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3 years ago

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A person is going to take a trip from town eight to town D to do this they travel 60 KM in .45 hour And then stop in town before

lunch it takes. 15 hour in the journey continues to town see over .5 hour and covers a distance of 30 KM they had to stop and pick up an item at the store that took .15 hour they restart their trip it took 1.5 hours to get to town which is 90 came away from town see what was a numerical value for the average speed of the trip

1 answer:

ANEK [815]3 years ago

4 0

Answer:

v = 72 km / h

Explanation:

The definition of average speed is the distance traveled between the time interval

v = Δx / Δt

let's find the distance traveled

x = 60 + 30

x = 90 km

time spent, all time must be included, travel time and when stopped

t = 0.45 + 0.15 + 0.5+ 0.15

t = 1.25 h

we substitute in the initial equation

v = 90 / 1.25

v = 72 km / h

in going from one city to the other

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Answer:

Explanation:

For free body diagram see attached sheet .

W is weight of steel girder acting at the middle point of its length . T is tension in the cable .

OB = √ ( 12² - 2² )

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OC = 11.83 / 2 = 5.915 m

Taking moment of tension T and weight W about point O

W x OC = T x OB

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T = 22 x 5.915 / 11.83

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T + R = W

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Momentum quiz for physics

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The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume e =

Svetllana [295]

Answer:

Explanation:

(a) Rigel, 2.7x10^32W, T = 11,000K

But L = 4pR²sT⁴

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(b)

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But L = 4pR²sT⁴

L = 2.1x10^23W, T = 10,000K, s= 5.67 x 10^-8, R= radius in meters

Procyon B parallax, p = 0.00284 arc sec

R² = 2.1x10^23/(4 x 0.00284 x 5.67x10^-8 x (10,000)⁴)

R² = 2.1x10^23/(6.441 x 10^6)

R² = 3.26036 x 10^16

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The radius of the sun a red star is 6.96 x 10^8meters which shows a certain level of resemblance with the size of a dwarf white star Procyon B.

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3 years ago

A student sits on a rotating stool holding two 3.09-kg masses. When his arms are extended horizontally, the masses are 1.08 m fr

schepotkina [342]

Answer:

a

The New angular speed is $w_f = 2.034 rad/s$

b

The Kinetic energy before the masses are pulled in is $KE_i = 3.101 \ J$

c

The Kinetic energy after the masses are pulled in is $KE_f = 8.192 \ J$

Explanation:

From the we are told that masses are 1.08 m from the axis of rotation, this means that

The radius $r =1.08m$

The mass is $m = 3.09\ kg$

The angular speed $w = 0.770 \ rad/sec$

The moment of inertia of the system excluding the two mass $I = 3.25 \ kg \cdot m^2$

New radius $r_{new} = 0.34m$

Generally the conservation of angular momentum can be mathematical represented as

$w_f = [\frac{I_i}{I_f} ]w_i .....(1)$

Where $w_f$ is the final angular speed

$w_i$ is the initial angular speed

$I_i$ is the initial moment of inertia

$I_f$ is the final moment of inertia

Moment of inertia is mathematically represented as

$I = m r^2$

Where I is the moment of inertia

m is the mass

r is the radius

So the Initial moment of inertia is given as

$I_i = moment \ of \ inertia \ of\ the \ two \ mass \ + 3.25 \ kg \cdot m^2$

$I_i = 2m r^2 + 3.25$

The multiplication by is because we are considering two masses

$I_i = 2 [(3.09)(1.08)^2] +3.25 = 10.46 kg \cdot m^2$

So the final moment of inertia is given as

$I_f = 2[(3.09)(0.34)^2] +3.25 = 3.96 \ kg \cdot m^2$

Substituting these values into equation 1

$w_f = [\frac{10.46}{3.96} ] * 0.77 = 2.034 \ rad/sec$

Generally Kinetic energy is mathematically represented in term of moment of inertia as

$KE = \frac{1}{2} * I * w^2$

Now considering the kinetic energy before the masses are pulled in,

$KE_i = \frac{1}{2} * I_i * w^2_i$

The Moment of inertia would be $I_i = 10.46 \ Kg \cdot m^2$

The Angular speed would be $w_i = 0.77 \ rad/s$

Now substituting these value into the equation above

$KE_i = \frac{1}{2} * (10.46) * (0.770)^2 = 3.101 J$

Now considering the kinetic energy after the masses are pulled in,

$KE_f = \frac{1}{2} * I_f * w^2_f$

The Moment of inertia would be $I_f = 3.96 \ Kg \cdot m^2$

The Angular speed would be $w_f = 2.034 \ rad/s$

Now substituting these value into the equation above

$KE_f= \frac{1}{2} *(3.96)(2.034)^2$

$= 8.192J$

8 0

3 years ago

A horizontal force of 14.0N is applied to a box of m=32.5kg with Vo=0. Ignoring friction, how far does the crate travel in 10.0s

Alex Ar [27]

I’m going to assume initial velocity is 0.

Use Newton’s second law:

F = m•a

F/m = a

14.0/32.5kg= 28/65 m/s^2

Use constant SUVAT acceleration formulae:

S- displacement - what we need to find out

U - initial velocity - 0

V

A - 28/65 m/s^2

T - 10 seconds

S = ut + 1/2at^2

Since u = 0

S = 1/2at^2

1/2• 28/65 • 10^2 = 21.5metres~

Answer is 21.5 metres

~Hoodini, here to help.

6 0

3 years ago