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3 years ago

Can someone please help me how to set this up?

2 answers:

5 0

The car has a 12 mile head start, going 80 mph, so his distance is:dcar=80∗t+12
The is going at 108 mph, so his distance is:dcop=108t
Setting them equal to each other we get:80t+12=108t⇒12=28t⇒t=1228=37
So 3/7 of an hour.about 25.7 mins.

Citrus2011 [14]3 years ago

3 0

Answer:

The time is 0.428 hr.

Explanation:

Given that,

Final speed = 108 mph

Initial speed = 80 mph

Distance = 12 miles

We need to calculate the time

Using formula of time

$v = \dfrac{d}{t}$

d = distance

t = time

Put the value into the formula

$108-80=\dfrac{12}{t}$

$t = \dfrac{12}{108-80}$

$t =0.428\ hr$

Hence, The time is 0.428 hr.

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A CD-ROM drive in a computer spins the 12-cm-diameter disks at 9500 rpm. Find acceleration in units of g that a speck of dust on

NeTakaya

The net force on an object is:
F = ma
The only force acting on the speck of dust as it is lays on the CD is centripetal force given by:
F = (mv²)/r
Equating the two
ma = (mv²)/r
We get:
a = v²/r
v is the linear velocity in this case; however, we can calculate only the angular velocity with the given data. Therefore, we must use:
v = ωr; where
ω = 2πf; f is the rotations per second
f = 9500 / 60
f = 950/6
ω = 2π(950/6)
ω = (950π)/3
v = (950π)/3 × 0.12
v = 38π
a = (38π)²/0.12
a = 1.2 × 10⁵ m/s²
To express this in terms of gravitational acceleration, we divide by the value of gravitational acceleration, 9.81
a = 1.2 × 10⁴ g

4 0

3 years ago

A typical flying insect applies an average force equal to twice its weight during each downward stroke while hovering. Take the

Lelu [443]

Answer:

Average power output of insect is 2.42W

Explanation:

Workdone by constant force during displacement is given by:

W= F× d cos theta

Where theta is angle between F and d.

Power output due to the force over the interval time is given by:

P= Workdone/change in time

Ginen:

Mass of insect,m= 7.0g= 7/1000 = 0.07kg

Downward force applied by insect,F= 2mg

Distance moved by the wing each stroke=1.5cm=1.5/100= 0.015m

W= F× d cos theta

Where theta=0° since force is in the same direction as the displacement.

W= 2mg×d

W= 2× 0.07 × 9.8 × 0.015

W= 0.02058J

Power output = W/ change in time

Since wings make 117strokes each second time interval is 1/117 = 8.5×10^-3seconds

Power= 0.02058/(8.5×10^-3)

Power= 2.42W

6 0

3 years ago

How is sound detected by the brain

ivann1987 [24]

begin with vibrations of the eardrum. Sound waves cause the eardrum to vibrate. Cells in the cochlea detect the vibrations and send a message to the brain.

3 0

3 years ago

Read 2 more answers

A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif

Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c) w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

μ = N*I*A

Where,

N: The number of turns

I : Current in coil

A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

vec ( τi ) = vec ( μi ) x vec ( B )

= 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

= ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

$\left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right] = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\$

- The initial torque ( τi ) is written as follows:

vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

Ui = - vec ( μi ) . vec ( B )

Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

Ui = - [(-0.000605556 + 0.00011)]

Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

Uf = - vec ( μf ) . vec ( B )

Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

ΔU = Uf - Ui

ΔU = -0.000252315 - 0.000495556

ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

Ui + Eki = Uf + Ekf