The earth’s magnetic field is located in the ____

An electric circuit has a 200 22 resistor and a current of 5 A. What is its voltage?

Wittaler [7]

Answer:

ccccccc. ccccccc ccc ccccccc c

8 0

3 years ago

The isotope 23893Np has a half-life of 2.0 days. If 4.00 grams are produced at noon on Monday, what will be the mass of neptuniu

viva [34]

Answer:

0.25 gram of neptunium is remaining

Explanation:

First we calculate the no. of half lives passed. For that we have formula:

n = t/T

where,

n = no. of half lives passed = ?

t = total time passed = 8 days (From Monday noon to Tuesday noon of following week)

T = Half Life Period = 2 days

Therefore,

n = 8 days/2 days

n = 4

Now, for the remaining mass of neptunium, we use the formula:

m = (mi)/(2)^n

where,

mi = initial mass of neptunium = 4.00 grams

m = remaining mass of neptunium = ?

Therefore,

m = 4 grams/2⁴

3 0

3 years ago

The intensity of sunlight reaching the earth is 1360 w/m2. Assuming all the sunligh is absorbed, what is the radiation pressure

krok68 [10]

(a) $1.15\cdot 10^9 N$

For an electromagnetic wave incident on a surface, the radiation pressure is given by (assuming all the radiation is absorbed)

$p=\frac{I}{c}$

where

I is the intensity

c is the speed of light

In this problem, $I=1360 W/m^2$ ; substituting this value, we find the radiation pressure:

$p=\frac{1360 W/m^2}{3\cdot 10^8 m/s}=4.5\cdot 10^{-6} Pa$

the force exerted on the Earth depends on the surface considered. Assuming that the sunlight hits half of the Earth's surface (the half illuminated by the Sun), we have to consider the area of a hemisphere, which is

$A=2pi R^2$

where

$R=6.37\cdot 10^6 m$

is the Earth's radius. Substituting,

$A=2\pi (6.37\cdot 10^6 m)^2=2.55\cdot 10^{14}m^2$

And so the force exerted by the sunlight is

$F=pA=(4.5\cdot 10^{-6} Pa)(2.55\cdot 10^{14} m^2)=1.15\cdot 10^9 N$

(b) $3.2\cdot 10^{-14}$

The gravitational force exerted by the Sun on the Earth is

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=1.99\cdot 10^{30}kg$ is the Sun's mass

$m=5.98\cdot 10^{24}kg$ is the Earth's mass

$r=1.49\cdot 10^{11} m$ is the distance between the Sun and the Earth

Substituting,

$F=(6.67\cdot 10^{-11} )\frac{(1.99\cdot 10^{30}kg)(5.98\cdot 10^{24} kg)}{(1.49\cdot 10^{11} m)^2}=3.58\cdot 10^{22}N$

And so, the radiation pressure force on Earth as a fraction of the sun's gravitational force on Earth is

$\frac{1.15\cdot 10^9 N}{3.58\cdot 10^{22}N}=3.2\cdot 10^{-14}$

6 0

4 years ago

Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti

Zigmanuir [339]

Answer:

10.1 m/s

Explanation:

By Newton's third law, the force on the squid and that due to the water expelled form an action reaction pair.

And by the law of conservation of momentum,

initial momentum of squid + expelled water = final momentum of squid + expelled water.

Now, the initial momentum of the system is zero.

So, 0 = final momentum of squid + expelled water

0 = MV + mv where M = mass of squid = 6.50kg, V = velocity of squid = 2.40m/s, m =mass of water in cavity = 1.55 kg and v = velocity of water expelled

So, MV + mv = 0

MV = -mv

v = -MV/m

= -6.50 kg × 2.40 m/s ÷ 1.55 kg

= -15.6 kgm/s ÷ 1.55 kg

= -10.1 m/s

So, speed must it expel this water to instantaneously achieve a speed of 2.40 m/s to escape the predator is 10.1 m/s

4 0

3 years ago

A massless spring with force constant ????=200N/m hangs from the ceiling. A 2.0-kg block is attached to the free end of the spri

Makovka662 [10]

Answer:

-0.4454 Joules

Explanation:

m = Mass of block = 2 kg

h = Height of extension = 17 cm = x

g = Acceleration due to gravity = 9.81 m/s²

Potential energy of the spring

$P=mgh\\\Rightarrow P=2\times 9.81\times 0.17\\\Rightarrow P=3.3354\ J$

The kinetic energy of the spring

$K=\frac{1}{2}mx^2\\\Rightarrow K=\frac{1}{2}\times 200\times 0.17^2\\\Rightarrow K=2.89\ J$

In this system as the potential and kinetic energy is conserved from work energy equivalence we get

$W=P-K\\\Rightarrow W=2.89-3.3354\\\Rightarrow W=-0.4454\ J$

The work done by friction is -0.4454 Joules

8 0

3 years ago

The earth’s magnetic field is located in the _____.