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Crazy boy [7]
3 years ago
11

A ball is kicked from the top of a building with a velocity of 50 m/s and lands 165 m away from the base of the building.

Physics
1 answer:
Akimi4 [234]3 years ago
3 0

Answer:

3.1 s

Explanation:

we can assume that this motion has a horizontal projectile motion.

projectile motion happens when  an object is subjected to only gravity. Gravity acts to influence the vertical motion of the projectile, it makes a  vertical acceleration.  

The horizontal motion of the projectile is the result of the constant velocity of the object.We assume no acceleration in the horizontal direction.  

applying motion equations

where

s - distance

u - initial velocity

a-acceleration

t - time

s = ut + (1/2)at²

→ 165 = 50t + 0 (as no acceleration is in horizontal direction)

t = 3.1 s

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which of the following is an example of a compound machine. a. a bicycle chain. b. a coffee mug. c. a paper cutter
Ann [662]
A. a bicycle chain is an example of a compound machine
8 0
3 years ago
A heat pump is to be used for heating a house in winter. The house is to be maintained at 70°F at all times. When the temperatur
Anna35 [415]

Answer:

\dot{W_{H} } = 4244.48 Btu/h

Explanation:

Temperature of the house, T_{H} = 70^{0} F

Convert to rankine, T_{H} = 70^{0}+ 460 = 530 R

Heat is extracted at 40°F i.e T_{L} = 40^{0}F  = 40 + 460 = 500 R

Calculate the coefficient of performance of the heat pump, COP

COP = \frac{T_{H} }{T_{H} - T_{L}  } \\COP = \frac{530 }{530 - 500  }\\ COP = \frac{530}{30} \\COP = 17.67

The minimum power required to run the heat pump is given by the formula:

\dot{W_{H} } = \frac{\dot{Q_{H} }}{COP} \\...............(*)

Where the heat losses from the house, \dot{Q_{H} } = 75,000 Btu/h

Substituting these values into * above

\dot{W_{H} } = \frac{75000}{17.67} \\ \dot{W_{H} } = 4244.48 Btu/h

3 0
3 years ago
The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area 4.50×10−9 m2 , a plate separ
Marizza181 [45]

The energy stored in the membrane is 6.44\cdot 10^{-14} J

Explanation:

The capacitance of a parallel-plate capacitor is given by

C=\frac{k\epsilon_0 A}{d}

where

k is the dielectric constant of the material

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the membrane in this problem, we have

k = 4.6

A=4.50\cdot 10^{-9} m^2

d=8.1\cdot 10^{-9} m

Substituting, we find its capacitance:

C=\frac{(4.6)(8.85\cdot 10^{-12})(4.50\cdot 10^{-9})}{8.1\cdot 10^{-9}}=2.26\cdot 10^{-11} F

Now we can find the energy stored: for a capacitor, it is given by

U=\frac{1}{2}CV^2

where

C=2.26\cdot 10^{-11} F is the capacitance

V=7.55\cdot 10^{-2} V is the potential difference

Substituting,

U=\frac{1}{2}(2.26\cdot 10^{-11} F)(7.55\cdot 10^{-2})^2=6.44\cdot 10^{-14} J

Learn more about capacitors:

brainly.com/question/10427437

brainly.com/question/8892837

brainly.com/question/9617400

#LearnwithBrainly

6 0
3 years ago
Fill-in-the-blank test questions measure ________; matching concepts with their definitions measures ________.
Ulleksa [173]

Test questions measure recall; matching concepts with their definitions measures recognition.

<u>Explanation: </u>

According to Psychology our brain remembers everything what we learn but the understanding and remembering the right answer for the right question needs training and understanding ability. So in order to enhance the ability of recalling and recognizing among the students, the concept of test questions and matching with definitions are used in curricular activities.

As the students will be learning different terms, definitions, methods and different subjects, they should be able to distinguish among different definitions as well as they should recall the things they have learnt. So the answers for the test questions will help to recall the topics learnt by the students while the matching concept will help the students to recognize each definition with their terms.

6 0
3 years ago
A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend
Viefleur [7K]

a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how

long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175

s this is the time to fall from the top; it would take the same time to travel

upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175

= 0.35s

b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice

to solve this problem the time it takes to fall the final 0.13 m is: time it

takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to

fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it

takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m

is then twice this, or 0.08s

5 0
3 years ago
Read 2 more answers
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