Gamma rays are high-frequency EM waves and radio waves are low-frequency. Which characteristics are the same for both waves?

A particle whose speed is 50 m/s moves along the line from A(2,1) to B(9,25). Find its velocity vector in the form of ai +bj.

Kitty [74]

To write it in the form ai + bj, we need to find a and b which are:

a = x component

b = y component

length of x = 9 – 2 = 7

length of y = 25 – 1 = 24

tan θ = 24 / 7

θ = 73.74°

a = (50 m/s) cos 73.74

a = 14 m/s

b = (50 m/s) sin 73.74

b = 48 m/s

Hence,

14i + 48j

3 0

3 years ago

What are the three elements that can be used to make a magnet?

Bess [88]

Answer:

Iron, cobalt, and nickel are the only three naturally occurring elements that are magnetic.

Explanation:

hope it helps

correct me if I'm wrong thank you

brainliest please

7 0

3 years ago

Read 2 more answers

If Hubble's constant is 73 km/s/Mpc then the age of the universe is 13.4 billion years, assuming a uniform expansion. Suppose it

ki77a [65]

Answer:

It would decrease the calculated age of the universe

Explanation:

Since the age of the universe is the reciprocal of Hubble's constant, it's therefore if Hubble's constant is increased the age decreases but if the Hubble's constant is decreased, the age of universe increases. Therefore, the age of universe and Hubble's constant are inversely proportional. Conclusively, any attempt to increase Hubble's constant would imply the calculated age of the universe decreases.

3 0

3 years ago

Can someone explain and slove question 4

Sonbull [250]

A. NaOH
B. PbNO3
C. FeSO4
D. P2O3
E. CSe2
F. HC2H3O2
G. HCIO
H. H2SO4

6 0

3 years ago

Two small spheres assumed to be identical conductors are placed at 30 cm from each other on a horizontal axis. the first S1 is l

charle [14.2K]

a) The electric force exerted by S1 on S2 is 21.58μN.

In this case we are talking about two different types of charges, a positive charge and a negative charge, therefore, they are sensing a force of attraction.

The magnitude of the force is determined by using the following formula:

$F_{e}=k_{e}\frac{|q_{1}||q_{2}|}{r^{2}}$

where:

= Electric force [N]

= Electric constant ()

= First charge [C]

= Second charge [C]

r = distance between the two charges

So, in this case, the force can be calculated like this:

$F_{e}=(8.99x10^{9}N\frac{m^{2}}{C^{2}})\frac{|12x10^{-9}C||18x10^{-9}C|}{(30x10^{-2}m)^{2}}$

So the force will be equal to:

$F=21.58x10^{-6}N$

which is the same as:

$F=21.58 \micro N$

b) The electric field created by S1 at the level of S2 is $1.20 \frac{kN}{C}$

The electric field tells us how many Newtons of force can be applied on a given point in space per unit of charge caused by an existing electric charge. From the concept, we can take the following formula for the electric field.

$E_{S1}=\frac{F_{e}}{q_{2}}$

where:

= electric field generated by the first sphere.

$E_{S1}=\frac{1.20 x10^{-6}N}{18x10^{-9}C}$

which yields:

$E_{S1}=1.20x10^{3} \frac{N}{C}$

$E_{S1}=1.20 \frac{kN}{C}$

When talking about electric fields, we know what their direction is if we suppose the electric field is always affecting a positive charge in the given point in space. In this case, since S1 is positive, we can asume the electric field is in a direction away from S1.

c)

The electric potential created by S1 at the level or S2 is 360V

Electric potential is defined to be the amount of energy you will have at a given point per electric charge. This electric potential can be found by using the following formula:

V=Er

Where V is the electric potential and it is given in volts.

Volts are defined to be 1 Joule per Coulomb. Energy by electric charge.

So we can use the data found in the previous sections to find the electric potential:

$V=(1.20x10^{3} \frac{N}{C})(30x10^{-2}m)$

V=360V

d) The force exerted by S2 on S1 will be the same in magnitude as the force exerted by S1 on S2 but oposite in direction. This is because the force will depend on the two charges, and the distance between them, so:

The electric force exerted by S1 on S2 is 21.58μN.

The magnitude of the force is determined by using the following formula:

$F_{e}=k_{e}\frac{|q_{1}||q_{2}|}{r^{2}}$

$F_{e}=(8.99x10^{9}N\frac{m^{2}}{C^{2}})\frac{|12x10^{-9}C||18x10^{-9}C|}{(30x10^{-2}m)^{2}}$

So the force will be equal to:

$F=21.58x10^{-6}N$

which is the same as:

$F=21.58 \micro N$

e) The electric field generated by S1 in the middle of S1 and S2 is $4.79 \frac{kN}{C}$

In order to find the electric field generated by S1, we can make use of the following formula

$E=k_{e} \frac{q_{1}}{r_{1}^{2}}$

$E=(8.99x10^{9} N\frac{m^{2}}{C^{2}})(\frac{12x10^{-9}C}{(15x10^{-2}m)^{2}})$

which yields:

$E=4.79 \frac{kN}{C}$

f) The electric field in the middle of S1 and S2 is $11.99 \frac{kN}{C}$

In order to find the electric field generated by two different charges at a given point is found by using the following formula:

$E=k_{e} \sum \frac{q_{i}}{r_{i}^{2}}$

where:

$q_{i}$ = each of the charges in the system

$r_{i}$ = the distance between each of the charges and the point we are analyzing.

Since the electric field is a vector, we need to take into account the individual electric fields' directions. In this case we suppose we have a positive test charge between the two charges. We can see that the positive test charge will sense a force in the same direction independently on if the force is excerted by the positive charge or the negative charge. Therefore both electric fields will have the same direction. We'll suppose the electric fields will be positive then, so:

$E=(8.99x10^{9} N\frac{m^{2}}{C^{2}})[\frac{12x10^{-9}C}{(15x10^{-2}m)^{2}}+\frac{18x10^{-9}C}{(15x10^{-2}m)^{2}}]$

which yields:

$E=11.99 \frac{kN}{C}$

g) The electric potential in the middle of S1 and S2 is 1.80 kV

Since we know what the electric field is from the previous question, we can make use of the same formula we used before to find the electric potential in the middle of S1 and S2

So let's take the formula:

V=Er

So we can use the data found in the previous sections to find the electric potential:

$V=(11.99x10^{3} \frac{N}{C})(15x10^{-2}m)$