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IgorLugansk [536]

3 years ago

12

An observer sitting on shore sees a canoe traveling 5.0 m/s east, and a sailboat traveling 15.0 m/s west. What is the velocity o

f the sailboat as observed on the canoe (relative to the canoe)?

1 answer:

Dovator [93]3 years ago

3 0

Answer:

Vs/c = 20 [m/s]

Explanation:

This is a problem of relative velocities, as velocity is a vector we can use addition or subtraction of vectors for the solution.

We are asked for the velocity of the sailboat with respect to an observer located in the canoe.

$V_{s/c}=V_{sailboat}-V_{canoe}\\V_{s/c}=15-(-5)\\V_{s/c}=20[m/s]$

Why is the speed of the canoe negative?, it is negative because the canoe moves in the opposite direction to the sailboat.

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The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t

dem82 [27]

Answer:

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

now for the time period of moon around the earth we can say

$T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}$

here we know that

$T_1 = 0.08 year$

$r_1 = 0.0027 AU$

$M_e$ = mass of earth

Now if the same formula is used for revolution of Earth around the sun

$T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}$

here we know that

$r_2 = 1 AU$

$T_2 = 1 year$

$M_s$ = mass of Sun

now we have

$\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}$

$\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}$

$12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}$

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

4 0

3 years ago

Why are you more likely to get a concussion when hit by a field hockey ball than a

Aleksandr-060686 [28]

Answer:

because it can be hard

Explanation:

I said that because they be on bed rest

6 0

3 years ago

If a bag holds 70.874 grams ,how many itemsare in the bag?

Alina [70]

Answer:

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Explanation:

7 0

3 years ago

Using your Periodic Table, which element below has the smallest atomic radius? A.) Sodium, B.) Chlorine, C.) Phosphorus, D.) Iro

egoroff_w [7]

Explanation:

Chlorine is the smallest atomic radius

6 0

3 years ago

Read 2 more answers

A 60 kg block slides along the top of a 100 kg block with an acceleration of 2.0 m/s2 when a horizontal force F of 340 N is appl

Taya2010 [7]

Answer:

The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

Explanation:

Suppose we find the coefficient of friction and the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.

Given that,

Mass of block = 60 kg

Acceleration = 2.0 m/s²

Mass = 100 kg

Horizontal force = 340 N

Let the frictional force be f.

We need to calculate the frictional force

Using balance equation

$F-f=ma$

Put the value into the formula

$340-f=60\times2.0$

$f=340-60\times2.0$

$f=220\ N$

We need to calculate the coefficient of friction

Using formula of friction force

$f= \mu mg$

$\mu=\dfrac{f}{mg}$

$\mu=\dfrac{220}{60\times9.8}$

$\mu =0.37$

We need to calculate the acceleration of the 100 kg block

Using formula of newton's law

$F = ma$

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

$a=\dfrac{220}{100}$

$a=2.2\ m/s^2$

Hence, The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

3 0

3 years ago