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Vsevolod [243]
2 years ago
6

On a straight road (taken to be in the +x direction) you drive for an hour at 50 km per hour, then quickly speed up to 90 km per

hour and drive for an additional two hours.Required:a. How far do you go?b. What is your average x component of velocity?c. Why isn't vavg,x equal to the arithmetic average of your initial and final values of vx, (50+90)/2 = 70 km per hour?
Physics
1 answer:
Rufina [12.5K]2 years ago
5 0

Answer:

A.) 230 km

B.) 76.67 km/h

Explanation:

Given that On a straight road (taken to be in the +x direction) you drive for an hour at 50 km per hour, then quickly speed up to 90 km per hour and drive for an additional two hours.

A.) How far do you go?

When driving for an hour, the distance covered will be

Distance = speed × time

Distance = 50 × 1 = 50 km

When driving for additional 2 hours, the distance covered will be

Distance = 90 × 2 = 180 km

Total distance = 180 + 50 = 230 km

b. What is your average x component of velocity?

Average Velocity = total distance/ total time

Average velocity = 230/3

Average velocity = 76.67 km/h

c. Why isn't vavg,x equal to the arithmetic average of your initial and final values of vx, (50+90)/2 = 70 km per hour

They are not equal because of the displacement is the same as distance travelled.

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The problem is about magnitude of the displacement vector of the lady bug and its directions. so the magnitude of the displacement of the lady bug is 16 in / 2 because it started from the center so the magnitude is 8 in. and the direction is the rotation of the turn table which is 60 degrees
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1 year ago
An engineer is designing a process for a new transistor. She uses a vacuum chamber to bombard a thin layer of silicon with ions
adelina 88 [10]

Answer:

E=5.7\times 10^{-3}\ V/m

Explanation:

Given that

m_p=5.18\times 10^{-26}\ kg

re= 46 cm

Vp= 180 m/s

We know that

E=\dfrac{\Delta V}{r}

\Delta V=\dfrac{1}{4}\dfrac{m_pv_p^2}{e}

So

E=\dfrac{1}{4}\dfrac{m_pv_p^2}{e.r_e}

Now by putting the all given values in the questions

E=\dfrac{1}{4}\times \dfrac{5.18\times 10^{-26}\times 180^2}{1.6\times 10^{-19}\times 0.46}

E=5.7\times 10^{-3}\ V/m

So the average electric field is E=5.7\times 10^{-3}\ V/m.

6 0
2 years ago
A monoatomic ideal gas undergoes an isothermal expansion at 300 K, as the volume increased from 0.010 m^3 to 0.040 m^3. The fina
Natali [406]

Answer:

A) 0.0 kJ

Explanation:

Change in the internal energy of the gas is a state function

which means it will not depends on the process but it will depends on the initial and final state

Also we know that internal energy is a function of temperature only

so here the process is given as isothermal process in which temperature will remain constant always

here we know that

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2 years ago
What is the internal energy (to the nearest joule) of 10 moles of Oxygen at 100 K?
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Answer:

U = 12,205.5 J

Explanation:

In order to calculate the internal energy of an ideal gas, you take into account the following formula:

U=\frac{3}{2}nRT        (1)

U: internal energy

R: ideal gas constant = 8.135 J(mol.K)

n: number of moles = 10 mol

T: temperature of the gas = 100K

You replace the values of the parameters in the equation (1):

U=\frac{3}{2}(10mol)(8.135\frac{J}{mol.K})(100K)=12,205.5J

The total internal energy of 10 mol of Oxygen at 100K is 12,205.5 J

6 0
2 years ago
A particle of charge 2.3 ✕ 10−8 C experiences an upward force of magnitude 4.6 ✕ 10−6 N when it is placed in a particular point
Marysya12 [62]
<h2>Answer:</h2>

(a) +2 x 10² N/C (upwards)

(b) -2.2μN or -2.2 x 10⁻⁶N (downwards)

<h2>Explanation:</h2>

The force (F) acting on a particle of charge (Q) at a particular point is related to its electric field (E) by the following;

F = Q x E   ----------------------(i)

This means that the force acting on the charged particle is the product of its charge and the electric field at that point.

<em>(a) Given</em>;

Q = charge of the particle = 2.3 x 10⁻⁸ C

F = force acting on the particle = 4.6 x 10⁻⁶N

<em>Substitute these values into equation (i) as follows;</em>

=> 4.6 x 10⁻⁶  = 2.3 x 10⁻⁸ x E

=> E = 4.6 x 10⁻⁶ ÷ 2.3 x 10⁻⁸

=> E =  2 x 10² N/C

Since the value is positive, the electric field is directed upwards.

Therefore, the electric field at that point is +2 x 10² N/C

<em>(b) If a charge of q = -1.1 x 10⁻⁸ is placed there, the force (F) acting is calculated as follows;</em>

<em>Substitute Q = q into equation (i) as follows;</em>

F = q x E

<em>Substitute the value of q and E = 2 x 10² N/C into the equation above as follows;</em>

F = -1.1 x 10⁻⁸ x 2 x 10²

F = -2.2 x 10⁻⁶ N

F = -2.2μN

Since the value of the force, F, is negative, it means it is directed downwards.

Therefore the force on the charge is -2.2μN

3 0
2 years ago
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