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katovenus [111]

1 year ago

10

imagine you are outside enjoying the warm sunshine with friends. as you briefly glance up at the sun, the part of the sun that y

ou can see directly is called its:

1 answer:

Mariulka [41]1 year ago

6 0

Photosphere

bc it is

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What is the charge on a hypothetical ion with 35 protons and 37 electrons?

irinina [24]

Charge= Protons- electrons
Charge= 35p-37e= -2

This Ion will have a charge of -2<span>. </span>

5 0

3 years ago

A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1

Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain $d=2.46\times10^{3}\ m$

Height of island $h=1.80\times10^{3}\ m$

Distance of the enemy ship from the mountain $d'=6.10\times10^{2}\ m$

Initial velocity $v=2.55\times10^{2}\ m/s$

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

$v_{x}=v\cos\theta$

Put the value into the formula

$v_{x}=2.55\times10^{2}\cos74.9$

$v_{x}=66.42\ m/s$

We need to calculate the vertical component of initial velocity

Using formula of vertical component

$v_{y}=v\sin\theta$

Put the value into the formula

$v_{y}=2.55\times10^{2}\sin74.9$

$v_{y}=246.19\ m/s$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v_{x}}$

$t=\dfrac{2.46\times10^{3}}{66.42}$

$t=37.03\ sec$

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

$H= v_{y}t-\dfrac{1}{2}gt^2$

Put the value in the equation

$H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2$

$H=2397.4\ m$

We need to calculate the distance close to the peak

Using formula of distance

$H'=H-h$

Put the value into the formula

$H'=2397.4-1800$

$H'=597.4\ m$

Hence, The distance close to the peak is 597.4 m.

6 0

3 years ago

Capacitor 2 has half the capacitance and twice the potential difference as capacitor 1. What is the ratio (U_{\rm C})_1/\,(U_{\r

IrinaK [193]

Answer:

1/2

Explanation:

The energy stored in a capacitor is given by

$U=\frac{1}{2}CV^2$

where

C is the capacitance

V is the potential difference

Calling $C_1$ the capacitance of capacitor 1 and $V_1$ its potential difference, the energy stored in capacitor 1 is

$U=\frac{1}{2}C_1 V_1^2$

For capacitor 2, we have:

- The capacitance is half that of capacitor 1: $C_2 = \frac{C_1}{2}$

- The voltage is twice the voltage of capacitor 1: $V_2 = 2 V_1$

so the energy stored in capacitor 2 is

$U_2 = \frac{1}{2}C_2 V_2^2 = \frac{1}{2}\frac{C_1}{2}(2V_1)^2 = C_1 V_1^2$

So the ratio between the two energies is

$\frac{U_1}{U_2}=\frac{\frac{1}{2}C_1 V_1^2}{C_1 V_1^2}=\frac{1}{2}$

4 0

3 years ago

The weight of an object on the surface of mass is 1850 Newton and its mass is 500kg.Find the acceleration due to gravity on mars

Fudgin [204]

F = m.a

a = F/m

a = 1850/500

a = 3.7 m/s²

3 0

2 years ago

Which frequencies can humans can hear but not mice

Blizzard [7]

Of approximately 1 to 4 kHz6 and approximately 16 kHz for mice.

6 0

3 years ago