A block m1 rests on a surface. A second block m2 sits on top of the first block. A horizontal force F applied to the bottom block pulls both blocks at constant velocity. Here m1 = m2 = m.(a)
What is the normal force exerted by the surface on the bottom block? (Use the following as necessary: m and g as necessary.)

Answer:

Explanation:

Cart A is moving to the right with constant speed i.e. net acceleration is zero

because acceleration is change in velocity in given time

Cart B is moving towards right with gradually speed up so there is net acceleration which helps to increase the velocity s

This indicates the net force acting on the cart towards right

For cart C there is gradual slow down of cart which indicates cart is decelerating and a net force is acting towards which opposes its motion.

Answer:

P = 1.99 10⁸ Pa

Explanation:

The definition of the bulk module is

      B = - P / (ΔV / V)

The negative sign is included for which balk module is positive, P is the pressure and V that volume

They tell us that the variation in volume is 9.05%, that is

    ΔV / V = ​​9.0Δ5 / 100 = 0.0905

    P = - B DV / V

    P = 2.2 10⁹ (0.0905)

    P = 1.99 10⁸ Pa

Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = = = 3.61cm = 0.036m

r₂ = = = 5cm = 0.05m

electric potential V =

change in potential ΔV =

ΔV = , where 2.00μC

ΔV =

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ ×

ΔV= 2.789×10⁵

= ΔV × q₃

ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s

The speed of the charge q₃ when it starts from rest at y = 2 cm and reaches y = 4 cm is; v = 14.89 m/s

We are given;

Charge 1; q₁ = 2.00 μC = 2 × 10⁻⁶ C

Charge 2; q₂ = 2.00 μC = 2 × 10⁻⁶ C

Distance of charge 1 along x = 3 cm = 3 × 10⁻² m

Distance of charge 2 along x = -3 cm = -3 × 10⁻² m

Charge 3; q₃ = +4.00 μC  = 4 × 10⁻⁶ C

mass; m = 0.01 g

distance of charge 3 along y = 4 cm = 4 × 10⁻² m

q₃ starts from rest at y = 2 × 10⁻² m and reaches y = 4 × 10⁻² m.

Thus;

Distance of charge 1 from the initial position of q₃;

r₁ = √((3 × 10⁻²)² + ((2 × 10⁻²)²)

r₁ = 0.0361 m

Distance of charge 2 from the final position of q₃;

r₂ = √((3 × 10⁻²)² + ((4 × 10⁻²)²)

r₂ = 0.05 m

Now, formula for electric potential is;

V = kq/r

Where k = 9 × 10⁹ N.m²/s²

Thus,change in potential is;

ΔV = V₁ - V₂

Now, Net V₁ = 2kq₁/r₁

Net V₂ = 2kq₂/r₂

Thus;

ΔV = 2kq₁/r₁ - 2kq₂/r₂

ΔV = (2 × 9 × 10⁹)[(2 × 10⁻⁶/0.0361) - (2 × 10⁻⁶/0.05)]

ΔV = 277229.92 V

Now, from conservation of energy;

½mv² = q₃ΔV

Thus;

½ × 0.01 × v² = 4 × 10⁻⁶ × 277229.92

v² = 2 × 4 × 10⁻⁶ × 277229.92/0.01

v = √(221.783936)

v = 14.89 m/s

Read more about point charges at;brainly.com/question/13914561

Answer: wavelength =3.52m

Explanation:

,λ=c/μ

where c=speed of the light,λ=wave length, μ=frequncy

c=3x10^8m/s

And

μ=83.5/MHz =85.3x10^6Hz==85.3x10^6Hz=

=85.3x10^6s-1

λ=c/μ

=3x10^8m/s/85.3x10^6s-1

=3.51699883

=3.52m

Answer:

Part a)

Part b)

Yes it is the expected value of electric field at the surface of an atom

Part c)

Explanation:

Since negative charge of electrons in uniformly distributed in the atom while positive charge is concentrated at the nucleus

So the electric field due to positive charge of the nucleus is given as

now charge due to electrons inside a radius "r" is given as

now we will have electric field given as

now net electric field is given as

Part b)

At the surface of an atom

Yes it is the expected value of electric field at the surface of an atom

Part c)

If Z = 92

R = 0.10 nm

so we will have

Answer:

a ut will move faster than the large object was moving initially

Answer: It will move faster than the large object was moving initially.

Explanation: