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
Oksi-84 [34.3K]
3 years ago
15

*this is a multiple select question*

Physics
1 answer:
dalvyx [7]3 years ago
6 0
When you fly through the trososphere into the stratosphere you would expect to have the temperature remain constant and the temperature would stop decreasing
You might be interested in
I NEED THIS ASAP
Stells [14]

Answer:

i believe its E.......

7 0
3 years ago
Read 2 more answers
The electric potential at the dot in the figure is 3160 V. What is charge q?
PSYCHO15rus [73]

Hi there!

Recall the equation for electric potential of a point charge:

V = \frac{kQ}{r}

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

Q = Charge (C)
r = distance (m)

We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.

Upper right charge's potential:

V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V

Lower left charge's potential:

V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V

Add the two, and subtract from the total EP at the point:

3160 + 1123.75 - 2247.5 = 2036.25


The remaining charge must have a potential of 2036.25 V, so:

2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}


5 0
2 years ago
The type of function that describes the amplitude of damped oscillatory motion is _______. The type of function that describes t
Salsk061 [2.6K]

Answer:

exponential

Explanation:

type of function that describes the amplitude of damped oscillatory motion is exponential because as we know that here function is

y = A × e^{\frac{-bt}{2m}}  × cos(ωt + ∅ )    ..................................... ( 1 )          

here function A × e^{\frac{-bt}{2m}}   is amplitude

as per equation ( 1 )it is exponential

so that we can say that amplitude of damped oscillatory motion is exponential

8 0
3 years ago
A transport plane travelling at a steady speed of 132 ms and an altitude of 113 m, releases a parcel when it is directly above a
Crank

Answer:

what time you thinking. about coming down to take a break and ok I will get to?. that was a right answer?

7 0
3 years ago
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

4 0
3 years ago
Other questions:
  • In a ballistics test, a 28-g bullet pierces a sand bag that is 30 cm thick. If the initial bullet velocity was 55 m/s and it eme
    11·1 answer
  • Which scientific design has the fewest limitations?
    12·1 answer
  • A train travels 225 kilometers in 2.5 hours . what's the trains average speed
    9·2 answers
  • Explain how wind acts as an agent of erosion and deposition
    10·1 answer
  • Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the o
    9·1 answer
  • Damage to internal organs will present as pain associated with a particular surface area of the body. Why would something like i
    13·1 answer
  • Why did Rita's hands get hot when she rubbed them ?
    6·2 answers
  • Which person will experience what happens to sounds due to the Doppler effect?
    6·1 answer
  • A dwarf planet has a mass of 0.0045 times that of the Earth and a diameter on average 0.19 times that of the Earth. What is the
    12·1 answer
  • When a certain amount of heat is supplied to 1KG of insulated aluminium. The temperature of the aluminium rises by 1°C
    7·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!