If you are swimming upstream (i.e., against the current), at what speed does your friend on the shore see you moving?

Answer:

7.166 hours = 430 minutes.

Explanation:

As both trains are approaching each other on the same track, their relative speed is the sum of their individual speeds. Hence, the time until they intersect (and inevitably collide) is determined by how long it takes for speeds of 65 mph and 55 mph to cover the total distance of 860 miles. One train will cover part of the distance, while the other will cover the remainder. To calculate the required time, we can apply the formula:

1 hour ---> 120 miles

X ----> 860 miles; hence X = (860 miles * 1 hour)/120 miles = 43/6 hours = 7.16666 hours. To convert this into minutes, recall that 1 hour equals 60 minutes; therefore, 43/6 hours * 60 minutes/hour = 430 minutes.

Respuesta:

La magnitud de la aceleración resultante es 2.2

Explicación:

La masa (m) del velero es 2000 kg

La fuerza que actúa sobre el velero debido a la marea del océano es = 3000N

Hacia el este significa que se da a lo largo de la dirección positiva del eje x

Entonces = 3000N y = 0

La fuerza del viento que actúa sobre el velero es 6000N dirigida hacia el noroeste, lo que significa a un ángulo de 45 grados sobre el eje negativo x

Luego

= -(6000N) cos 45 grados = -4242.6 N

= (6000N) cos 45 grados = 4242.6 N

Por lo tanto, la fuerza neta que actúa sobre el velero en la dirección x es

= - 3000 N + 4242.6 N

= - 3000 N +4242.6 N

= 1242.6N

La fuerza neta que actúa sobre el velero en la dirección y es

= 0+ 4242.6N

= 4242.6N La magnitud de la fuerza resultante =

Usando el teorema de Pitágoras de 1243 N y 4243 N

4420.8 NF = ma

= 2.2

Response:

1 slit width = 0.158 mm, slit separation = 0.633 mm, distance between diffraction maxima = 12.7 mm

Explanation:

This involves defining several terms:

λ, the wavelength of the beam

D, the distance from the plane to the slit

x, the distance between minima in the diffraction pattern (in single slit setups)

w, the fringe width for double slit setups

1. In a double slit experiment, the fringe width is also recognized as the distance between Maxima.

Thus, w=λD/d, leading to d=λD/w when rearranged.

So, d= (632.8 x 10^-9 x 1 x 10^3)/1

which equals d= 0.633mm.

For single slit diffraction, minima are defined by a*sinΘ= m*x

where a is the slit width and m is an integer.

For small angles, it simplifies to Θ= (x/D) = (λ/a), allowing us to solve for a:

a = λD/a

a = (632.8x10^-9 x 1)/4

yielding a= 0.158mm.

2. A grating with 20 slits/mm gives d= 1/20mm= 0.05x10^-3.

To find y (the distance between maxima), apply y= λL/d:

Finally, y= (632.8 x 10^-9 x 1)/0.05x10^-3, which results in y= 12.7mm.

A basketball player maintains a steady pace of 2.5 m/s while throwing a basketball vertically at 6.0 m/s. How far does the player advance before getting the ball back? Air resistance is negligible. I was unsure which formula to apply to this scenario. Is there any relevance to an angle? First, we determine the duration to reach peak height. The total time for the flight will be double the ascent duration. According to Newton's equations of motion: v = u + at. At the highest point, v = 0, where u is 6 m/s. Thus, the equation becomes 0 = 6 - 9.81t, leading us to t = 0.61 seconds. Therefore, the total flight time equals 1.22 seconds as the player runs towards the ball at a horizontal speed of 2.5 m/s. The distance traveled can be calculated using distance = speed × time, resulting in distance = 2.5 m/s * 1.22, yielding a final distance of 6.11m.

The correct choice is D!

Clarification: