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Serga [27]
3 years ago
5

A 912-kg car is being driven down a straight, level road at a constant speed of 31.5 m/s. When the driver sees a police cruiser

ahead, she removes her foot from the accelerator. After 8.00 is, the speed of the car is 24.6 m/s, which is the posted speed limit. What is the magnitude of the average net force acting on the car during the 8.00 s interval
Physics
1 answer:
sergejj [24]3 years ago
8 0

Answer:

786.6 N

Explanation:

mass of car, m = 912 kg

initial velocity of car, u = 31.5 m/s

final velocity of car, v = 24.6 m/ s

time, t = 8 s

Let a be the acceleration of the car

Use first equation of motion

v = u + a t

24.6 = 31.5 + a x 8

a = - 0.8625 m/s^2

Force, F = mass x acceleration

F = 912 x 0.8625

F = 786.6 N

Thus, the force on the car is 786.6 N.

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Given the thermochemical equations X2+3Y2⟶2XY3ΔH1=−370 kJ X2+2Z2⟶2XZ2ΔH2=−120 kJ 2Y2+Z2⟶2Y2ZΔH3=−270 kJ Calculate the change in
Alchen [17]

Answer : The change in enthalpy of the reaction is, -310 kJ

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main reaction is,

4XY_3+7Z_2\rightarrow 6Y_2Z+4XZ_2    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) X_2+3Y_2\rightarrow 2XY_3     \Delta H_1=-370kJ

(2) X_2+2Z_2\rightarrow 2XZ_2    \Delta H_2=-120kJ

(3) 2Y_2+Z_2\rightarrow 2Y_2Z    \Delta H_3=-270kJ

Now we will reverse the reaction 1 and multiply reaction 1 by 2, reaction 2 by 2 and reaction 3 by 3 then adding all the equations, we get :

(1) 4XY_3\rightarrow 2X_2+6Y_2     \Delta H_1=2\times (+370kJ)=740kJ

(2) 2X_2+4Z_2\rightarrow 4XZ_2    \Delta H_2=2\times (-120kJ)=-240kJ

(3) 6Y_2+3Z_2\rightarrow 6Y_2Z    \Delta H_3=3\times (-270kJ)=-810kJ

The expression for enthalpy of formation of CH_4 will be,

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+740kJ)+(-240kJ)+(-810kJ)

\Delta H=-310kJ

Therefore, the change in enthalpy of the reaction is, -310 kJ

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3 years ago
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3 years ago
What are all the different energy transformations and what are examples of them being used? PLEASE HELP ;)
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6 0
3 years ago
1. Her angular speed increases because by pulling in her arms she creates a net torque in the direction of rotation. 2. Her angu
choli [55]

Answer:

3. Her angular speed increases because her angular momentum is the same but her moment of inertia decreases

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II ωi the intial angular momentum of the skater. Her angular momentum changes to If ωf after pulling her arms in.

It must be noted that If is less than II, then it is because her arms now go round not far from the rotation axis which brings down the mementos of inertia.

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Note: the mathematical representations are better written on the attached diagram.

5 0
3 years ago
Read 2 more answers
an object is moving with a speed of 35 m/s and has a kinetic energy of 1500 J, what is the mass of the object?
Elena-2011 [213]
You'd use the equation kinetic energy=mass*0.5*speed^2
So you'd rearrange this to get mass =kinetic energy /0.5 *speed^2
Which is mass= 1500J/0.5*35^2
=2.44897959183673469........kg
5 0
3 years ago
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