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3 years ago

11

How are hurricanes and thunderstorms similar? A. Both cover an area as large as 650 km. B. Both involve the formation of large c

umulonimbus clouds. C. Both form when warm air rises, then cools quickly to form clouds.

2 answers:

Karo-lina-s [1.5K]3 years ago

6 0

Both form when warm air rises

Iteru [2.4K]3 years ago

5 0

Both form when warm air rises:)

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A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50

laila [671]

Answer:

The ball is dropped at a height of 9.71 m above the top of the window.

Explanation:

<u>Given:</u>

Height of the window=1.5 m
Time taken by ball to cover the window height=0.15

Now using equation of motion in one dimension we have

$s=ut+\dfrac{at^2}{2}$

Let u be the velocity of the ball when it reaches the top of the window

then

$1.5=0.15u+\dfrac{9.8\times0.15^2}{2}\\u=3.96\ \rm m/s\\$

Now u is the final velocity of the ball with respect to the top of the building

so let t be the time taken for it to reach the top of the window with this velocity

$3.96=gt\\t=0.4\ \rm s\\$

Let h be the height above the top of the window

$h=\dfrac{gt^2}{2}\\\\h=\dfrac{9.8\times 0.4^2}{2}\\h=9.71\ \rm m$

3 0

3 years ago

A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio

natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

$\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}$

For the maximum velocity of the ball at the top of the vertical circular motion,

$v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}$

where g is the acceleration due to gravity,

Substituting the known values,

$\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}$

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

8 0

1 year ago

Sally and Sam are in a spaceship that comes to within 14,000 km of the asteroid Ceres. Determine the force Sally experiences, in

tatiyna

Sam and Sally are traveling aboard a spacecraft that approaches the asteroid Ceres within 14,000 kilometers. Sally will experience 1.989 × 10⁻¹¹ N of force.

<h3>What is the gravitational force?</h3>

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.

Given data

Mass of asteroid ,m₁ = 8.7 1020 kg

Mass of sally,m₂ = 67 kg

Gravitational constant,G = 6.6 × 10⁻¹¹ kg⁻² m²

Distance of seperation,R = 14,000 km

$\rm F = G\frac{m_1m_2}{R^ 2} \\\\ F = 6.6 \times 10^{-11 }\times \frac{8.71020 \times 67 }{(14,000)^2} \\\\F = 1.989 \times 10^{-11 } \ N$

Hence, the force Sally experiences will be 1.989 × 10⁻¹¹ N.

To learn more about the gravitational force, refer to the link;

brainly.com/question/24783651

#SPJ1

4 0

1 year ago

A car starts from rest, speeds up with constant acceleration, and travels 400 meters in 10 seconds. What was the acceleration of

PIT_PIT [208]

Answer:

Acceleration of the car will be $a=8m/sec^2$

Explanation:

We have given that car starts from rest so initial velocity of the car u = 0 m/sec

And car traveled 400 m in 10 sec

So distance traveled by car s = 400 m

Time taken to compete this distance t = 10 sec

We have to find the acceleration of the car

From second equation of motion we know that $s=ut+\frac{1}{2}at^2$

So $400=0\times 10+\frac{1}{2}\times a\times 10^2$

$a=8m/sec^2$

So acceleration of the car will be $a=8m/sec^2$

7 0

3 years ago

A pair of in-phase stereo speakers is placed side by side, 0.764 m apart. You stand directly in front of one of the speakers, 2.

Goryan [66]

Answer:

f = 2858.33 Hz

Explanation:

given,

distance between speaker (A) and the person = 2.34 m

Distance between speaker (B) and the person is AB =

= $\sqrt{(0.764)^2 + (2.34)^2}$

= 2.46 m

path difference d = BP - AP

= 2.46 - 2.34 m

= 0.12 m

now, λ = 0.12

speed of sound = 343 m/s

$f= \dfrac{c}{\lambda}$

$f= \dfrac{343}{0.12}$

f = 2858.33 Hz

the lowest frequency that will produce constructive interference is equal to

f = 2858.33 Hz

7 0

3 years ago