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ycow [4]
2 years ago
10

a 1.2 kg rocket is launched from ground level with initial velocity of 12 m/s. What is the maximum height the rocket can reach?

Physics
1 answer:
Tanzania [10]2 years ago
5 0

As the rocket is launched from the ground its height will go on increasing till it stops

So the height of rocket will be maximum when its speed becomes zero

so here we can use energy conservation theory

PE = KE

mgH = \frac{1}{2}mv^2

9.8*H = \frac{1}{2}*12^2

H = 7.35 m

So it will reach upto height 7.35 m

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A single-phase transformer has a turns ratio of 20,000/5,000. If a DC voltage of 50 V is applied to the primary winding, what wi
Kipish [7]

Sneaky question.

The secondary voltage will be zero.

Secondary voltage is the result of CHANGES in the primary voltage. That means the primary is energized with AC. The transformer in this question is energized with DC, which doesn't change. So there is no secondary voltage, this transformer doesn't work, and what you get out of it is: Smoke !

6 0
3 years ago
If two runners take the same amount of time to run a mile, they have the same __________.
Svet_ta [14]
I am going to say velocity because you have the same amount of acceleration going in a certain direction.
8 0
3 years ago
Read 2 more answers
What Is the acceleration of a toy car that starts from rest and has a speed of 0.12 m/s after 0.1?
VikaD [51]

Given: Initial velocity of toy car (u ) = 0

Final velocity of toy car (v) = 0.12 m/s

Required time (t) = 0.1 s

To find: The acceleration of the toy car.

Let the acceleration of the toy car be (a)

Formula Used: 1st kinematic equation of motion

                              v = u + at ---------------------------(i)

Here, all alphabets are in their usual meanings

Now, from equation (i), we shall calculate the value of 'a'.

so,                          a = (v - u) /t

or,                          a = (0.12 m/s - 0) / 0.1s

or,                          a = 1.2 m/s²

Hence, the required acceleration of the toy car will be 1.2 m/s².

6 0
3 years ago
Please Help Me!
balandron [24]

The image distance when a boy holds a toy soldier in front of a concave mirror, with a focal length of 0.45 m. is -0.56 m.

<h3>What is image distance?</h3>

This is the distance between the image formed and the focus when an object is placed in front of a plane mirror.

To calculate the image distance, we use the formula below.

Formula:

  • 1/f = 1/u+1/v........... Equation 1

Where:

  • f = Focal length of the mirror
  • v = Image distance
  • u = object distance

From the question,

Given:

  • f = 0.45 m
  • u = 0.25 m

Substitute these values into equation 1 and solve for the image distance

  • 1/0.45 = 1/0.25 + 1/v
  • 2.22 = 4+1/v
  • 1/v = 2.22-4
  • 1/v = -1.78
  • v = 1/(-1.78)
  • v = -0.56 m

Hence, The image distance is -0.56 m.

Learn more about image distance here: brainly.com/question/17273444

5 0
1 year ago
Consider f(x) = -4x2 + 24x + 3. Determine whether the function has a maximum or minimum value. Then find the
murzikaleks [220]

Answer:

The function has a maximum in x=3

The maximum is:

f(3) = 39

Explanation:

Find the first derivative of the function for the inflection point, then equal to zero and solve for x

f(x)' = -4*2x + 24=0

-4*2x + 24=0

8x=24

x=3

Now find the second derivative of the function and evaluate at x = 3.

If f (3) ''< 0 the function has a maximum

If f (3) '' >0 the function has a minimum

f(x)''= 8

Note that:

f(3)''= -8

the function has a maximum in x=3

The maximum is:

f(3)=-4(3)^2+24(3) + 3\\\\f(3) = 39

4 0
3 years ago
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