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1 year ago

6

The tip of the fan blade is 0.61 m from the center of the fan. The fan turns at a constant speed and completes 2 rotation every

1.0s

1 answer:

Goryan [66]1 year ago

5 0

Answer:

What is the centripetal acceleration of the tip of the fan blade?

6.0 m/s2

48 m/s2

53 m/s2

96 m/s2

Answer is 96

Explanation:

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A basketball is tossed upwards with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, start fraction, start text, m

Hatshy [7]

Answer:

The last one

Explanation:

4 0

1 year ago

What is the net torque on the bar shown in (Figure 1) about the axis indicated by the dot? Suppose that F=6.0N

AysviL [449]

Answer:

net torque on the system is 2.5 Nm

Explanation:

As we know that net torque on the system is given as

$\tau = r_1F_1 - r_2F_2$

now we have

$r_1 = 75 cm$

$F_1 = 6 N$

$r_2 = 25 cm$

$F_2 = 8 N$

now we have

$\tau = 0.75 (6) - 0.25(8)$

$\tau = 2.5 Nm$

6 0

1 year ago

If c1=c2=4.00μf and c4=8.00μf, what must the capacitance c3 be if the network is to store 2.70×10−3 j of electrical energy?

Akimi4 [234]

Missing detail in the text: total voltage of the circuit $\Delta V = 46.0 V$
Missing figure: https://www.physicsforums.com/attachments/prob-24-68-jpg.190851/

Solution:

1) The energy stored in a circuit of capacitors is given by
$U= \frac{1}{2} C_{eq} (\Delta V)^2$
where $C_{eq}$ is the equivalent capacitance of the circuit. We can find the value for $C_{eq}$ by using $\Delta V=46.0 V$ and the energy of the system, $U=2.7\cdot 10^{-3} J$
$C_{eq}= \frac{2U}{(\Delta V)^2}=2.55\cdot 10^{-6} F=2.55\mu F$

2) Then, let's calculate the equivalente capacitance of C1 and C2. The two capacitors are in series, so their equivalente capacitance is given by
$\frac{1}{C_{12}}= \frac{1}{C_1}+ \frac{1}{C_2}= \frac{1}{4 \mu F} + \frac{1}{4 \mu F}$
from which we find $C_{12}=2 \mu F$

3) Then let's find $C_{123}$ , the equivalent capacitance of $C_{12}$ and C3. $C_{123}$ is in series with C4, therefore we can write
$\frac{1}{C_{eq}}= \frac{1}{C_{123}}+ \frac{1}{C_4}$
Since we already know $C_4=8 \mu F$ and $C_{eq}=2.55 \mu F$ , we find
$C_{123}=3.70 \mu F$

4) Finally, we can find $C_{3}$ , because it is in parallel with $C_{12}$ , and the equivalent capacitance of the two must be equal to $C_{123}$ :
$C_{123}=C_{12}+C_3$
So, using $C_{123}=3.70 \mu F$ and $C_{12}=2 \mu F$ , we find
$C_3=1.70 \mu F$

7 0

1 year ago

You’ve been given the challenge of balancing a uniform, rigid meter-stick with mass M = 95 g on a pivot. Stacked on the 0-cm end

Mariulka [41]

Answer: d = 4750n/3.1+95n

Explanation:

Using the principle of moment to solve the question.

Sum of clockwise moments = sum of anti clockwise moments

Since there are n identical coins with mass 3.1g placed at point 0cm, 1 coin will have mass of 3.1/n grams

Taking moment about the pivot,

Mass 3.1/n grams will move anti-clockwisely while the mass 95g will move in the clockwise direction.

Since its a meter rule (100cm) the distance from the center mass(95g) to the pivot will be 50-d (check attachment for diagram).

To get 'd'

We have 3.1/n × d = 95 × (50-d)

3.1d/n = 4750-95d

3.1d = 4750n-95dn

3.1d+95dn=4750n

d(3.1+95n) = 4750n

d = 4750n/3.1+95n

6 0

2 years ago

A cyclist traveling at 5m/s uniformly accelerates up to 10 m/s in 2 seconds. Each tire of the bike has a 35 cm radius, and a sma

konstantin123 [22]

Answer:

$a=2.5\ m/s^2$

Explanation:

Given that,

Initial speed, u = 5 m/s

Final speed, v = 10 m/s

Time, t = 2 s

The radius of the tire of the bike, r = 35 cm

We need to find the angular acceleration of the pebble during those two seconds. It can be calculated as follows.

$a=\dfrac{v-u}t{}\\\\a=\dfrac{10-5}{2}\\\\a=2.5\ m/s^2$

So, the required angular acceleration of the pebble is equal to $2.5\ m/s^2$ .

8 0

1 year ago