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2 years ago

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during a journey, a car travels at 40 km in 2.5 hours, next 62 km in 3 hours, then took a break for 30 minutes, again travelled

the last 120 km in 3.2 hours. calculate the average speed of the car during the journey.

1 answer:

wlad13 [49]2 years ago

6 0

45mph is the answer if you do the math right

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A 12.0N force with a fixed orientation does work on a

kvasek [131]

Answer:

(a) $\theta=62.31^{\circ}$

(b) $\theta=117.68^{\circ}$

Explanation:

It is given that,

Force acting on the particle, F = 12 N

Displacement of the particle, $d=(2.00i -4.00j+3.00k)\ m$

Magnitude of displacement, $d=\sqrt{2^2+4^2+3^2}= 5.38\ m$

(a) If the change in the kinetic energy of the particle is +30 J. The work done by the particle is given by :

$W=Fd\ cos\theta$

$\theta$ is the angle between force and the displacement

According to work energy theorem, the charge in kinetic energy of the particle is equal to the work done.

So,

$cos\theta=\dfrac{W}{Fd}$

$cos\theta=\dfrac{+30\ J}{12\times 5.38}$

$\theta=62.31^{\circ}$

(b) If the change in the kinetic energy of the particle is (-30) J. The work done by the particle is given by :

$cos\theta=\dfrac{W}{Fd}$

$cos\theta=\dfrac{-30\ J}{12\times 5.38}$

$\theta=117.68^{\circ}$

Hence, this is the required solution.

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2 years ago

A force is described by its BLANK and by the direction in which it acts

Andre45 [30]

Strength/magnitude would both work

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2 years ago

Calculate the kinetic energy, in joules of a 1160-kg automobile moving at 19.0 m/s.

Rudiy27

Answer:

Explanation:

K.E=1/2 mv²

m=1160 kg

v=19.0 m/s

so k.e=1/2*1160*(19.0)²

K.E=1/2*1160*361

K.E=1/2*418760

K.E=209380=2.0*10^5 j

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2 years ago

A physics student looks into a microscope and observes that small particles suspended in water are moving about in an irregular

Dominik [7]

Answer:

d. the actual motion is regular, but the speeds of particles are too large to observe the regular motion

Explanation:

The speeds of the particles are very large and comparatively the average free path is very small . Therefore time taken in covering the free path ( path between two consecutive collision with medium particles ) is very small . Hence the st line path covered by particles between two collision is less likely to be visible. Hence motion appears irregular or zig-zag.

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2 years ago

An automobile battery has an emf of 12.6 V and an internal resistance of 0.0600 . The headlights together present equivalent res

murzikaleks [220]

Answer:

(a) $V=11.86\ V$

(b) $V=9.76\ V$

Explanation:

<u>Electric Circuits</u>

Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:

$V=R.I$

(a) The electromagnetic force of the battery is $\varepsilon =12.6\ V$ and its internal resistance is $R_i=0.06\ \Omega$ . Knowing the equivalent resistance of the headlights is $R_e=5.2\ \Omega$ , we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:

$\varepsilon=i.R_i+i.R_e=i.(R_i+R_e)$

Solving for i

$\displaystyle i=\frac{\varepsilon}{ R_i+R_e}=\frac{12}{0.06+5.2}=2.28\ A$

i=2.28\ A

The potential difference across the headlight bulbs is

$V=\varepsilon -i.R_i=12\ V-2.28\ A\cdot 0.06\ \Omega=11.86\ V$

$V=11.86\ V$

(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is

$V=\varepsilon -i.R_i=12\ V-37.28\ A\cdot 0.06\ \Omega=9.76\ V$

5 0

2 years ago