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3 years ago

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a 1200 kg trailer is hitched to a 1400 kg car. the car and trailer are traveling at 72 km.h when the driver applies the brakes o

n both the car and the trailer. knowing that the braking forces exerted on the car and the trailer are 5000 N and 4000 N respectively, determine (a) the distance traveled by the car and trailer before they come to a stop and (b) the horizontal component of the force exerted by the trailer hitch

1 answer:

Sergio [31]3 years ago

5 0

Answer:

a) 8.67m

b) 1000N

Explanation:

(a) To find the distance you use the second Newton Law for both car and trailer, in order to calculate the dis-acceleration of the system:

$F=ma\\\\a_=\frac{F}{m}=\frac{5000N+4000N}{1400kg+1200kg}=3.46\frac{m}{s^2}$

once you have this value, you use the the following kinematic equation to calculate the distance traveled by both car and trailer:

$v^2=v_o^2-2ax\\\\x=\frac{-v^2+v_o^2}{2a}$

v: final velocity=0

vo: initial velocity = 72km/h = 60 m/s

by replacing the values of these parameters you obtain for x:

$x=\frac{-0m/s+60m/s}{2(3.46m/s^2)}\\\\x=8.67m$

(b) The horizontal component of the force exerted by the trailer hitch is given by:

$F_T=5000N-4000N=1000N$

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The driver of a car traveling at 22.8 m/s applies the brakes and undergoes a constant deceleration of 2.95 m/s 2 . How many revo

a_sh-v [17]

Answer:

70 revolutions

Explanation:

We can start by the time it takes for the driver to come from 22.8m/s to full rest:

$t = \Delta v/a = (22.8 - 0)/2.95 = 7.73 s$

The tire angular velocity before stopping is:

$\omega_0 = v/r = 22.8 / 0.2 = 114 rad/s$

Also its angular decceleration:

$\alpha = a / r = 2.95/0.2 = 14.75 rad/s^2$

Using the following equation motion we can findout the angle it makes during the deceleration:

$\omega^2 - \omega_0^2 = 2\alpha\Delta \theta$

where $\omega$ = 0 m/s is the final angular velocity of the car when it stops, $\omega_0$ = 114rad/s is the initial angular velocity of the car $\alpha$ = 14.75 rad/s2 is the deceleration of the can, and $\Delta \theta$ is the angular distance traveled, which we care looking for:

$-114^2 = 2*(-14.75)*\Delta \theta$

$\Delta \theta = 440rad$ or 440/2π = 70 revelutions

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3 years ago

Two electric charges, held a distance, dd, apart experience an electric force of magnitude, FF, between them. If one of the char

lorasvet [3.4K]

Answer:

$F'=2F$

Explanation:

The Coulomb's law states that the magnitude of the electrostatic force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them:

$F=\frac{kq_1q_2}{d^2}$

In this case, we have $q_1'=2q_1$ :

$F'=\frac{kq'_1q_2}{d^2}\\F'=\frac{k(2q_1)q_2}{d^2}\\F'=2\frac{kq_1q_2}{d^2}\\F'=2F$

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A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in

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Answer:

a) The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) The electric force is $2.0\times 10^{-6}$ newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

$E = \frac{F_{e}}{q}$ (1)

Where:

$E$ - Electric field, measured in newtons per coulomb.

$F_{e}$ - Electric force, measured in newtons.

$q$ - Electric charge, measured in coulombs.

If we know that $F_{e} = 4.0\times 10^{-6}\,N$ and $q = 2.0\times 10^{-8}\,C$ , then the electric field at that point is:

$E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}$

$E = 2.0\times 10^{2}\,\frac{N}{C}$

The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) If we know that $E = 2.0\times 10^{2}\,\frac{N}{C}$ and $q = 1.0\times 10^{-8}\,C$ , then the electric force is:

$F_{e} = E\cdot q$

$F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)$

$F_{e} = 2.0\times 10^{-6}\,N$

The electric force is $2.0\times 10^{-6}$ newtons.

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Answer:

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Explanation:

who need to add the exponents only

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