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2 years ago

6

What would you expect to cause a drop in air pressure?

2 answers:

givi [52]2 years ago

7 0

D. The air temperature staying the same

Murrr4er [49]2 years ago

6 0

Answer:

D

Explanation:

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Find the sum of the vectors:11 km N ,11km E

vladimir1956 [14]

The resultant vector is 11√2 km due north east.

<h3><u>Explanation:</u></h3>

The vector is a type of quantity which has both magnitude and direction. This quantities when expressed needs to specify both magnitude and direction.

We need to calculate the magnitude and direction separately.

Here firstly for the magnitude,

The magnitudes are both 11 km and they are at right angles to each other.

So, the resultant magnitude = √(11² +11²) km

=11√2 km

Now for the direction, one vector is due north and the other is due east.

So the resultant vector is due north east.

So the final vector is 11√2 km due North-East.

6 0

2 years ago

10. Someone takes 11 minutes to walk up a hill 120m high. His weight is 550N.

belka [17]

Answer:

okay with you if you want to

8 0

1 year ago

The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various

Schach [20]

Answer:

$q = -461532.5 \ C$

Explanation:

From the question we are told that

The electric filed is $E = 102 \ N/C$

Generally according to Gauss law

=> $E A = \frac{q}{\epsilon_o }$

Given that the electric field is pointing downward , the equation become

$- E A = \frac{q}{\epsilon_o }$

Here $q$ is the excess charge on the surface of the earth

$A$ is the surface area of the of the earth which is mathematically represented as

$A = 4\pi r^2$

Where r is the radius of the earth which has a value $r = 6.3781*10^6 m$

substituting values

$A = 4 * 3.142 * (6.3781*10^6 \ m)^2$

$A =5.1128 *10^{14} \ m^2$

So

$q = -E * A * \epsilon _o$

Here $\epsilon_o$ s the permitivity of free space with value

$\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

$q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}$

$q = -461532.5 \ C$

6 0

3 years ago

which type of wave spreading do you think causes faster energy loss-two-dimensional or three-dimensional? explain.

sdas [7]

Three dimensional would loose faster

3 0

2 years ago

Read 2 more answers

A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary

Snezhnost [94]

Explanation:

Let $N_p\ and\ N_s$ are the number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p}{N_s}=\dfrac{1}{3}$

A resistor R connected to the secondary dissipates a power $P_s=100\ W$

For a transformer, $\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}$

$V_s=(\dfrac{N_s}{N_p})V_p$

$V_s=3V_p$ ...............(1)

The power dissipated through the secondary coil is :

$P_s=\dfrac{V_s^2}{R}$

$100\ W=\dfrac{V_s^2}{R}$

$V_p^2=\dfrac{100R}{9}$ .............(2)

Let $N_p'\ and\ N_s'$ are the new number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p'}{N_s'}=\dfrac{1}{24}$

New voltage is :

$V_s'=(\dfrac{N_s'}{N_p'})V_p'$

$V_s'=24V_p$ ...............(3)

So, new power dissipated is $P_s'$

$P_s'=\dfrac{V_s'^2}{R}$

$P_s'=\dfrac{(24V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}$

$P_s'=6400\ Watts$

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0

2 years ago