Ariel's puppy weighs 15 lb. Heather's puppy weighs 50 N. Whose puppy weighs more?

Please Help!!! I WILL GIVE BRAINLIEST!!!! An electron is in motion at 4.0 × 10^6 m/s horizontally when it enters a region of spa

dybincka [34]

Answer:

a. 9.52 cm b. 4.34 × 10⁶ m/s

Explanation:

a. The horizontal distance traveled by the electron when it hits the plate.

The electric force F on the electron due to the electric field E of mass, m is

F = -eE = ma

a = -eE/m where a = acceleration of electron

The vertical distance moved by the electron is given by

Δy = ut +1/2at²

u = initial vertical velocity = 0. and take the top plate as y = 0 and bottom plate as y

So,

0 - y = 0 × t + 1/2at²

-y = 1/2at²

substituting a = -eE/m

-y = 1/2(-eE/m)t²

y = eEt²/2m

making t subject of the formula,

t = √(2ym/eE) where t is the time it takes to reach the bottom plate.

Since E = 4.0 × 10² N/C, y = distance between plates = 2.0 cm = 0.02 m, m = 9.109 × 10⁻³¹kg and e = 1.602 × 10⁻¹⁹ C

t = √[(2 × 0.02 m × 9.109 × 10⁻³¹kg)/(1.602 × 10⁻¹⁹ C × 4.0 × 10² N/C)]

t = √[(0.36436 × 10⁻³¹kgm)/(6.408 × 10⁻¹⁷ N)]

t = √[(0.0569 × 10⁻¹⁴kgm/N)t

t = 0.238 × 10⁻⁷ s

t = 23.8 × 10⁻⁹ s

t = 23.8 ns

The horizontal distance moved when it hits the plates x = vt where v = initial horizontal velocity = 4.0 × 10⁶ m/s

x = 4.0 × 10⁶ m/s × 23.8 × 10⁻⁹ s

= 0.0952 m

= 9.52 cm

b. The velocity of the electron as it strikes the plate.

To find the velocity of the electron as it strikes the plates, we calculate its final vertical velocity V as it strikes the plate. This is gotten from

v' = u + at since u = 0,

v' = at

= -eEt/m

= -(1.602 × 10⁻¹⁹ C × 4.0 × 10² N/C × 0.238 × 10⁻⁷ s)/9.109 × 10⁻³¹kg

= -1.525 × 10⁻²⁴ Ns/9.109 × 10⁻³¹kg

= -0.167 × 10⁷ m/s

= -1.67 × 10⁶ m/s

So, the resultant velocity as it strikes the plate v = √(v'² + v²)

= √((-1.67 × 10⁶ m/s)² + (4 × 10⁶ m/s)²)

= √(2.7889 + 16) × 10⁶ m/s

= √18.7889 × 10⁶ m/s

= 4.335 × 10⁶ m/s

≅ 4.34 × 10⁶ m/s

6 0

4 years ago

A wave is incident on the surface of a mirror at an angle of 41° with the normal. What can you say about its angle of reflection

bazaltina [42]

Hold on 63 deg angle my friend

7 0

4 years ago

Read 2 more answers

A polarized light that has an intensity I0 = 52.0 W/m² is incident on four polarizing disks whose planes are parallel and center

IceJOKER [234]

Answer:

The transmitted intensity through all polarizers = 34.73

Explanation:

Given :

Incident intensity = $52 W/m^2$

Angle between the transmission axis and polarizer optic axis = 18°

According to the malus law, when unpolarized or polarized light passes through polarizing disk, the intensity of the transmitted light is directly proportional to the square of the cosine of angle between the transmission axis and polarizer optic axis.

∴ $I = I' cos^2\alpha$

Where $I=$ transmitted intensity, $I'=$ incident intensity, $\alpha =$ angle between the transmission axis and polarizer optic axis.

Here, there are four polarizing disks so that.

from first disk,

∴ $I$ ₁ $= 52 (cos)^2$ 18°

= $52$ × $0.904$

= $47.01$

Now $I$ ₁ behave as an incident light for second polarizer so we only multiply

$cos^218$ term

so we write,

∴ $I$ ₂ = $47.01$ × $0.904$

$= 42.495$

From third polarizer,

∴ $I$ ₃ = $42.495$ × $0.904$

$= 38.415$

From forth polarizer,

∴ $I$ ₄ = $38.415$ × $0.904$

$= 34.73$

Therefor, the the transmitted intensity through forth polarizer = 34.73 $W/m^2$ .

5 0

3 years ago

A pebble is thrown into the air with a velocity of 19/m at an angle of 36 with respect to the horizontal.

kow [346]

Answer:

The maximum height the pebble reaches is approximately;

A. 6.4 m

Explanation:

The question is with regards to projectile motion of an object

The given parameters are;

The initial velocity of the pebble, u = 19 m/s

The angle the projectile path of the pebble makes with the horizontal, θ = 36°

The maximum height of a projectile, $h_{max}$ , is given by the following equation;

$h_{max} = \dfrac{\left (u \times sin(\theta) \right)^2}{2 \cdot g}$

Therefore, substituting the known values for the pebble, we have;

$h_{max} = \dfrac{\left (19 \times sin(36 ^{\circ}) \right)^2}{2 \times 9.8} = 6.3633894140470403035477570509439$

Therefore, the maximum height of the pebble projectile, $h_{max}$ ≈ 6.4 m.

3 0

3 years ago

A box slides across a frozen pond toward the left shore as two children standing on opposite shores pull on the box with ropes.

erik [133]

Answer:

given,

children on left shore pulls the rope by 3 N

Children on the right shore pulls the rope by 2 N

speed of the box = 3 m/s

a) Power = Force x Velocity

Power = 3.0 N * 3

Power = 9 Watt

Power = 2.0 N x 3

Power = 6 Watt

b) Net Force = 3.0 N - 2.0 N = 1.0 N

Net Power = 1.0 x 3 Watt

Net Power = 3 Watt

c) net force is acting on the box and at an instant velocity is given so, when the body moves it will accelerate and velocity will change so, the value of part A and B will also change

5 0

3 years ago