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UkoKoshka [18]
4 years ago
9

As a train accelerates uniformly, it passes successive 5-kilometer markers while traveling at velocities 10 m/s and 25 m/s. what

is the velocity in km/h when it passes the next marker? round to nearest tenth of a km/h.
Physics
1 answer:
gregori [183]4 years ago
8 0
<span>122.0 km/hr. First let’s make sure all of our units are in the base meter form: i.e. convert 5km to 5000m. (We will convert back to km later). The first thing to do is look at the equation relating velocity, acceleration, and distance: Vf^2 = Vi^2 + 2*a*d, where Vf is final velocity, Vi is initial velocity, a is acceleration, and d is distance. 25^2 = 10^2 + 2*a*5000 =?> 625 = 100 +10000a => a= 0.0525m/s^2. Now that we have acceleration, we can use the same equation again with different numbers.: Vf^2 = Vi^2 + 2*a*d = 25^2 + 2*0. 0525m*5000 = 625 + 525 =1150 => Vf^2 = 1150 => 33.9m/s. Convert to km/hour: 33.9m/s * 1km/1000m *60s/1min * 60min/ 1 hr = 122.0 km/hr.</span>
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A horizontal 745 N merry-go-round of radius
Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is  544J

Explanation:

Given :

Weight w = 745 N

Radius r =  1.45 m

Force =  56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62  = ?

Solution:

Step 1:  Finding the Mass of merry-go-round

m = \frac{ weight}{g}

m = \frac{745}{9.81 }

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I =0.5 \times m \times r^2

Substituting the values

Moment of Inertia of solid cylinder I  

=>0.5 \times 76.02 \times (1.45)^2

=> 0.5 \times 76.02\times 2.1025

=> 79.91 kg.m^2

Step 3: Finding the Torque applied T

Torque applied T = F \times r

Substituting the values

T = 56.3  \times 1.45

T = 81.635 N.m

 Step 4: Finding the Angular acceleration

Angular acceleration ,\alpha  = \frac{Torque}{Inertia}

Substituting the values,

\alpha  = \frac{81.635}{79.91}

\alpha = 1.021 rad/s^2

 Step 4: Finding the Final angular velocity

Final angular velocity ,\omega = \alpha \times  t

Substituting the values,

\omega = 1.021 \times  3.62

\omega = 3.69 rad/s

Now KE (100% rotational) after 3.62s is:

KE = 0.5 \times I \times \omega^2

KE =0.5 \times 79.91 \times 3.69^2

KE = 544J

6 0
4 years ago
Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm
Burka [1]

Answer:

The atmospheric pressure is 0.97622\times10^{5}\ Pa.

Explanation:

Given that,

Atmospheric pressure P_{atm}= 1.013\times10^{5}\ Pa    

drop height h'= 27.1 mm

Density of mercury \rho= 13.59 g/cm^3

We need to calculate the height

Using formula of pressure

p = \rho g h

Put the value into the formula

1.013\times10^{5}=13.59\times10^{3}\times9.8\times h

h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}

h=0.76\ m

We need to calculate the new height

h''=h - h'

h''=0.76-27.1\times10^{-3}

h''=0.76-0.027

h''=0.733\ m

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

P=\rho g h

Put the value into the formula

P= 13.59\times10^{3}\times9.8\times0.733

P=0.97622\times10^{5}\ Pa

Hence, The atmospheric pressure is 0.97622\times10^{5}\ Pa.

7 0
3 years ago
A lawn mower engine running for 20 m i n does 4, 5 6 0, 0 0 0 J of work. What is the power output of the engine?
VladimirAG [237]

Answer:3800\ W

Explanation:

Given

Lawn mover running for t=20\ min

and does W=4560\times 10^3\ J

We know Power is rate of work i.e.

P=\frac{\text{Work}}{\text{time}}

P=\frac{4560\times 10^3}{20\times 60}

P=3800\ W

Thus Power output is 3800\ W

4 0
3 years ago
And I need help with seven and eight only I will appreciate it
Mumz [18]

Answer:

7] Force = mass × acceleration

Force = 2 × 5

<u>Force = 10 N</u>

<u></u>

8] Velocity = acceleration due to gravity × time taken

Velocity = 9.8 × 12

<u>Velocity = 117.6 m/s</u>

8 0
2 years ago
Calculate the amount of heat required to completely sublime 40.0 g of solid dry ice (co2) at its sublimation temperature. the he
Mashcka [7]

The amount of heat required for the sublimation of 40.0 g of solid dry ice CO₂ is <u>29.3 kJ.</u>

One mole of CO₂ has a mass of 44.0095 g.

Calculate the number of moles  n in 40.0 g of CO₂ .

n =\frac{40.0 g}{44.0095 g/mol}  = 0.9089 mol

Heat of sublimation is the amount of heat required by 1 mole of a substance to convert itself from solid state to a vapor state at constant temperature and pressure.

1 mole of CO₂ requires 32.3 kJ of energy to sublimate.

Therefore, the heat required to sublimate 0.9089 mol of CO₂ is given by,

Q = (32.3 kJ/mol)(0.9089 mol) =29.3 kJ

Thus, the heat required to sublimate 40.0 g of CO₂ is <u>29.3 kJ</u>.


8 0
3 years ago
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