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
Ira Lisetskai [31]
2 years ago
5

99 percent of the earth's atmosphere by mass is made up of only elements. These elements are

Physics
2 answers:
Alja [10]2 years ago
8 0
99% of the earth´s atmosphere by mass is made up of only these elements: Nitrogen and oxygen.

Answer: B) Nitrogen and oxygen.
diamong [38]2 years ago
7 0

The correct answer is B) Nitrogen and oxygen

Explanation:

The atmosphere refers to the gaseous layer that covers a body such as a planet. The composition of the atmosphere varies according to the planet or body and therefore the elements found on it are different. In the case of our planet, the atmosphere is mainly composed of nitrogen and oxygen as these are 99% of the atmosphere, while other elements such as carbon dioxide represent the 1% missing. Also, the atmosphere is divided into different layers according to heigh this includes the troposphere, stratosphere, among others. Thus, 99 percent of the Earth's atmosphere is made up only of nitrogen and oxygen.

You might be interested in
What provides the force on the person on the passenger seat?
frutty [35]

When the car speeds up, slows down, or goes around a curve,
passengers need a force applied to them to make them do the
same thing, otherwise they won't keep up with the car. 

The force on the passenger is applied by means of friction between
the upholstery and the seat of his pants, and also by the seat-back
or his seat-belt.

4 0
3 years ago
Read 2 more answers
A psychologist is interested in exploring the effect tutorial support on students academic performance and assign students in to
NikAS [45]

Answer:

The dependent variable is academic performance

The independent variable is the presence/absence of tutorial support

The control group are students who did not get the tutorial support.

The experimental group were students that got the tutorial support

Explanation:

In every experiment, there is a dependent and independent variable as well as an experimental and a control group.

The experimental group receive the treatment while the control group do not receive the treatment. The independent variable is manipulated and its impact on the dependent variable is evaluated.

The control group are students who did not receive the tutorial support while the experimental group are students that received the tutorial support.

The dependent variable in this case is academic performance. Its outcome depends on the presence or absence of tutorial support (independent variable).

5 0
3 years ago
A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.
Hoochie [10]

Answer:

(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

Because the hockey puck  moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s

Calculation of the acceleration of the  hockey puck

We apply the Formula (1)

vf=v₀+a*t      v₀=5.8 m/s ,  vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t   Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t²   ,  d=15.1 m ,  v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²  

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the  frictional force (Ff)

We apply Newton's second law

∑F = m*a    Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the  hockey puck in the attached graphic

∑Fx = m*a     m= 115g * 10⁻³ Kg/g = 0.115g    ,  a= -1.12 m/s²

-Ff = 0.115*(-1.115)  We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck  Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N =  1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

8 0
3 years ago
Explain what happens during stages A and B.
fgiga [73]
Is there a picture?
3 0
2 years ago
A push of magnitude P acts on a box of weight W as shown in the figure. The push is directed at an angle θ below the horizontal,
Kipish [7]

Answer:

its on wheels and they are supposed to make it eas

Explanation:

3 0
2 years ago
Other questions:
  • A garden hose having with an internal diameter of 1.1 cm is connected to a (stationary) lawn sprinkler that consists merely of a
    11·1 answer
  • The mass of a single gold atom is 3.27×10-22 grams. how many gold atoms would there be in 236 milligrams of gold?
    5·2 answers
  • Which of the following will cause an increase in gas pressure in a closed container?
    12·2 answers
  • The Sl temperature scale used in science is the
    6·2 answers
  • The SI unit of force is the
    5·1 answer
  • To find amplitude, measure
    12·2 answers
  • The sum of the interior angles of a quadrilateral is _______. 180° 90° 270° 360°
    14·2 answers
  • Two carts (m1 = m2 = 0.400 kg) are placed on an aluminum track. The first cart is pushed with the initial velocity of 1.5 m/s to
    8·1 answer
  • Velocities of two bodies A and B are given in vectors notation as va =i+2j-3k and Vb=3i+2j-k what will be the relative velocity
    12·1 answer
  • Compare the momentum of a 7160 kg truck moving at 5.00 m/s to the
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!