1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Nuetrik [128]
3 years ago
13

Here's another question!

Physics
2 answers:
KATRIN_1 [288]3 years ago
6 0

Answer:

<u>Organ</u>

Explanation:

Tissues make up organs which in turn forms the organ system.

Talja [164]3 years ago
6 0

Organ its the correct answer
You might be interested in
A recipe for a sweet tea calls for 3 cups of sugar when you place a sugar into the tea it doesn't dissolve. How could you use di
barxatty [35]

Answer: You could dissolve it by heating it back up, then just cooling it down again.

Hope that helps!

6 0
3 years ago
Read 2 more answers
30 points - how do I do 2 b and c?
Delvig [45]
1. its must be B and 2. must be C
7 0
3 years ago
A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region
zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with  due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

K_{1} = K_{2} + W_{f}

Where:

K_{1}, K_{2} are the initial and final translational kinetic energies of the tobbogan, measured in joules.

W_{f} - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})

Where:

f - Friction force, measured in newtons.

\Delta s - Distance travelled by the toboggan in the rough region, measured in meters.

m - Mass of the toboggan, measured in kilograms.

v_{1}, v_{2} - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}

If m = 375\,kg, v_{1} = 4.50\,\frac{m}{s}, v_{2} = 1.20\,\frac{m}{s} and \Delta s = 5.40 \,m, then:

f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}

f = 653.125\,N

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%

\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%

\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%

\%K_{loss} = 92.889\,\%

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%

\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%

\%v_{loss} = 73.333\,\%

The speed of the toboggan is reduced in 73.333 %.

5 0
3 years ago
The USDA recommends that women consume about 2000 Calories per day and men consume about 2500 Calories per day. How much average
professor190 [17]

Answer:

   1500 W to 2200 W

Explanation:

Every person does work in his daily day to day life. A person needs energy in order to perform work. And the energy consumed by an individual while performing a daily work is directly responsible to the mount of oxygen consumed by the person.

The USDA is the federal agency which looks after the food and agriculture matters of the US government. It deals with and formulates different policies and laws for the country and it s people. It recommends about 2000 calories per day for women and for men it recommends about 2500 calorie per days of food intake.

Accordingly, the average power required by a human body for doing regular work is in the range of 1500 W to 2200 W.

4 0
3 years ago
A ball is launched horizontally from a height of 172 m above the ground. its initial horizontal velocity is 16m/s. how long does
Romashka-Z-Leto [24]

Answer:

172*16 times it and that is your answer

8 0
3 years ago
Other questions:
  • Is Their A Video On The Pysical Thingie
    8·1 answer
  • What is the free-fall acceleration at the surface of the jupiter?
    5·1 answer
  • What did the greeks call the mineral they found and why?
    6·1 answer
  • What makes a clinical thermometer suitable for measuring small changes in body temperature? *
    9·1 answer
  • What is fundamental unit?​
    8·1 answer
  • Pls answer……………………..
    15·1 answer
  • A motor is used to pump water from a depth of 5m to fill a volume of 10cubic meter in 5 min . If 50 percent of the power is wast
    11·1 answer
  • Use F = 1/T as your basis:
    12·1 answer
  • During the course of a concert, the temperature of a pipe organ gradually increases. As a result, the frequency of the sound fro
    15·1 answer
  • Help pls it’s about physics I need all of them
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!