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qwelly [4]
3 years ago
5

The tension in the horizontal towrope pulling a water-skier is 250 N while the skier moves due west a distance of 50 m. How much

work does the towrope do on
the water-skier?
Physics
1 answer:
icang [17]3 years ago
5 0

Answer:

W = 250(50) = 12500 J

Explanation:

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A lady bug is sitting on the bottom of a can while you twirl it overhead on a string that is 65.0
MA_775_DIABLO [31]

The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

\omega=\frac{2\pi}{T}

where

T is the period of revolution

The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

Therefore, the angular speed is

\omega=\frac{2\pi}{1 s}=6.28 rad/s

Now we can find the linear speed of the ladybug, which is given by

v=\omega r

where:

\omega=6.28 rad/s is the angular speed

r = 65.0 cm = 0.65 m is the distance of the ladybug from the axis of rotation

Substituting, we find

v=(6.28)(0.65)=4.1 m/s

Learn more about angular speed:

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

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7 0
3 years ago
A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a
aleksley [76]

Answer:

 T₂ =602  °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\

T₂ = 875 K

T₂ =875- 273 °C

T₂ =602  °C

5 0
3 years ago
A ball is dropped from the top of a building. it initially moves at 4.0 m/s. after 0.5 seconds, it moves at 3.8 m/s. what force
Dima020 [189]

Air resistance (meeting space requirements)

7 0
2 years ago
Someone please help
saul85 [17]

Based on the attached image:

  • The name of the longitude line that passes through point A is the International Date Line
  • The longitude 180° is experiencing solar noon because the rays of the sun are parallel to it.
  • The longitude for 6 pm is 90° W, 12 midnight is 0°, and 6 am is 90° E
  • Longitude 120° is B
  • Solar time at Point B is 4 pm
  • The location will correspond to any point on the same latitude as A

<h3>What are lines of longitude?</h3>

Lines of longitude are imaginary lines which run along the earth from the North pole. to the South pole.

Longitude lines divide the earth into semi-circles.

Longitude lines are known as meridians and each meridian measures one arc degree of longitude.

Considering the attached image:

  • The name of the longitude line that passes through point A is the International Date Line
  • The longitude 180° is experiencing solar noon because the rays of the sun are parallel to it.
  • The longitude for 6 pm is 90° W, 12 midnight is 0°, and 6 am is 90° E
  • Longitude 120° is B
  • Solar time at Point B is 4 pm
  • the location will correspond to any point on the same latitude as A

In conclusion, longitude lines are imaginary lines and run from North to South on the earth.

Learn more about lines of longitude at: brainly.com/question/1939015

#SPJ1

8 0
1 year ago
A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb
NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

\sum \tau = F*d

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

F*(27-6)= 6*600

F = \frac{6*600}{21}

F= 171.42 lb

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

4 0
3 years ago
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