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Ludmilka [50]
3 years ago
12

Please help I just need help for question 4. And 5. (They’re at the top)

Physics
1 answer:
olya-2409 [2.1K]3 years ago
7 0

Explanation:

4.he invented the atomic model with all different electrons proton and neutrons

5. this was important as this was developed and is still used in the modern day

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An electric heating element has a resistance of 16 Ω and is connected to a voltage of 120 V. How much current will flow in this
sergeinik [125]

This is weird.

All three 'choices' are true.

Line um up. (a) shows how to solve the problem. (b) does it. and (c) is the answer.

7 0
4 years ago
A 1,000 kg car is travelling at 6.5 m/s to the North. A 3,500 kg truck is travelling South at the same velocity. What is the tot
Molodets [167]

Answer:

16250 kgm/s due south

Explanation:

Applying,

M = mv................. Equation 1

Where M = momentum, m = mass, v = velocity.

From the car,

Given: m = 1000 kg, v = 6.5 m/s

Substitute these values into equation 1

M = 1000(6.5)

M = 6500 kgm/s

For the truck,

Given: m = 3500 kg, v = 6.5 m/s

Substitute these values into equation 1

M' = 3500(6.5)

M' = 22750 kgm/s.

Assuming South to be negative direction,

From the question,

Total momentum of the two vehicles = (6500-22750)

Total momentum of the two vehicles = -16250 kgm/s

Hence the total momentum of the two vehicles is 16250 kgm/s due south

3 0
3 years ago
A compass needle acts as a(n)
beks73 [17]

Explanation:

A compass needle acts as a Magnet

6 0
3 years ago
Please Help On These 2 Questions!!!!! I severely need help!!!!!!
lys-0071 [83]
3. In a uniform electric field, the equation for the magnitude of the magnetic field is E=(V/d). V= voltage d= distance. If the magnetic field magnitude is constant , as stated in your problem, then the voltage must stay the same otherwise the value of "E" would change". And the problem already told us the "E" is uniform and so, not changing. Does that make sense?

4a. If the magnetic field lines are equally spaced apart, in other words share the same density. Then we know that the magnitude of the magnetic field is unchanging. This is because the density of of the magnetic field lines(how many are in a certain area) is related to the magnitude being expressed by the electric field. Greater magnitude is expressed by the presence of more lines (higher line density) 

4b. The electric potential is measured in Volts(V) and is uniform along  the same equipotential line. What is an equipotential line(gray)? It is a line drawn perpendicular(forms a right angle with) to the magnetic field lines(black) to show the changes in electric potential. One space where electric potential will always be the same because it will always be equal to 0 Volts is exactly in between a positive and negative charges of equal charge value I have pointed to this line with a purple arrow in my picture.

I really hope this makes sense to you and that my pictures help! :)

3 0
3 years ago
Can someone solve this problem and explain to me how you got it​
evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒F\alpha\frac{q1*q2}{r^{2}}

∴F=k\frac{q1*q2}{r^{2}}

where F is the force of attraction or repulsion

k is Coulumb's constant=9*10^{9}Nm^{2}C^{-2}

q1 and q2 are the magnitude of the charges

r is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

F=9*10^{9}*\frac{0.041*0.029}{12^{2}}

F=74312.5N

Question6:

Given: q1=3*10^{-18}C, r=5 m, F=4*10^{-11}N

therefore by Coulumb's law,

4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}

⇒q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C

4 0
4 years ago
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