Every magnet must have what at its ends?

An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the accelerati

lubasha [3.4K]

Answer:

The correct option is

a. v = $g (1-e^{-bt})/b$

Explanation:

Time at which the object start fall t = 0

The acceleration a is given by a = g - bV

Where V = Speed of the object

Speed V² = u² + 2·a·h

However with the drag force the object will approach terminal velocity as t becomes progressively larger whereby v will stop increasing

Option a. is the only option that has limiting value of v which is in the range of g as t increases ∴ option a. is the correct option.

v = $g (1-e^{-bt})/b$ as t increases $(1-e^{-bt})$ → 1 s and v→ g/b m/s

6 0

3 years ago

Calculate the minimum frequency of ultrasound (in Hz) that will allow you to see details as small as 0.193 mm in human tissue. (

STatiana [176]

Answer:

f = 7.97 x 10⁶ Hz = 7.97 MHz

Explanation:

The speed of a wave is given by the following formula:

$v = f\lambda$

where,

v = speed of the ultrasound wave through human tissue = 1540 m/s

f = frequency of ultrasound wave required = ?

λ = wavelength of ultrasound waves = smallest detail required = 0.193 mm

λ = 0.193 mm = 1.93 x 10⁻⁴ m

Therefore,

$1540\ m/s = f(1.93\ x\ 10^{-4}\ m)\\f = \frac{1540\ m/s}{1.93\ x\ 10^{-4}\ m}$

8 0

3 years ago

A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×10^{4} 4 m/s^{2} 2 , and 1.85 ms (1

Paha777 [63]

Answer:

u = - 38.85 m/s^-1

Explanation:

given data:

acceleration = 2.10*10^4 m/s^2

time = 1.85*10^{-3} s

final velocity = 0 m/s

from equation of motion we have following relation

v = u +at

0 = u + 2.10*10^4 *1.85*10^{-3}

0 = u + (21 *1.85)

0 = u + 38.85

u = - 38.85 m/s^-1

negative sign indicate that the ball bounce in opposite directon

6 0

3 years ago

Un montañero de 65kg de masa ha ascendido a la cima del Everest, la montaña más alta del mundo de 8848m de altura sobre el nivel

andrew-mc [135]

Answer:

El trabajo realizado para subir los últimos 500 metros es 318727,5 joules.

Explanation:

Por la definición de trabajo sabemos que el montañero debió contrarrestar trabajo causado por la gravedad terrestre. Si asumimos que el cambio de la altura es muy pequeño en comparación con el radio del planeta (6371 kilómetros vs. 0,5 kilómetros), entonces podemos considerar que la aceleración gravitacional es constante y la ecuación de trabajo ( $\Delta W$ ), medido en joules, que reducida a: