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3 years ago

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A student conducts an experiment in which a cart is pulled by a variable applied force during a 2 s time interval. In trial 1, t

he student exerts the force on a cart of mass M. In trial 2, the student exerts the force on a cart of mass 3M. In trial 3, the student exerts the force on a cart of 5M. In which trial will the cart experience the greatest change in momentum from 0 s to 2 s?

1 answer:

fredd [130]3 years ago

5 0

Answer:

The answer is "Including all three studies of 0s to 2s, that shift in momentum is equal".

Explanation:

Its shift in momentum doesn't really depend on the magnitude of its cars since the forces or time are similar throughout all vehicles.

Let's look at the speed of the car

$F = m a\\\\a =\frac{F}{m}$

We use movies and find lips

$\to v = v_0 + a t\\\\\to v = v_0 + (\frac{F}{m}) t$

The moment is defined by

$\to p = m v$

The moment change

$\Delta p = m v - m v_0$

Let's replace the speeds in this equation

$\Delta p = m (v_0 + \frac{F}{m t}) - m v_0\\\\\Delta p = m v_0 + F t - m v_0\\\\\Delta p = F t$

They see that shift is not directly proportional to the mass of cars since the force and time were the same across all cars.

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Two lamps rated 60W; 240V and 100W, 240Vrespectively are connected in series to a 240V power source. Calculate;

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Answer:

See the answers below.

Explanation:

The total power of the circuit is equal to the sum of the powers of each lamp.

$P=60+100\\P=160 [W]$

Now we have a voltage source equal to 240 [V], so by means of the following equation we can find the current circulating in the circuit.

$P=V*I$

where:

P = power [W]

V = voltage [V]

I = current [amp]

$I = P/V\\I=160/240\\I=0.67 [amp]$

So this is the answer for c) I = 0.67 [amp]

We know that the voltage of each lamp is 240 [V]. Therefore using ohm's law which is equal to the product of resistance by current we can find the voltage of each lamp.

a)

$V=I*R$

where:

V = voltage [V]

I = current [amp]

R = resistance [ohms]

Therefore we replace this equation in the first to have the current as a function of the resistance and not the voltage.

$P=V*I\\and\\V = I*R\\P = (I*R)*I\\P=I^{2}*R$

$60 = (0.67)^{2}*R\\R_{60}=133.66[ohm] \\and\\100=(0.67)^{2} *R\\R_{100}=100/(0.66^{2} )\\R_{100}=225 [ohm]$

b)

The effective resistance of a series circuit is equal to the sum of the resistors connected in series.

$R = 133.66 + 225\\R = 358.67 [ohms]$

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Technician A says that the oil pan on a two-stroke engine holds more oil than a typical four-stroke engine. Technician B says th

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Answer:

Technician A is right, because the lubrication system in a two-stroke engine requires mixing with the fuel, therefore oil is constantly burned, while a 4-stroke engine has a separate lubrication system.

Technician B is wrong, usually the 2-stroke engines are small and the cooling system is that the cylinder container is constructed with a geometry that allows to increase the area of heat transfer called fins, this allows to release heat quickly .

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A force platform is a tool used to analyze the performance of athletes measuring the vertical force that the athlete exerts on t

Phoenix [80]

Answer:

a. $I=981.34 N*s$

b. $v_f=3.96 m/s$

c. $v_{f1}=3.63m/s$

d. $y_f=0.673m$

Explanation:

Given: $m=67kg$ , $h=0.720m$ , $0$

a.

$I=\int\limits^{t_1}_{t_2} {F(t)} \, dt$

$F(t)=9200*t-11500t^2$

$I=\int\limits^{0.8s}_{0s}{9200*t-11500*t^2} \, dt$

$I=4600*t^2-3833.3*t^3|(0.80,0)$

$I=2944-1962.66=981.35$

$I=981.34 N*s$

b.

$v_f^2=v_i^2+a*y'$

Starting from the rest

$v_f^2=0+2*9.8m/s^2*0.80s$

$v_f^2=15.68$

$v_f=\sqrt{15.68m^2/s^2}=3.96 m/s$

c.

$I_{total}=p_f$

$I_1-m*g*d=m*v_{f1}-m*v_f$

$981.34-67kg*9.8m/s^2*0.720=67.0kg*v_{f1}-67.0kg*(-3.96m/s)$

Solve to vf

$v_{f1}=3.63m/s$

d.

$v_f^2=v_i^2+2*a*y_f'$

$y_f'=v_i/2*a =(3.63m/s)^2/2*9.8m/s^2$

$y_f=0.673m$

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A mass executes SHM at the end of a light spring. (a) What fraction of the total energy of the system is potential and what frac

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Answer:

Explained

Explanation:

A) The total energy of the system is defined by the energy at maximum amplitude, which we'll call A. At that point, the energy of the system is

E = 1/2×m×A^2;

since energy is conserved, this is also the total amount of energy that the system ever has.

So at x=1/2A,

the potential energy of the system is 1/8×m×A^2

which is one-fourth of the system's total energy. Therefore, the remaining three-fourths is kinetic.

B) (i) Doubling the maximum amplitude will quadruple the total energy:

$E= \frac{1}{2}m(2A)^2$

(ii) Doubling the maximum amplitude will double the maximum velocity

$\frac{1}{2}m(2A)^2= \frac{1}{2}mV^2$

(iii) Doubling the maximum amplitude will double the maximum acceleration: m×a = -k(2A)

(iv) Doubling the maximum amplitude leaves the period unchanged:

$T= 2\pi\sqrt{\frac{m}{k} }$

(neither m nor k has changed).

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