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aliya0001 [1]
3 years ago
7

What variables show a direct relationship? Check all that apply. the speed of a car and the distance traveled the elevation abov

e sea level and the air temperature the number of students in a cafeteria and the amount of food consumed the mass of an icicle and the time it takes to melt the distance a planet is from the Sun and that planet’s temperature the mass of a space shuttle and its acceleration through space
Physics
2 answers:
ZanzabumX [31]3 years ago
5 0
1, 3, and 4. I just took the quiz.
Llana [10]3 years ago
3 0
The variables that show a direct relationship are :
- The speed of a car and the distance traveled
- Number of students in  a cafeteria and the amount of food consumed
- The distance a planet is from the sun and that planet's temperature
- The mass of a space shuttle and its acceleration through space

In direct relationship, when one factor is increased/decreased , it will directly cause the other factor to be increased/decreased
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Do sound waves always travel in straight lines explain
babunello [35]
They travel like waves. Just throw rock at lake you will see wave. When it bumps to barrier barrier reflects some part of it . Not like a line lika a wave
3 0
2 years ago
What is the speed of sound in air at 50°F (in ft/s)?
gizmo_the_mogwai [7]

Answer:

Speed of air = 1106.38 ft/s

Explanation:

Speed of sound in air with temperature

v_{air}=331.3\sqrt{1+\frac{T}{273.15}} \\

Here speed is in m/s and T is in celcius scale.

T = 50°F

T=(50-32)\times \frac{5}{9}=10^0C \\

Substituting

v_{air}=331.3\sqrt{1+\frac{10}{273.15}}=337.31m/s \\

Now we need to convert m/s in to ft/s.

1 m = 3.28 ft

Substituting

v_{air}=337.31\times 3.28=1106.38ft/s \\

Speed of air = 1106.38 ft/s

6 0
2 years ago
A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse
Elanso [62]

Answer:

v_{f} = 0.51 \frac{m}{s}

Explanation:

We apply Newton's second law at the crate :

∑F = m*a (Formula 1)

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

m : mass in kilograms (kg)

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

Data:

m=90kg :  crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight  (W)

W= m*g

W= 90kg*9.8 m/s²

W= 882 N

Friction force : Ff

Ff= μk*N Formula (2)   

μk: coefficient of kinetic friction

N : Normal force (N)  

Problem development

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-W = 0

N = W

N = 882 N

We replace the  data in the formula (2)

Ff= μk*N  = 0.351* 882 N

Ff=  309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax    , ax=0

282 N - 309.58 N = 90*a  

a=  (282 N - 309.58 N ) / (90)

a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the  data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}

v_{f} = 0.51 \frac{m}{s}

8 0
3 years ago
A proton is held at rest in a uniform electric field. When it is released, the proton will lose?
nydimaria [60]

A proton is held at rest in a uniform electric field. When it is released, the proton will lose its kinetic energy.

Kinetic energy

The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity. The body holds onto the kinetic energy it acquired during its acceleration until its speed changes. The body exerts the same amount of effort when slowing down from its current pace to a condition of rest. Formally, kinetic energy is any term that includes a derivative with respect to time in the Lagrangian of a system.

To learn more about kinetic energy refer here:

brainly.com/question/11301578

#SPJ4

5 0
1 year ago
A 5c charge experiences a force of 40n when put at a certain location in space. What is the electric field at that location?
stepladder [879]
The 'strength' of the electric field is the force on 1C of charge at that point.

At this 'certain location', the field is 40/5 = 8 newtons per coulomb = <u>8 volts</u>
3 0
3 years ago
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