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
BARSIC [14]
3 years ago
10

Write the equation for eccentricity

Physics
1 answer:
Svetradugi [14.3K]3 years ago
3 0
<span>e=ca{\displaystyle e={\frac {c}{a}}}.</span>

You might be interested in
4. A net force of 15 N is exerted on a book to cause it to accelerate at a rate of 5 m/s2. Determine the mass of the
kenny6666 [7]

Answer:

<h2>3 kg </h2>

Explanation:

The mass of an object given it's force and acceleration can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{15}{5}  = 3 \\

We have the final answer as

<h3>3 kg</h3>

Hope this helps you

4 0
2 years ago
What is the final velocity of a body if it is moving with 13 m/s in 300 seconds and its acceleration is 30 m/s 2.
Nutka1998 [239]

Answer:

9013m/s

Explanation:

acceleration= v- u/ t

=> at = v-u

=> v = at + u

=> v =30*300+13

= 9013m/s

6 0
2 years ago
(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

F=(1100 kg)(-2.23 m/s^2)=-2451 N

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

8 0
3 years ago
Which is more reliable—using a manual stop watch or using light gates?<br> Explain.
Kitty [74]
Light gates are more reliable. When using a manual stop watch, it is difficult to stop it at an exact time. A light gate is able to detect when an object passes through a 'gate' with the infrared transmitter and receiver. 
7 0
3 years ago
When a star is in the main-sequence stage of its life, it is fusing __________ into ________
gizmo_the_mogwai [7]
Hydrogen into helium
7 0
3 years ago
Other questions:
  • Which is the satellite Terra able to do?
    6·2 answers
  • ok so for 99 points(not multiple choice)what is one way to separate salt from grains NO MULTIPLE CHOICE
    11·2 answers
  • Which two factors does the power of a machine depend on
    8·2 answers
  • Do the negative ions tend to be metals or nonmetals?
    12·1 answer
  • If a fluid has high viscosity, does it flow faster or slower than a fluid with less viscosity?
    8·2 answers
  • According to the kinetic theory, all matter is composed of _______.
    11·2 answers
  • The word back is up there!!!
    10·1 answer
  • find the gravitational force between the two bodies of unit mass each and separated by unit distance​
    8·1 answer
  • a sphere of mass 5kg and volume 2×10-5completely immersed in water find the buoyant force exerted water​
    9·1 answer
  • Which of the following conditions would always result in a net force of 0N?
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!