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
BlackZzzverrR [31]
4 years ago
13

14. A rocket is shot up into the air and then comes back down and hits the ground 9.2 second later.

Physics
1 answer:
sineoko [7]4 years ago
3 0

Answer:

105.8 m

46 m/s

Explanation:

From the time the rocket is launched to the time it reaches its maximum height:

v = 0 m/s

a = -10 m/s²

t = 9.2 s / 2 = 4.6 s

Find: Δy and v₀

Δy = vt − ½ at²

Δy = (0 m/s) (4.6 s) − ½ (-10 m/s²) (4.6 s)²

Δy = 105.8 m

v = at + v₀

0 m/s = (-10 m/s²) (4.6 s) + v₀

v₀ = 46 m/s

You might be interested in
I need help with this question how to solve it for Brass and Cooper
Ksenya-84 [330]

Take into account that density and relative density are given by:

\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

1 m^3 = 1000000 ml

1 kg = 1000 g

Then, you obtain the following results:

Brass:

\begin{gathered} 53.2g\cdot\frac{1kg}{1000g}=0.0532kg \\ 6ml\cdot\frac{1m^3}{1000000ml}=0.000006m^3 \\ \text{density}=\frac{0.0532kg}{0.000006m^3}\approx8866.67\frac{kg}{m^3} \\ \text{relative density=}\frac{(\frac{8866.66kg}{m^3})}{(1000\frac{kg}{m^3})}\approx8.87 \end{gathered}

Cooper:

\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}

3 0
1 year ago
What is the ability to complete extended periods of physical activity?
gogolik [260]
Endurance is the ability to complete extended periods of physical activity
8 0
3 years ago
Read 2 more answers
If an astronaut throws an object in space, the object’s speed will _____
BigorU [14]
The object's speed will not change.

In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
\sum F=ma
where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.
7 0
3 years ago
Which of these best describes the difference between energy and power? A) Power and energy have nothing in common B) Energy and
Sphinxa [80]

D.Power has a time component while energy does not. This is because power is the RATE at which work is performed.

8 0
3 years ago
A ball is attached to a string of length 3 m to make a pendulum. The pendulum is placed at a location that is away from the Eart
Musya8 [376]

1) 0.61 m/s^2

2) 13.9 s

Explanation:

1)

The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.

It can be calculated using the equation:

g=\frac{GM}{r^2} (1)

where

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

r is the distance of the object from the Earth's center

The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is

r=4R

where

R=6.37\cdot 10^6 m is the Earth's radius

Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):

g=\frac{GM}{(4R)^2}=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})0.}{(4\cdot 6.37\cdot 10^6)^2}=0.61 m/s^2

2)

The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity at the location of the pendulum

Note that the period of a pendulum does not depend on its mass.

For the pendulum in this problem, we have:

L = 3 m is its length

g=0.61 m/s^2 is the acceleration due to gravity (calculated in part 1)

Therefore, the period of the pendulum is:

T=2\pi \sqrt{\frac{3}{0.61}}=13.9 s

4 0
3 years ago
Other questions:
  • The takeoff speed for an airbus a320 jetliner is 82 m/s . velocity data measured during takeoff are as follows: t(s) vx(m/s) 0 0
    9·1 answer
  • What does combustion mean?what clues help u detemine the meaning?
    6·2 answers
  • Name two ways that friction is harmful and two ways that friction is helpful to you when riding a bicycle
    14·1 answer
  • Two small identical balls A and B are held a distance r apart on a frictionless surface, with i very large compared with the siz
    6·1 answer
  • Please help me with physics a very nice person
    10·1 answer
  • Which of the following statements about the
    12·1 answer
  • Use Kirchoff first law and second law to derise the expression for the total resistancs​
    11·1 answer
  • Light travels about 180 million kilometers in 10 minutes. How far does it travel in 1 minute? How far does it travel in 1 second
    13·1 answer
  • The cross country team goes on a 10 mile run after school. They end the run outside of the locker room where they started fund t
    6·1 answer
  • Perform the following mathematical operation and report the answer to the correct number of significant figures 143.6/21.2
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!