All categories

3 years ago

I need help quick!

Physics

1 answer:

zvonat [6]3 years ago

7 0

Explanation:

At the moment, high tides occur twice a day. So if the moon orbits the earth twice as fast, high tides will occur twice as many, which means four times a day.

You might be interested in

If you increase the force on an object, its acceleration

Snezhnost [94]

Answer:

True!

Explanation:

if you increase the force on an object, its acceleration

pls mark me brainliest

5 0

4 years ago

is described as the sum of all the individual resistances in a series.1) Equivalent resistance2) Parallel circuit3) Series circu

irga5000 [103]

The sum of all resistances in the series circuit is known as Equivalent resistance.

$R=r_1+r_2+.\ldots.r_n$

where R is the equivalent resistance.

Hence, option 1 is the correct answer.

6 0

1 year ago

cupoosta [38]

Answer:

<h3>The answer is 1.92 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 2.5 g

Volume = 1.3 cm³

We have

$density = \frac{2.5}{1.3} \\ = 1.923076...$

We have the final answer as

Hope this helps you

5 0

3 years ago

A fisherman is watching water waves ripple past his boat. How could he determine the wavelength of the water waves?

natta225 [31]

Wavelength can be calculated using the following formula: wavelength = wave velocity/frequency. Wavelength usually is expressed in units of meters.

7 0

3 years ago

A 65.0 kg skier is moving at 6.85 m/s on a frictionless, horizontal snow-covered plateau when she encounters a rough patch 4.00

marusya05 [52]

Answer:

v = 4.58 m/s

Explanation:

In order to calculate the speed of the skier when she gets the bottom of the hill, you have to calculate the speed of the skier when she crosses the rough patch.

To calculate the velocity at the final of the rough patch you take into account that the work done by the friction surface is equal to the change in the kinetic energy of the skier:

$W_f=\Delta K\\\\-N\mu_kd=\frac{1}{2}mv^2-\frac{1}{2}mv_o^2=\frac{1}{2}m(v^2-v_o^2)$ (1)

Where the minus sign means that the work is against the motion of the skier.

Wf: friction force

m: mass of the skier = 65.0kg

N: normal force = mg

g: gravitational acceleration = 9.8m/s^2

d: distance of the rough patch = 4.00m

v: speed at the end of the rough patch = ?

vo: initial speed of the skier = 6.85m/s

μk: coefficient of kinetic friction = 0.330

You replace the expression for the normal force in the equation (1), and solve for v:

$-mg\mu_kd=\frac{1}{2}m(v^2-v_o^2)\\\\-g\mu_kd=\frac{1}{2}(v^2-v_o^2)\\\\v=\sqrt{-2g\mu_kd+v_o^2}\\\\v=\sqrt{-2(9.8m/s^2)(0.330)(4.00m)+(6.85m/s)^2}=8.53\frac{m}{s}=4.58\frac{m}{s}$

Then, the speed fot he skier at the bottom of the hill is 4.58m/s

3 0

4 years ago