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Lubov Fominskaja [6]
3 years ago
15

A car speeds up from 18.54 m/s to 29.52 m/s in 13.84 s. The acceleration of the car is:

Physics
2 answers:
valkas [14]3 years ago
5 0

Answer:

.7934m/s^{2}

Explanation:

Acceleration = change in velocity / change in time

A = 10.98m/s / 13.84s

A = .7934m/s^{2}

raketka [301]3 years ago
3 0

Answer:0.8 m/s^2

Explanation:

initial velocity(u)=18.54m/s

Final velocity(v)=29.52m/s

Time(t)=13.84 sec

Acceleration =(v-u)/t

acceleration =(29.52-18.54)/13.84

Acceleration =10.98/13.34

Acceleration=0.8 m/s^2

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Consider horizontal parallel plates with a fixed potential difference. The upper plate has a voltage difference of 30 V with the
BlackZzzverrR [31]

relation between potential difference and electric field is given as

E . d = \Delta V

so here we know that

d = 3 cm

\Delta V = 30 V

E \times 0.03 = 30

E = 1000 N/C

So now when plates are separated to 4 cm distance carefully

the potential difference between them will change but the electric field between them will remain constant

So at distance of 4 cm also the electric field will be E = 1000 N/C

5 0
3 years ago
What is the sources of error and suggestion on how to overcome it in the hooke's law experiment?
goblinko [34]
In physics, Hooke's law is written in equation as:

F = kx

It states that the force F exerted on the spring is directly proportional to the displacement x by a constant called spring constant k.

In the laboratory, this is done in an experiment through the apparatus shown in the attached figure. The object experimented here is the spring, and you are to find the spring constant. A known mass of object is attached below the spring. That object carries a force in the form of gravitational pull in terms of weight. When the spring stretches, the displacement is measured with the use of the ruler.

There are a number of sources of error for this experiment. First, the reading from the ruler by the reader may be inaccurate. That's why digital balances are much more reliable because it minimizes human error. Reading the measurement on the ruler is subjective especially when you don't read it on eye level. Second, the force of the object might also be inaccurate if you use an unreliable weighing scale. Lastly, the apparatus might not be properly calibrated.

6 0
3 years ago
Visualizing the body in the position is significant because all observers have a common point of reference when describing and d
FromTheMoon [43]

Visualizing the body in the anatomical position is significant because all observers have a common point of reference when describing and discussing its region.

<h3>What is the anatomical position?</h3>
  • Anatomical position, also known as conventional anatomical position, refers to the body's position while it is standing erect and looking forward, with each arm hanging on either side of the body and palms facing forward.
  • The legs are parallel, with the feet level on the floor and forward facing.
  • When explaining particular anatomical words and locations in human anatomy and physiology, the anatomical position is a standard point of reference.
  • Visualizing the body in its anatomical location is important because it provides a single point of reference for all observers when describing and discussing its region.

As the description itself says, visualizing the body in its anatomical location is important because it provides a single point of reference for all observers when describing and discussing its region.

Therefore, visualizing the body in the anatomical position is significant because all observers have a common point of reference when describing and discussing its region.

Know more about the anatomical position here:

brainly.com/question/5029605

#SPJ4

The question you are looking for is given here:

Visualizing the body in the _____________ is significant because all observers have a common point of reference when describing and discussing its region.

6 0
2 years ago
Using hookes law find the distance a spring with am elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it
siniylev [52]
484848484 48v88 848 48gv
6 0
3 years ago
A shopper weighs 4.00 kg of apples on a supermarket scale whose spring obeys Hooke's law and notes that the spring stretches a d
Irina-Kira [14]

Answer:

a) 0.040625 m

b) 5.02272 J

Explanation:

k = Spring constant

x = Stretched length

F = Force

a)

F=kx\\\Rightarrow k=\frac{F}{x}\\\Rightarrow k=\frac{4\times 9.81}{0.025}\\\Rightarrow k=1569.6\ N/m

F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{6.5\times 9.81}{1569.6}\\\Rightarrow x=0.040625\ m

Extension of the spring would be 0.040625 m

b) Work done in a spring

W=\frac{1}{2}kx^2\\\Rightarrow W=\frac{1}{2}\times 1569.6\times 0.08^2\\\Rightarrow W=5.02272\ J

The work done by the shopper to stretch this spring a total distance of 8.00 cm is 5.02272 J

4 0
3 years ago
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