Lightning results from _______

Alex Ar [27]

Yeah the hand is distal to the elbow

7 0

3 years ago

A 460 nm monochromatic light is split into two beams. The two beams arrive at a point on a screen with a path difference of 920n

VladimirAG [237]

Answer:

phase difference $\Delta \Phi = 12.566$

Explanation:

Phase difference is expressed as $\Delta \Phi$

$\Delta \Phi =\frac {2\pi}{\lambda} \Delta X$

Where,

$\Delta \Phi$ = phase difference

$\lambda$ wavelength = 460nm

$\Delta X$ path difference = 920 nm

$\Delta \Phi =\frac {2\pi}{460} *920 = 4\pi$

therefore

phase difference $\Delta \Phi = 12.566$

7 0

4 years ago

A boy is exerting a force of 70 N at a 50-degree angle on a lawn mower. He is accelerating at 1.8 m/s2. Round the answers to the

loris [4]

Answer:

The mass of the lawn mower is 25 kg

The normal force exerted on the lawn mower is 299 N

Explanation:

The boy is exerting a force of 70 N at a 50-degree angle on a lawn mower

We must to distribute the for into two component

→ Horizontal component in direction of motion

→ Vertical component acting down ward

The angle between the exerting force and the lawn mower which is

horizontal

→ The horizontal component = 70 cos(50) in direction of motion

→ The vertical component = 70 sin(50) down ward

Due to Newton's Law;

→ ∑ Force in direction of motion = mass × acceleration

→ F = 70 cos(50) , acceleration = 1.8 m/s²

Substitute these values in the rule above

→ 70 cos(50) = mass × 1.8

Divide both side by 1.8

→ mass = 25 kg

<em>The mass of the lawn mower is 25 kg</em>

The normal force R is exerted on the lawn mower is vertically upward

The weight of the lawn mower and the vertical component of the

exerted force acting downward

→ ∑ The vertical forces = 0

→ ∑ vertical forces = R - 70 sin(50) - mg, mg is the weight of the

lawn mower

→ R - 70 sin(50) - 25(9.8) = 0

→ R - 53.62 - 245 = 0

→ R - 289.62 = 0

Add 289.62 to both sides

→ R = 289.62 ≅ 299 N

<em>The normal force exerted on the lawn mower is 299 N</em>

7 0

4 years ago

Read 2 more answers

A car is driving at 85 km/h and the driver spots a stop sign ahead. What coefficient of friction is needed to stop the car at th

almond37 [142]

Answer:

μ = 0.0315

Explanation:

Since the car moves on a horizontal surface, if we sum forces equal to zero on the Y-axis, we can determine the value of the normal force exerted by the ground on the vehicle. This force is equal to the weight of the cart (product of its mass by gravity)

N = m*g (1)

The friction force is equal to the product of the normal force by the coefficient of friction.

F = μ*N (2)

This way replacing 1 in 2, we have:

F = μ*m*g (2)

Using the theorem of work and energy, which tells us that the sum of the potential and kinetic energies and the work done on a body is equal to the final kinetic energy of the body. We can determine an equation that relates the frictional force to the initial speed of the carriage, so we will determine the coefficient of friction.

$\frac{1}{2} *m*v_{i}^{2}-(F*d)= \frac{1}{2} *m*v_{f}^{2}$

where:

vf = final velocity = 0

vi = initial velocity = 85 [km/h] = 23.61 [m/s]

d = displacement = 900 [m]

F = friction force [N]

The final velocity is zero since when the vehicle has traveled 900 meters its velocity is zero.

Now replacing:

(1/2)*m*(23.61)^2 = μ*m*g*d

0.5*(23.61)^2 = μ*9,81*900

μ = 0.0315

5 0

3 years ago

A student measures the speed of sound by echo destiny classes hands and then measures the time to hear the echo his distance to

777dan777 [17]

Explanation:

∆x=300 m×2

∆t=1.5 s

v=∆x/∆t → v=2×300/1.5 = 400 m/s

6 0

3 years ago

Lightning results from ________.