All categories

3 years ago

It is winter in Puerto Rico. Compare the air temperatures beachgoers feel near the water

1 answer:

7 0

Large amounts of water do have a big impact on the weather: indeed, it takes less energy to warm/cool land than water.
Therefore, places near large amounts of water tend to have smaller differences in temperature between summer and winter than places far from waters.

Hence, during winter in Puerto Rico, alongside the coast, the temperature will be higher than in the innermost parts of the island.

You might be interested in

A block of wood mass 0.60kg is balanced on top of a vertical port 2.0m high. A 10gm bullet is fired horizontally into the block

anzhelika [568]

Answer:

Mass of bullet is m=0.01kg

Mass of the block is M=4kg

Coefficient=0.25,distance=20m

So, let the speed of the block just after the bullet embedded in it be V and v be the speed of bullet before striking the block,

By applying conservation of momentum,

mv=(m+M)V

M+m

Explanation:

please mark me as the brainliest answer and please follow me for more answers to your questions..

7 0

3 years ago

Which of the following are the advantages of a DBMS (database management system)?

soldier1979 [14.2K]

Answer:

Explanation:

6 0

3 years ago

Se golpea una pelota de golf de manera que su velocidad inicial forma un ángulo de 45° con la horizontal. La pelota alcanza el s

nordsb [41]

Answer:

42m/s

6.06s

Explanation:

To find the initial velocity and time in which the ball is fling over the ground you use the following formulas:

$x_{max}=\frac{v_o^2sin(2\theta)}{g}\\\\x_{max}=vt_{max}$

θ: angle = 45°

vo: initial velocity

g: gravitational constant = 9.8m/s^2

x_max: max distance = 180 m

t_max: max time

by replacing the values of the parameters and do vo the subject of the first formula you obtain:

$v_o=\sqrt{\frac{gx_{max}}{sin(2\theta)}}\\\\v_o=\sqrt{\frac{(9.8m/s^2)(180m)}{sin(2(45\°))}}=42\frac{m}{s}$

with this value of vo you calculate the max time:

$t_{max}=\frac{x_{max}}{v}=\frac{x_{max}}{v_ocos(45\°)}\\\\t_{max}=\frac{180m}{(42m/s)cos(45\°)}=6.06s$

hence, the initial velocity of the ball is 42m/s and the time in which the ball is in the air is 6.06s

- - - - - - - - - - - - -- - - - - - - - - - - - - -

TRANSLATION:

Para encontrar la velocidad inicial y el tiempo en el que la pelota está volando sobre el suelo, use las siguientes fórmulas:

θ: ángulo = 45 °

vo: velocidad inicial

g: constante gravitacional = 9.8m / s ^ 2

x_max: distancia máxima = 180 m

t_max: tiempo máximo

reemplazando los valores de los parámetros y haciendo el tema de la primera fórmula que obtiene:

con este valor de vo usted calcula el tiempo máximo:

por lo tanto, la velocidad inicial de la pelota es de 42 m / sy el tiempo en que la pelota está en el aire es de 6.06 s

4 0

2 years ago

An observer stands 24.7 m behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of

GaryK [48]

Explanation:

The given data is as follows.

Velocity of bullet, $c_{p}$ = 814.8 m/s

Observer distance from marksman, d = 24.7 m

Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.

t = $\frac{24.7}{343}$ (velocity in air = 343 m/s)

= 0.072 sec

Now, before the observer hears the report the distance traveled by the bullet is as follows.

$d_{b} = c_{b} \times t$

= $814.8 \times 0.072$

= 58.66

= 59 (approx)

Thus, we can conclude that each bullet will travel a distance of 59 m.

8 0

2 years ago

A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back

elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity= $9.8 m/s^2$

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates= $\mu mg=0.202\times 9.8\times 35$

Force by which box accelerates= $62.286 N$

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

$ma=\mu mg$

$a=\mu g=9.8\times 0.202=1.9796 m/s^2$

Hence, the truck can have maximum acceleration before the box slide= $1.9796 m/s^2$

3 0

3 years ago