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2 years ago

8

A motorcycle is moving at 18 m/s when its brakes are applied, bringing the cycle to rest in 4.7 s. To the nearest meter, how far

does the motorcycle travel while coming to a stop?

1 answer:

morpeh [17]2 years ago

4 0

Answer:

the motorcycle travels 42.4 meters until it stops.

Explanation:

Vi= 18 m/s

Vf= 0 m/s

t= 4.7 sec

Vf= Vi - a*t

deceleration:

a= Vi/t

a= 18m/s / 4.7 sec => a=-3.82 m/s²

x= Vi*t - (a * t²)/2

x= 42.4m

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2 years ago

8. An object accelerates 12.0 m/s2 when a force of 6.0 newtons is applied to it. What is the mass of the object? _______________

Sholpan [36]

Answer:

Mass of object is 0.5kg

Explanation:

Given the following data;

Force = 6N

Acceleration = 12m/s²

Mass =?

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

$F = ma$

Where;

F represents force.

m represents the mass of an object.

a represents acceleration.

Making mass (m) the subject, we have;

$Mass (m) = \frac{F}{a}$

Substituting into the equation;

$Mass (m) = \frac{6}{12}$

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2 years ago

A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr

Molodets [167]

Answer:

Explanation:

Given

$P_1=150 kPa$

$T_1=20^{\circ}C$

$V_1=0.5 m^3$

$T_2=140^{\circ}C$

$P_2=400 kPa$

R for Helium $R=2.076$

$c_v=3.115 kJ/kg-K$

mass of gas $m=\frac{P_1V_1}{RT_1}$

$m=\frac{150\times 0.5}{2.076\times 293}$

$m=0.123 kg$

Similarly $V_2$ can be found

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=0.264 m^3$

Work done $W=\int_{V_1}^{V_2}PdV$

$W=\frac{P_2V_2-P_1V_1}{n-1}$

$W=\frac{mR(T_2_T_1)}{n-1}$

Since it is a polytropic Process

therefore $PV^n=c$

$P_1V_1^n=P_2V_2^n$

$(\frac{V_1}{V_2})^n=\frac{P_2}{P_1}$

$(\frac{0.5}{0.264})^n=\frac{400}{150}$

$n=\frac{\ln 2.66}{\ln 1.893}$

$n=1.533$

$W=\frac{0.123\times 2.076(140-20)}{1.533-1}$

$W=57.48 kJ$

From Energy balance

$E_{in}-E_{out}=\Delta E_{system}$

Neglecting kinetic and Potential Energy change

$Q_{in}+W_{in}=change\ in\ Internal\ Energy$

Change in Internal Energy $\Delta U=u_2-u_1$

$\Delta U=mc_v(T_2-T_1)$

$\Delta U=0.123\times 3.115(140-20)$

$\Delta U=45.977 kJ$

$Q_{in}+57.48=45.977$

$Q_{in}=-11.50 kJ$

i.e. Heat is being removed

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3 years ago

Which of the following graph is used for determining the instantaneous velocity from the slope?

hoa [83]

Answer:

B. x - t graph

Explanation:

A position-time (x-t) graph is a graph of the position of an object against (versus) time.

Generally, the slope of the line of a position-time (x-t) graph is typically used to determine or calculate the velocity of an object.

An instantaneous velocity can be defined as the rate of change in position of an object in motion for a short-specified interval of time. Thus, an instantaneous velocity is a quantity that can be found by measuring the slope of a line that is tangent to a point on the graph.

Hence, the x - t graph also referred to as the position-time graph is used for determining the instantaneous velocity from the slope.

<u>For example;</u>

Given that the equation of motion is S(t) = 4t² + 2t + 10. Find the instantaneous velocity at t = 5 seconds.

Solution.

$S(t) = 4t^{2} + 2t + 10$

Differentiating the equation, we have;

$S(t) = 8t + 2$

Substituting the value of "t" into the equation, we have;

$S(5) = 8(5) + 2$

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S(5) = 42 m/s.

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3 years ago

Read 2 more answers