. Then, the brakes are applied and the bike i

Scilla [17]

Answer:

(a) t = 22.9 s

(b) α= - 0.467 rad/s²

Explanation:

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

We apply the equations of circular motion uniformly accelerated :

ωf ²= ω₀² + 2*α*θ Formula (1)

ωf= ω₀ + α*t Formula (2)

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 speed ( rad/s)

ωf : final angular speed ( rad/s)

Data

θ = 19.5 revolutions : angular displacement of each wheel or angle that the wheel has rotated in a given time interval

ω₀= 10.7 rad/s : initial angular speed of the Wheel ( rad/s)

ωf = 0 : final angular speed of the Whee( rad/s)

Calculating of the angular acceleration (α )

We replace data in the fómula (1),considering that 1 revolution is equal to 2π radians :

ωf ²= ω₀² + 2*α*θ

(0 )²= (10.7)² + 2*α*(19.5*2*π )

0= 114.49 + (245.04)*α

-114.49 = (245.04)*α

α= (-114.49) /(245.04)

α= -114.49 /(245.04)

α= -0.467 rad/s²

Time does it take for the bike to come to rest

We replace data in the formula (2)

ωf = ω₀ + α*t

0 = 10.7 + -0.467*t

-10.7 = - 0.467*t we multiply by (-1) both sides of the equation :

10.7 = 0.467*t

t = 10.7 / 0.467

t = 22.9 s

3 0

3 years ago

Find electric field at point p which is a distance l away from the both +q and -q

denis-greek [22]

Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

q is the charge
r is the distance
$E$ is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2= $\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

8 0

3 years ago