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
hram777 [196]
2 years ago
11

A rocket starting from its launch pad is subjected to a uniform acceleration of 100 meters/second2. Determine the time needed to

reach the final velocity of 1,000 meters/second.
Physics
1 answer:
balandron [24]2 years ago
5 0
The velocity is the integral of acceleration.  If acceleration is 100 m/s^2 then velocity is:

v= \int\limits^{}_{}100 \, dt=100t

So to know the velocity at any time, t, we just put t in seconds into this equation.  To know at what time we get to a certain velocity, we set this equation equal to that velocity and solve for t:

100t = 1000 \\  \\  t= \frac{1000}{100} =10s

 
You might be interested in
A long, straight metal rod has a radius of 5.75 cm and a charge per unit length of 33.3 nC/m. Find the electric field at the fol
PIT_PIT [208]

Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a)  at 4.75 cm

b)  at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula E = \dfrac{kq }{d}

E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}

E = 1461.95 N/C

c) The electric field E is calculated as:

E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}

E = 239.76 N/C

7 0
3 years ago
Please !!! I really need help !!! How do I understand these ?!!!!
trasher [3.6K]

Answer

The answer for the first one I think is false.

The second one would be true i think. I hope i got it right and have a wonderful day

8 0
3 years ago
Read 2 more answers
A block of mass m is pushed a horizontal distance D from position A to position B, along a horizontal plane with friction coeffi
Wewaii [24]

Answer:

The total work done by friction is -2 · μ · m · g · D

Explanation:

Hi there!

The work done by a force is calculated as follows:

W = F · d · cos θ

Where:

W = work.

F = force that does the work.

d = displacement.

θ = angle between the displacement and the force.

If the force is horizontal, as in this case, cos θ = 1

The friction force is calculated as follows:

Ffr = μ · m · g

Where:

μ = friction coefficient.

m = mass of the object.

g = acceleration due to gravity.

Then, in this case, the work done by friction when pushing the block from A to B will be:

W AB = -Ffr · D

W AB = - μ · m · g · D

Notice that the friction force is negative because it is opposite to the pushing force P.

When the block is pushed from B to A, the work done by friction will be:

W BA = Ffr · (-D)

W BA = -μ · m · g · D

Now, the displacement is negative and the friction force is positive (in the opposite direction to -P).

The total work done by friction will be:

W AB + W BA = - μ · m · g · D  - μ · m · g · D  = -2 μ · m · g · D

5 0
3 years ago
Why the effect of gravitational force is more in liquid than in solid?​
LekaFEV [45]
  • Gravitational force depends only on mass and distance, not on the state of matter.
  • The forces of attraction between molecules in matter are electromagnetic in nature, not gravitational.
  • These attractive forces are stronger in a solid than in a liquid than in a gas.
  • Gravitational forces between molecules is completely negligible compared to the em forces.

So, key answer is inter-molecular forces of solids is stronger than liquids.

4 0
3 years ago
Read 2 more answers
Question 1(Multiple Choice Worth 4 points) The star named Canopus has a declination of approximately –52°. Which of these statem
Y_Kistochka [10]

Answer:

It is 52° below the celestial equator.

Explanation:

The declination is the angle in degrees measured north (+) or south (-) of the an imaginary line called the celestial equator.

The celestial equator is a projection of the earth's equator on the celestial sphere. imaginary

The star named Canopus has a declination of approximately –52°.

Since the angle is negative, this shows that it is south or below the celestial equator and at 52° south of the celestial equator.

Thus, the star named Caponus is 52° below the celestial equator.

8 0
3 years ago
Other questions:
  • When an electron falls from a higher to a lower energy level in an atom, the photon released has a wavelength of 121.6 nm. What
    9·1 answer
  • The formula for calculating power is work divided by time (power = work ÷ time). What are two ways of stating the same relations
    8·2 answers
  • Show the equation of simple pendulum to be dimensionally consistent
    7·1 answer
  • The speed of light is 3.0 x 108 m/s. How far does light travel in 2.4 minutes? Identify all variables, write a formula, show all
    13·1 answer
  • A golf ball thrown at an angle of 20 relative to the ground at a speed of 40 m/s. if the range of the motion is 55m. what is the
    14·1 answer
  • A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p
    7·1 answer
  • What is the difference between arteries and artiole?
    7·1 answer
  • State a situation in which force is applied on a body, but no work is done​
    12·2 answers
  • Find the magnitude of the buoyant force on the block. The acceleration of gravity is 9.8 m/s 2 . Answer in units of N.
    7·1 answer
  • Indicate all ways in which the graph used in question 4 would change if the Normal Force applied to the object was increased. Be
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!