Which statement about ocean surface currents is false?

If you see a lightning stroke and then, 15 seconds later, hear the thunder, the lightning is about ____ miles away

makvit [3.9K]

Answer:

3.2 miles

Explanation:

We can assume that the light coming from the lightning reach instantaneously our eyes (because light travels very fast, $3\cdot 10^8 m/s$ ), so the time between the lightining and the hearing of the thunder is equal to the time that sound takes to each us.

Sound travels in air approximately at speed

v = 343 m/s

So the distance it covers in a time of t = 15 seconds is

$d=vt=(3403 m/s)(15 s)=5145 m$

And using the conversion from meters to miles,

1 mile = 1609 meters

we find

$d=\frac{5145 m}{1609 m/miles} = 3.2 miles$

3 0

4 years ago

A4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects encounter the same const

Evgesh-ka [11]

Answer:

Both objects travel the same distance.

(c) is correct option

Explanation:

Given that,

Mass of first object = 4.0 kg

Speed of first object = 2.0 m/s

Mass of second object = 1.0 kg

Speed of second object = 4.0 m/s

We need to calculate the stopping distance

For first particle

Using equation of motion

$v^2=u^2+2as$

Where, v = final velocity

u = initial velocity

s = distance

Put the value in the equation

$0= u^2-2as_{1}$

$s_{1}=\dfrac{u^2}{2a}$ ....(I)

Using newton law

$a=\dfrac{F}{m}$

Now, put the value of a in equation (I)

$s_{1}=\dfrac{8}{F}$

Now, For second object

Using equation of motion

$v^2=u^2+2as$

Put the value in the equation

$0= u^2-2as_{2}$

$s_{2}=\dfrac{u^2}{2a}$ ....(I)

Using newton law

$F = ma$

$a=\dfrac{F}{m}$

Now, put the value of a in equation (I)

$s_{2}=\dfrac{8}{F}$

Hence, Both objects travel the same distance.

6 0

3 years ago

-what meal do you eat after doing those exercises?

TiliK225 [7]

I eat dicks and and I also masturbate

6 0

3 years ago

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Suppose you have a pendulum clock which keeps correct time on Earth(acceleration due to gravity = 1.6 m/s2). For ever hour inter

kaheart [24]

The moon clock is A) (9.8/1.6)h compared to 1 hour on Earth

Explanation:

The period of a simple pendulum is given by the equation

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity

In this problem, we want to compare the period of the pendulum on Earth with its period on the Moon. The period of the pendulum on Earth is

$T_e=2\pi \sqrt{\frac{L}{g_e}}$

where

$g_e = 9.8 m/s^2$ is the acceleration of gravity on Earth

The period of the pendulum on the Moon is

$T_m=2\pi \sqrt{\frac{L}{g_m}}$

where

$g_m = 1.6 m/s^2$ is the acceleration of gravity on the Moon

Calculating the ratio of the period on the Moon to the period on the Earth, we find

$\frac{T_m}{T_e}=\frac{g_e}{g_m}=\frac{9.8}{1.6}$

Therefore, for every hour interval on Earth, the Moon clock will display a time of

A) (9.8/1.6)h

#LearnwithBrainly

6 0

3 years ago

A 5 kg block moves with a constant speed of 10 ms to the right on a smooth surface where frictional forces are considered to be

nirvana33 [79]

Answer:

Work done, W = 19.6 J

Explanation:

It is given that,

Mass of the block, m = 5 kg

Speed of the block, v = 10 m/s

The coefficient of kinetic friction between the block and the rough section is 0.2

Distance covered by the block, d = 2 m

As the block passes through the rough part, some of the energy gets lost and this energy is equal to the work done by the kinetic energy.

$W=\mu_kmgd$

$W=0.2\times 5\times 9.8\times 2$

W = 19.6 J

So, the change in the kinetic energy of the block as it passes through the rough section is 19.6 J. Hence, this is the required solution.

5 0

3 years ago

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