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mafiozo [28]
3 years ago
12

An electron and a proton have the same kinetic energy upon entering a region of constant magnetic field and their velocity vecto

rs are perpendicular to the magnetic field. Suppose the magnetic field is strong enough to allow the particles to circle in the field. Note: you'll need to look up the masses for an electron and proton. 1) What is the ratio of the radii of their circular paths rp/re?
Physics
1 answer:
kupik [55]3 years ago
4 0

Answer: rp/re= me/mp= 544 * 10^-6.

Explanation: To calculate this problem we have to consider the circular movement by the electron and proton inside a magnetic field.

Then the dynamic equation for the circular movement is given by:

Fcentripetal= m*ω^2.r

q*v*B=m*ω^2.r

we write this for each particle then we have the following:

q*v*B=me* ω^2*re

q*v*B=mp* ω^2*rp

rp/re=me/mp=9.1*10^-31/1.67*10^-27=544*10^-6

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You apply a horizontal force of 25N to push a shopping cart across the parking lot at a constant velocity. a) what is the net fo
AlekseyPX

(a) The net force on the shopping cart is zero.

(b) The the force of friction on the shopping cart is 25 N.

(c) When same force is applied to the shopping cart on a wet surface, it will move faster.

<h3>Net force on the shopping cart</h3>

The net force on the shopping cart is calculated as follows;

F(net) = F - Ff

where;

  • F is the applied force
  • Ff is the frictional force

ma = F - Ff

where;

  • a is acceleration of the cart
  • m is mass of the cart

at a constant velocity, a = 0

0 = F - Ff

F(net)  = 0

F = Ff = 25 N

Net force is zero, and frictional force is equal to applied force.

<h3>On wet surface</h3>

Coefficient of kinetic friction of solid surface is greater than that of wet surface.

Since frictional force limit motion, when the frictional force is smaller, the object tends to move faster.

Thus, the cart will move faster on a wet surface due to decrease in friction.

Learn more about frictional force here: brainly.com/question/24386803

#SPJ1

4 0
2 years ago
Why do people bleed .-.
SashulF [63]
Because the liver helps to regulate blood clotting, people<span> with liver disease (for example, hepatitis or cirrhosis) also have a tendency to </span>bleed<span> easily. Most commonly, easy or excessive bruising occurs because the skin and blood vessels are fragile.

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3 0
3 years ago
Read 2 more answers
A bowling ball is traveling at 3 m/s. It starts to roll up a ramp. How high above the ground will the ball be when it stops roll
Zepler [3.9K]

Answer:

FREE FREE FREE FREEE FREEEEEEEEEEE FREEEEEEEEEEEEEEEE FREEEEEEEEEEEEEEEEEE

Explanation:

FREE FREE FREE FREEE FREEEEEEEEEEE FREEEEEEEEEEEEEEEE FREEEEEEEEEEEEEEEEEE

8 0
3 years ago
Read 2 more answers
Determine the thrust that a boat with a volume of 1.2m³ receives when it is stranded at sea. The density of seawater is 1020kg /
puteri [66]

Answer:

The maximum possible up-thrust on the boat is 11,995.2 N

Explanation:

According to Archimedes' principle, the thrust received by an object immersed a fluid is equal to the weight of the fluid displaced;

The given parameter of the boat in sea water are;

The volume of the boat = 1.2 m³

The density of seawater = 1020 kg/m³

Density = Mass/Volume

Therefore, Mass = Density × Volume

The maximum volume of water that the boat displaces = 1.2 m³

The mass of the water displaced by the boat = (Density of seawater) × (Volume of seawater displaced)

∴ The maximum possible mass of the water displaced by the boat = 1.2 m³ × 1020 kg/m³ = 1224 kg

The maximum possible mass of the water displaced by the boat, m = 1224 kg

Weight = Mass, m × g

Where;

g = The acceleration due to gravity = 9.8 m/s²

The up-thrust on the boat = The weight of the seawater displaced

∴ The maximum possible up-thrust on the boat = m × g = 1224 kg × 9.8 m/s² = 11,995.2 N

The maximum possible up-thrust on the boat = 11,995.2 N.

3 0
3 years ago
If two equal charges are separated by a certain distance, the force of repulsion is F. Given F1, if each charge in F1 is doubled
Alisiya [41]

Answer:F_2=16\times F_1

Explanation:

Given

Force of repulsion between two charge particle is given by force F

Electrostatic force is given by

F=\frac{kq_1q_2}{r^2}

where q_1 and q_2 is the charges  of particle

r=distance between charge particle

=\frac{kq^2}{r^2}when charges are doubled and distance is reduced to half

i.e. q become 2 q and r becomes 0.5 r

F_2=\frac{k(2q)^2}{(0.5r)^2}

F_2=\frac{kq^2}{r^2}\times 4\times 4

F_2=16\times F_1

         

6 0
3 years ago
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