Ask question

Ask question

All categories

3 years ago

12

A sailboat moves north for a distance of 10.00 km when blown by a wind 30° east of south with a force of 5.00 × 104 N. How much

work was done by the wind?

1 answer:

Snowcat [4.5K]3 years ago

8 0

2.50 × 108 J 4.33 × 108 J 5.00 × 108 J 0.00 J

You might be interested in

This is a small rocky body that orbits the sun, and when close to the sun exhibit a small tail of ice and gas.

Over [174]

This is a comet. It is also called a dirty snowballs.

3 0

3 years ago

Only 5 question plz answer

Aleks04 [339]

It is subduction;) good luck on the others my man

4 0

3 years ago

Read 2 more answers

Due Ma<br> duart<br> ded<br> out<br> 25 N<br> 35 N<br> 1-03

Anuta_ua [19.1K]

Answer:

what's that all about

hehehwhe

Explanation:

dgbjjjedgkigdssfhkkoyddwrhkoyeqaxghjjhasghffhjiopjtewqetujjgda

8 0

3 years ago

50 grams of ice cubes at -15°C are used to chill a water at 30°C with mass mH20 = 200 g. Assume that the water is kept in a foam

Arada [10]

Answer : The final temperature is, $25.0^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ice = $2.09J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ice = 50 g

$m_2$ = mass of water = 200 g

$T_f$ = final temperature = ?

$T_1$ = initial temperature of ice = $-15^oC$

$T_2$ = initial temperature of water = $30^oC$

Now put all the given values in the above formula, we get:

$50g\times 2.09J/g^oC\times (T_f-(-15))^oC=-200g\times 4.184J/g^oC\times (T_f-30)^oC$

$T_f=25.0^oC$

Therefore, the final temperature is, $25.0^oC$

5 0

3 years ago

You are trying to hear your friend give directions to new store in town. But from your distance (1 point) of 15 m you only hear

Marat540 [252]

Answer:

option D

Explanation:

given,

Intensity of sound = 20 dB

distance = 15 m

intensity of sound is increased to = 50 dB

distance between the sound level = ?

Using relation

$L_2 = L_1 - |20(log \dfrac{r_2}{r_1})|$

L₁ = 20 dB L₂ = 50 dB r₁ = 15 m r₂ = ?

$log (\dfrac{r_2}{r_1}) = \dfrac{L_1 -L_2}{20}$

$\dfrac{r_2}{r_1}= 10^{\dfrac{|L_1 -L_2|}{20}}$

$r_2 =r_1 10^{\dfrac{|L_1 -L_2|}{20}}$

$r_2 =15 \times 10^{\dfrac{|20-50|}{20}}$

$r_2 =15 \times 10^{-1.5}$

r₂ = 0.47 m

r₂ = 47 cm

hence, the correct answer is option D

7 0

3 years ago