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kicyunya [14]
3 years ago
6

Sound travels fastest through A) space. B) cool air. C) warm air. D) a metal spoon.

Physics
2 answers:
Mademuasel [1]3 years ago
7 0
A metal spoon. Because metal travels the fastest
steposvetlana [31]3 years ago
4 0

I took the test the answer is Metal Spoon because sound always travels fastest through solids!

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What net force acting on a 14 KG wagon produces an acceleration of 1.5 MS^2?
valentina_108 [34]

21N

Explanation:

Given parameters:

Mass of wagon = 14kg

Acceleration = 1.5m/s²

Unknown:

Net force on wagon = ?

Solution:

Force is a pull or push on a body that causes a body to change its state. It is expressed as:  

                 Force = mass  x acceleration

  Force on wagon = 14 x 1.5 = 21N

Learn more:

Force brainly.com/question/10470406

#learnwithBrainly

7 0
2 years ago
A solenoid 25.0 cmcm long and with a cross-sectional area of 0.550 cm^2 contains 460 turns of wire and carries a current of 90.0
ankoles [38]

Answer:

a.  B = 0.20T

b.  u = 17230.6 J/m³

c.  E = 0.236J

d.  L = 5.84*10^-5 H

Explanation:

a. In order to calculate the magnetic field in the solenoid you use the following formula:

B=\frac{\mu_o n i}{L}               (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

n: turns of the solenoid = 460

L: length of the solenoid = 25.0cm = 0.25m

i: current  = 90.0A

You replace the values of the parameters in the equation (1):

B=\frac{(4\pi*10^{-7}T/A)(460)(90.0A)}{0.25m}=0.20T

The magnetic field in the solenoid is 0.20T

b. The magnetic permeability of air is approximately equal to the magnetic permeability of vacuum. To calculate the energy density in the solenoid you use:

u=\frac{B^2}{2\mu_o}=\frac{(0.20T)^2}{2(4\pi*10^{-7}T/A)}=17230.6\frac{J}{m^3}

The energy density is 17230.6 J/m³

c. The total energy contained in the solenoid is:

E=uV           (2)

V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:

V=AL

A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2

V=(5.5*10^{-5}m^2)(0.25m)=1.375*10^{-5}m^3

Then, the energy contained in the solenoid is:

E=(17230.6J/m^3)(1.375*10^{-5}m^3)=0.236J

The energy contained is 0.236J

d. The inductance of the solenoid is calculated as follow:

L=\frac{\mu_o N^2 A}{L}=\frac{(4\pi*10^{-7}T/A)(460)^2(5.5*10^{-5}m^2)}{0.25m}\\\\L=5.84*10^{-5}H

The inductance of the solenoid is 5.84*10^-5 H

3 0
3 years ago
A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal
Setler [38]

Answer

given,

mass of the ball = 3 kg

swing in vertical circle with radius = 2 m

   work done by the gravity = ?          

   work done by the tension = ?            

Work done by the gravity = - m g Δh            

 Δ h = 2 + 2 = 4 m                                                                

Work done by the gravity =- 3 \times 9.8 \times 4

                                           = -117.6 J                  

work done by gravity is equal to -117.6 J            

Work done by tension will be equal to zero.        

Zero because tension is always perpendicular to velocity

work done by tension is equal to 0 J                          

7 0
3 years ago
HELP DUE TONIGHT<br><br> What is friction? How does friction affect the motion of an object?
hram777 [196]

Answer:

Friction always acts in the direction opposing motion. This means if friction is present, it counteracts and cancels some of the force causing the motion (if the object is being accelerated).

Explanation:

Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other.

3 0
3 years ago
Water flows through a 2.5cm diameter pipe at a rate of
My name is Ann [436]

Answer:

From the Bernoulli energy principle,

ΔP + 1/2ρΔv² = 0 -------------------------- eqn 1

where

ΔP = pressure drop = P2 - P1 = (1 - 0.25)x10⁵ N/m =7.5 x 10⁴N/m

Δv²= velocity change = v₂² - v₁²

ρ = water density = 1kg/m3

Recall volumetric flow rate, Q=A v = constant

A = cross sectional area = πr²=πd²/4

d=pipe diameter at point 2 = 2.5cm = 0.025m and Q =0.20m³/min = 0.00333m³/s

So A= 0.000491m²

we can get v2 = Q/A = 6.79m/s

From eqn 1, v₁² = 2(P2 - P1)/ρ + v₂²

v₁² =  (2 x 7.5 x 10⁴)/1000 + 6.79²

v₁² = 196

v₁ = 14m/s

we can now get the area of the constriction point 2, A₁ = Q/v₁

A₁ = 0.000238m² and the diameter now will be d₁

d₁² = 4 x A₁ / π

     = 4 x 0.000238/3.14 = 0.000303m²

d₁ = √0.000303 = 0.0174m

Therefore, the diameter of  a constriction in the pipe  at the new pressure = 0.0174m = 1.74cm

6 0
2 years ago
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