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Alekssandra [29.7K]

2 years ago

6

A car tire rotates with an average angular speed of 29 rad/s. In what time interval will the tire rotate 3.5 times?

1 answer:

meriva2 years ago

8 0

Answer:

"0.758315 sec" is the appropriate choice.

Explanation:

The given values are:

Angular speed,

= 29 rad/s

or,

= $\frac{29}{2 \pi} \ rps$

Tire rotates,

= 3.5 times

Now,

The time interval will be:

⇒ $\Delta t=\frac{3.5}{\frac{29}{2 \pi} }$

⇒ $=\frac{2 \pi\times 3.5}{29}$

⇒ $=\frac{21.99}{29}$

⇒ $=0.758315 \ sec$

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Why do scientists often take several measurements of the same object, average them together, and use the average value as the me

DaniilM [7]

Answer:

To minimize the practical errors and improve the accuracy.

Explanation:

Scientists take several measurements and use the average of all the measured values to minimize the practical errors due to unavoidable circumstances and be close to the true value.

True value is the actual value of a measured parameter without any error, it is the value which is proposed in the design of a certain object.

Variations may occur in the measurement due to wear and tear, dust and air pressure. It is almost impossible to create ideal surrounding for the measurement procedure.

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3 years ago

a student moves a box across the floor by exerting 23.3 N of force and doing 47.2 J of work on the box. How far does the student

kozerog [31]

Work = Force x Distance
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3 years ago

What are the conditions under which the resultant of three coplanar forces is zero?<br>

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When three or more coplanar forces are acting at a point and the vector diagram closes, there is no resultant. The forces acting at the point are in equilibrium.

4 0

2 years ago

Two tiny particles having charges 20.0 μC and 8.00 μC are separated by a distance of 20.0 cm What are the magnitude and directio

Alecsey [184]

Answer:

The magnitude and direction of electric field midway between these two charges is $10.8\times10^{5}\ N/C$ along AB.

Explanation:

Given that,

First charge $q_{1}= 20\mu C$

second charge $q_{2}= 8\mu C$

Distance = 20 cm

We need to calculate the electric field

For first charge,

Using formula of electric field

$E_{1}= \dfrac{kq_{1}}{r^2}$

Put the valueinto the formula

$E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}$

$E_{1}=18\times10^{5}\ N/C$

Direction of electric field along AB

We need to calculate the electric field

For second charge,

Using formula of electric field

$E_{2}= \dfrac{kq_{2}}{r^2}$

Put the valueinto the formula

$E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}$

$E_{2}=7.2\times10^{5}\ N/C$

Direction of electric field along AO

We need to calculate the net electric field at midpoint

$E_{net}=E_{1}-E_{2}$

$E_{net}=(18-7.2)\times10^{5}\ N/C$

$E_{net}=10.8\times10^{5}\ N/C$

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Hence, The magnitude and direction of electric field midway between these two charges is $10.8\times10^{5}\ N/C$ along AB.

8 0

3 years ago

To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch.

ki77a [65]

Answer:

0.0133A

Explanation:

Since we have two sections, for the Inductor region there would be a current $i_1$ . In the case of resistance 2, it will cross a current $i_2$

Defined this we proceed to obtain our equations,

For $i_1$ ,

$\frac{di_1}{dt}+i_1R_1 = V$

$I_1 = \frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})$

For $i_2$ ,

$I_2R_2 =V$

$I_2 = \frac{V}{R_2}$

The current in the entire battery is equivalent to,

$i_t = I_1+I_2$

$i_t = \frac{V}{R_2}+\frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})$

Our values are,

$V=1V$

$R_1 = 95\Omega$

$L= 1.5*10^{-2}H$

$R_2 =360\Omega$

Replacing in the current for t= 0.4m/s

$i=\frac{1}{360}+\frac{1}{95}(1-e^{-\frac{95*0.4}{1.5*10^{-2}}})$

$i= 0.0133A$

$i_1 = 0.01052A$

3 0

2 years ago