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3 years ago

14

A mosquito flying at 3 m/s that encounters a breeze blowing at 3 m/s in the same direction has a speed of

1 answer:

Vika [28.1K]3 years ago

4 0

Answer:

6m/s

Explanation:

The mosquito will have a speed of 6 m/s.

if the speed of the mosquito and breeze was in opposite direction, we will subtracted them from each other and the result will be the speed of mosquito moving in the direction of the higher speed (between the breeze and mosquito). while if they are both moving in the same direction the speed of the mosquito will be the sum of both the speed of the mosquito and breeze thus the mosquito and breeze are moving in the same direction so we just sum it up (3m/s + 3m/s = 6m/s).

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Consider the interactions involved when you use a TV remote control to change the channel. Classify each interaction as long ran

makvit [3.9K]

Explanation:

Following are two interactions that are generally involved when we use a TV remote control to change the channel :

1. Figure touches remote buttons, and its a short range interaction.

2. Now remote sends signal to Television, this is a long range interaction.

7 0

3 years ago

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon

LiRa [457]

Answer:

241.8 N.

Explanation:

The force on branch provides a reaction to the ape's weight force plus the centripetal force needed to keep the gibbon in a circular motion of radius 0.60 m.

Centripetal force = mv^2/r

F = mg + mv²/r

F = m(g + v²/r)

where,

m = mass

= 9 kg

g = acceleration due to gravity

= 9.8 m/s²

v = 3.2 m/s

r = 0.60 m

F = 9 * (9.8 + 3.2²/0.60)

= 241.8 N.

3 0

3 years ago

Capacitor C1 is initially charged to V1 and capacitor C2 is initially charged to V2. The capacitors are then connected to each o

o-na [289]

Answer:

<em>20.08 Volts</em>

Explanation:

<u>Parallel Connection of Capacitors</u>

The voltage across any two elements connected in parallel is the same. If the elements are capacitors, then each voltage is

$\displaystyle V_1=\frac{Q_1}{C_1}$

$\displaystyle V_2=\frac{Q_2}{C_2}$

They are both the same after connecting them, thus

$\displaystyle \frac{Q_2}{C_2}=\frac{Q_1}{C_1}$

Or, equivalently

$\displaystyle Q_2=\frac{C_2Q_1}{C_1}$

The total charge of both capacitors is

$\displaystyle Q_t=Q_1\left(1+\frac{C_2}{C_1}\right)$

We can compute the total charge by using the initial conditions where both capacitors were disconnected:

$Q_t=V_{10}C_1+V_{20}C_2=25\cdot 24+13\cdot 11=743\ \mu C$

Now we compute Q1 from the equation above

$\displaystyle Q_1=\frac{Q_t}{\left(1+\frac{C_2}{C_1}\right)}=\frac{743}{\left(1+\frac{13}{24}\right)}=481.95\ \mu C$

The final voltage of any of the capacitors is

$\displaystyle V_1=V_2=\frac{481.95}{24}=20.08\ V$

7 0

3 years ago

What color is seen when the red light is on

Dennis_Churaev [7]

Answer:

Red is seen

Explanation:

8 0

3 years ago

How much heat must be removed from 200 pounds of blanched vegetables at 189 degrees F to freeze them to a temperature of 22 degr

Virty [35]

To solve this problem it is necessary to apply the concepts related to heat exchange in the vegetable and water.

By definition the exchange of heat is given by

$Q =mc\Delta T$

where,

m = mass

c = specific heat

$\Delta T$ = Change in temperature

Therefore the total heat exchange is given as

$\Delta Q = Q_w+Q_v$

$\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)$

Our values are given as,

Total mass is $M_T$ = 200lb ,however the mass of solid vegetable and water is given as,

$m_v= 0.4*200lb = 80lb$

$m_w=0.6*200lb=120lb$

$T_i = 183\°F\\T_f = 22\°F\\c_w = 1Btu/lbF\\c_v = 0.24Btu/lbF$

Replacing at our equation we have,

$\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)$

$\Delta Q = (120)(1)(183-22)+(80)(0.24)(183-22)$

$\Delta Q = 22411.2Btu$

Therefore the heat removed is 22411.2 Btu

6 0

3 years ago