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
4vir4ik [10]
3 years ago
10

A 1000 kg weather rocket is launched straight up.The rocket motor provides a constant acceleration for 16 s, then the motor stop

s. The rocket altitude 20 s after launch is 5100 m. You can ignore any effects of air resistance. What is the rocket's speed as it passes through a cloud 5100 m above the ground?
Not sure what i'm doing.
Physics
1 answer:
ohaa [14]3 years ago
3 0
I’d say 10200m as the answer not sure though
You might be interested in
A father racing his son has 1/3 the kinetic energy of the son, who has 1/4 the mass of the father. The father speeds up by 1.5 m
Feliz [49]

Explanation:

Let the speeds of father and son are v_f\ and\ v_s. The kinetic energies of father and son are K_f\ and\ K_s. The mass of father and son are  m_f\ and\ m_s

(a) According to given conditions, K_f=\dfrac{1}{3}K_s

And m_s=\dfrac{1}{4}m_f

Kinetic energy of father is given by :

K_f=\dfrac{1}{2}m_fv_f^2.............(1)

Kinetic energy of son is given by :

K_s=\dfrac{1}{2}m_sv_s^2...........(2)

From equation (1), (2) we get :

\dfrac{v_f^2}{v_s^2}=\dfrac{1}{12}..............(3)

If the speed of father is speed up by 1.5 m/s, so the ratio of kinetic energies is given by :

\dfrac{K_f}{K_s}=\dfrac{1/2m_f(v_f+1.5)^2}{1/2m_sv_s^2}

v_s^2=4(v_f+1.5)^2

Using equation (3) in above equation, we get :

v_f=\dfrac{1.5}{\sqrt3-1}=2.04\ m/s

(b) Put the value of v_f in equation (3) as :

v_s=7.09\ m/s

Hence, this is the required solution.

8 0
4 years ago
A 150kg person stands on a compression spring with spring constant 10000n/m and nominal length of 0.50.what is the total length
Ivahew [28]

Answer:

<em>The total length of the spring would be 0.65 m</em>

Explanation:

The Concept

Hooke's law evaluates the increment of  spring in relation to the force acting on the body. Hooke's law states that for a spring undergoing deformation, the  force applied is directly proportional to the deformation experienced by the spring. Hooke's law is represented thus;

F = k x ..................1

where F is the force applied to the spring

k is the spring constant

x is the spring stretch or extension

Step by Step Calculations

We have to obtain x before adding it to the nominal length, We make x the subject formula in equation 1;

x = F/k

but F = m x g

so, x = (m x g)/k

given that, the mass of the person m =150 kg

g is the acceleration due to gravity = 9.81 m/s^{2}

k is the spring constant = 10000 N/m

then x = (9.81 m/s^{2} x 150 kg)/10000 N/m

x = 0.14715 m

the extension experienced by the spring after the compression is 0.14715 m

The total length of the spring would be;

L = 0.14715 m + 0.5 m = 0.64715

L ≈  0.65 m

Therefore the total length of the spring would be 0.65 m

4 0
3 years ago
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
-BARSIC- [3]

Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

7 0
3 years ago
The Reynolds number, rho VD/mu, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionle
Agata [3.3K]

Answer:

Re = 1 10⁴

Explanation:

Reynolds number is

         Re = ρ v D /μ

The units of each term are

       ρ = [kg / m³]

       v = [m / s]

      D = [m]

      μ = [Pa s]

The pressure

      Pa = [N / m²] = [Kg m / s²] 1 / [m²] = [kg / m s²]

      μ = [Pa s] = [kg / m s²] [s] = [kg / m s]

We substitute the units in the equation

      Re = [kg / m³] [m / s] [m] / [kg / m s]

      Re = [kg / m s] / [m s / kg]

      RE = [ ]

Reynolds number is a scalar

Let's evaluate for the given point

Where the data for methane are:

viscosity       μ = 11.2 10⁻⁶ Pa s

the density  ρ = 0.656 kg / m³

       D = 2 in (2.54 10⁻² m / 1 in) = 5.08 10⁻² m

       Re = 0.656 4 2 5.08 10⁻² /11.2 10⁻⁶

       Re = 1.19 10⁴

4 0
3 years ago
A force F of magnitude 2x^3 is applied to stop a particle moving with an initial velocity of v0. The particle travels from x=0 t
3241004551 [841]

Answer:

Explanation:

Given that

F=2x³

Work is given as

The range of x is from x=0 to x=D

W=-∫f(x)dx

Then,

W=-∫2x³dx from x=0 to x=D

W=- 2x⁴/4 from x=0 to x=D

W=-2(D⁴/4-0/4)

W=-D⁴/2

W=1/2D⁴

The correct answer is F

5 0
3 years ago
Other questions:
  • A(n) 0.155 kg baseball is thrown with a speed of 18.4 m/s. It is hit straight back at the pitcher with a final speed of 18.4 m/s
    7·1 answer
  • Two wooden boxes of equal mass but different density are held beneath the surface of a large container of water. Box A has a sma
    15·1 answer
  • A 2.36 kg block resting on a frictionless surface is attached to an ideal spring with spring constant k = 260 Nm . A force is ap
    15·1 answer
  • Imagine you are on an airplane. Your plane has flown from 30 degrees North to 42 degrees North. How far have you flown?
    8·1 answer
  • A 30-g bullet traveling horizontally at 300 m/s strikes a 1.0-kg block which is attached to a horizontal spring with a force con
    9·1 answer
  • 2. A car is sitting at the top of a hill that is 14 m high. The car has a mass of 53 kg. The car has
    7·1 answer
  • In any vector space au=bu implies a=b ? Trou or False​
    12·2 answers
  • Can I have the answer please
    7·1 answer
  • Help, please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
    11·1 answer
  • During a NASCAR race a car goes 58 m/s around a curved section of track that has a radius of 260 m. What is the car's accelerati
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!