Write the nuclear equation for the release of the beta particles by Pb-210.

A lifeguard on a beach observes that waves have a speed of 2.60 m/s and a distance of 2.50 m between wave crests. What is the pe

Alja [10]

Answer:

T = 0.96 seconds

Explanation:

Given that,

The speed of wave, v = 2.6 m/s

The distance between wave crests, $\lambda=2.5\ m$

We need to find the period of the wave motion. Let T be the period. So,

$v=\dfrac{\lambda}{T}\\\\T=\dfrac{\lambda}{v}\\\\T=\dfrac{2.5}{2.6}\\\\T=0.96\ s$

So, the period of the wave motion is equal to 0.96 seconds.

4 0

3 years ago

Question 1(Multiple Choice Worth 4 points)

ArbitrLikvidat [17]

Question 1: Amount of energy
Question 2: Freezing ice
Question 3: Chemical change
Question 4: More energy is given off by the new bonds formed than is needed to break the original bonds.
Question 5: The resulting mass would be 35 grams.

4 0

3 years ago

Which situation has the greatest pressure? A. Atmospheric pressure at sea level, equal to 1.01 X 105 N/m2 B. The pressure that a

Alexeev081 [22]

Answer:

The correct option is B

Explanation:

In part A pressure is given as $1.01*10^5 N/m$

In Part B we can calculate the pressure using the formula

$P = \frac{F}{A}$

Where F is the force which in this case is weight which can be obtained as follow

W = Mg

Substituting 70kg for M and $9.8m/s^2$ for g

$W = 70 *9.8$

$=686N$

A is the area with a value $50cm^2 = \frac{50}{10000} = 5.0*10^{-3}m^2$

So substituting into the above equation

$P = \frac{686}{5.0*10^{-3}}$

$= 1.37 *10^5 N/m^2$

for part c

A $= 22*28 cm^2= \frac{22*28}{10000}m^2$

$=6.16*10^{-2}m^2$

F $= 2.5 * 9.8$

$= 24.525N$

Therefore $P = \frac{24.252}{6.16*10^{-2}}$

$= 298 N/m^2$

From this we see that the value of P for part B is the greatest

5 0

3 years ago

Add the following forces: 20 Newtons south, 30 newtons north, 40 newtons east, and 50 newtons west

Roman55 [17]

The north and south forces have no effect on the east and west forces.
We can handle the two groups separately.

   (20N south) + (30N north)

   = (20N south) + (20N north + 10N north)

   = 10N north .

   (40N east) + (50N west)

   = (40N east) + (40N west + 10N west)

   = 10N west .

5 0

3 years ago

Suppose a thick nimbostratus cloud contains ice crystals and cloud droplets all about the same size. Which precipitation process

Ronch [10]

Answer:The most important process would be the ice crystal process

Explanation:

Ice crystal process also called Bergeron process requires numerous small water drops that are supercooled, which is a common feature in clouds between about 0° and -20°C or below, along with a small number of ice crystals. Therefore because the collision-coalescence process requires that cloud droplets be of varying size so that drops will fill at different speeds, the most important process would be the ice crystal process.

7 0

3 years ago