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Dmitry_Shevchenko [17]

4 years ago

5

A(n) 30 kg boy rides a roller coaster. The acceleration of gravity is 9.8 m/s 2 . With what force does he press against the seat

when the car moving at 7.3 m/s goes over a crest whose radius of curvature is 15 m? Answer in units of N.

1 answer:

forsale [732]4 years ago

5 0

Answer:

Force, F = 187.42 N

Explanation:

It is given that,

Mass of boy, m = 30 kg

Acceleration due to gravity, $a=9.8\ m/s^2$

Radius of curvature of the roller coaster, r = 15 m

Speed of the car, v = 7.3 m/s

The force acting on the boy are force of gravity and the centripetal force. The net force acting on him is as follows :

$F=mg-\dfrac{mv^2}{r}$

$F=m(g-\dfrac{v^2}{r})$

$F=30\times (9.8-\dfrac{(7.3)^2}{15})$

F = 187.42 N

So, he press against the seat with a force is 187.42 N. Hence, this is the required solution.

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To calibrate your calorimeter cup, you first put 45 mL of cold water in the cup, and measure its temperature to be 24.7 °C. You

drek231 [11]

Answer : The heat change of the cold water in Joules is, $1.6\times 10^3J$

Explanation :

First we have to calculate the mass of cold water.

As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.

$Density=\frac{Mass}{Volume}$

$Mass=Density\times Volume=1g/mL\times 45mL=45g$

Now we have to calculate the heat change of cold water.

Formula used :

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat change of cold water = ?

m = mass of cold water = 45 g

c = specific heat of water = $4.184J/g^oC$

$T_1$ = initial temperature of cold water = $24.7^oC$

$T_2$ = final temperature = $33.4^oC$

Now put all the given value in the above formula, we get:

$Q=45g\times 4.184J/g^oC\times (33.4-24.7)^oC$

$Q=1638.036J=1.6\times 10^3J$

Therefore, the heat change of cold water is $1.6\times 10^3J$

4 0

3 years ago

Why are peer reviews important?

riadik2000 [5.3K]

Peer review is important because it is used by scientists to decided which results should be published in a scientific journal

8 0

3 years ago

Read 2 more answers

This question is related to inertia:

luda_lava [24]

The way I do it is suddenly, in the same sort of way that magicians try to pull a table cloth off a table when there's things on the table cloth.The sudden approach acts as an impulse of force and starts to accelerate the roll. But, the piece (assuming it has perforations) is off the roll before the roll can move, due to inertia. Then the roll will acclerate, move, slow down and stop. However, in accelerating, the roll will unravel. The bigger the impulse the more it will unravel.+++++++++++++++++++++++++++++++++++++++If on the other hand, the piece of paper is held firmly, and the roll is pulled, then the impulse is presumably given to the paper and the hand whose inertia is a lot more than that of the roll. So, I think I'd actually go for choice c)+++++++++++++++++++++++++++++++++++++This assumes that the roll is free to rotate.I think that a similar idea is behind the design and use of a "ballistic galvanometer". The charge is passed through the galvanometer quickly, as a current pulse. Then the needle starts to deflect, and the deflection is arranged to depend on the total charge that has passed through in the time of the current pulse.

3 0

3 years ago

. Each valve A, B, and C, when open, releases water into a tank at its own constant rate. With all three valves open, the tank f

olasank [31]

Answer:

Time taken by A and B is 1.2 hr.

Explanation:

Given that

Time taken by tank when all(A+B+C) are open = 1 hr

Time taken by tank when A+C are open = 1.5 hr

Time taken by tank when B+C are open = 2 hr

If we treat as filling of tank is a work then

Work = time x rate

Lets take work is 1 unit

1 = 1(1/a+1/b+1/c) ---------1

1 = 1.5(1/a+1/c) ----------2

1 = 2(1/b+1/c) --------3

From equation 1 and 3

1=1(1/a+1/2)

a=2

Form equation 2

1 = 1.5(1/2+1/c)

c=6

From equation 3

1 = 2(1/b+1/6)

b=3

So time taken by

A is alone to fill tank is 2 hr

B is alone to fill tank is 3 hr

C is alone to fill tank is 6 hr

So $time\ taken\ by\ A\ and\ B =\dfrac{2\times 3}{2+3} hr$

Time taken by A and B is 1.2 hr.

7 0

3 years ago

Air enters a turbine operating at steady state at 8 bar, 1600 K and expands to 0.8 bar. The turbine is well insulated, and kinet

kobusy [5.1K]

Answer:

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

Explanation:

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}$

Where

Temperature at inlet of turbine

Temperature at exit of turbine

Pressure at exit of turbine

Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

$m_i = m_0 = mm(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W$

Where,

m = mass

m(i) = mass at inlet

m(o)= Mass at outlet

h(i)= Enthalpy at inlet

h(o)= Enthalpy at outlet

W = Work done

Q = Heat transferred

v(i) = Velocity at inlet

v(o)= Velocity at outlet

Z(i)= Height at inlet

Z(o)= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

$h_i = h_0 + WW = h_i -h_0$

Using the relation T-P we can find the final temperature:

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}\\$

$\frac{T_2}{1600K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}\\ = 828.716K$

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

$W = h_i -h_0W = C_p (T_1-T_2)W = 1.005(1600 - 828.716)W = 775.140kJ/Kg$

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

4 0

3 years ago