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sammy [17]
3 years ago
13

What do radio waves and microwaves have in Common?

Physics
1 answer:
DerKrebs [107]3 years ago
3 0

What do radio waves and microwaves have in Common?

<u>A) Both are at the ideological the spectrum that has the lowest frequency. </u>

B) Both are at the spectrum that has the shortest wavelengths.

C) Both have higher frequencies than visible light.

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You have a set of calipers that can measure thicknesses of a few inches with an uncertainty of 0:005 inches. I mesure the thickn
solmaris [256]

To solve this problem we will apply the concepts related to the calculation of significant figures under tolerance levels. We will also take into account that the number of significant digits at the end of an answer must not be greater than the number of significant digits of a number:

PART A) The thickness of the 52 cards is

T = 0.590 \pm 0.005 in

The thickness, t, of 1 card can be:

t = \frac{0.590}{52} \pm \frac{0.005}{52} in

t = 0.0113461\pm 0.000096 in

The thickness of 52 cards has 3 significant figures and the uncertainty has 1 significant digit. So

the significant figures of the thickness of one card and the uncertainty should also be 3 and 1 respectively

t = 0.01134 \pm 0.005 in

t = 0.0113 \pm 0.0001 in

Therefore the thickness of one card is \mathbf {t = 0.0113 \pm 0.0001 in }

PART B) One card has uncertainty of 0.0001 in if measured using 1 deck.

The number of decks, n, required to create the uncertainty of 0.00002 in is

n = \frac{0.0001}{0.00002}

n = 5

Therefore, 5 decks are required to measure the thickness of one card with an uncertainty of 0.00002 in

6 0
3 years ago
A flat (unbanked) curve on a highway has a radius of 240.0 m m . A car rounds the curve at a speed of 26.0 m/s m/s . Part A What
Grace [21]

Answer:

a) u_s=0.375

b )v=14.4 m/s

Explanation:

1 Concepts and Principles

Particle in Equilibrium: If a particle maintains a constant velocity (so that a = 0), which could include a velocity of zero, the forces on the particle balance and Newton's second law reduces to:  

∑F=0                 (1)

2- Particle in Uniform Circular Motion:

If a particle moves in a circle or a circular arc of radius Rat constant speed v, the particle is said to be in uniform circular motion. It then experiences a net centripetal force F and a centripetal acceleration a_c. The magnitude of this force is:

F=ma_c

 =m*v^2/R            (2)

where m is the mass of the particle F and a_c are directed toward the center of curvature of the particle's path.  

<em>3- The magnitude of the static frictional force between a static object and a surface is given by:</em>  

f_s=u_s*n              (3)

where u_s is the coefficient of kinetic friction between the object and the surface and n is the magnitude of the normal force.  

Given Data

R (radius of the car's path) = 170 m

v (speed of the car) = 25 m/s  

Required Data

- In part (a), we are asked to find the coefficient of static friction u_s that will prevent the car from sliding.  

- In part (b), we are asked to find the speed of the car v if the coefficient of static friction is one-third u_s found in part (a).  

Solution:

see the attachment pic

Since the car is not accelerating vertically, we can model it as a particle in equilibrium in the vertical direction and apply Equation (1)

∑F_y=n-mg

       = mg               (4)  

We model the car as a particle in uniform circular motion in the horizontal direction. The force that enables the car to remain in its circular path is the force of static friction at the point of contact between road and tires. Apply Equation (2) to the horizontal direction:

∑F_x=f_s

       =m*v^2/R

Substitute for f_s from Equation (3):  

u_s=m*v^2/R

Substitute for n from Equation (4):  

u_s*mg=m*v^2/R

  u_s*g = v^2/R

Solve for u_s:

u_s=  v^2/Rg                          (5)

Substitute numerical values:  

u_s=0.375

(b)  

The new coefficient of static friction between the tires and the pavement is:

u_s'=u_s/3

Substitute u_s/3 for u_s in Equation (5):

u_s/3=v^2/Rg  

Solve for v:  

v=14.4 m/s

8 0
3 years ago
How would the force of Earth's gravity change if a satellite were to orbit at a distance that was only 1/2 of its original orbit
GarryVolchara [31]
B. The force of gravity would increase


4 0
3 years ago
20. Q: How long will it take for an apple falling from a 29.4m-tall tree to hit the ground?
saul85 [17]

Answer:

2.45s

Explanation:

is explanation needed too?

7 0
1 year ago
A block of lead has dimensions of 4.50cm by 5.20cm by 6.00cm. The block weighs 1587g. From this information, calculate the densi
postnew [5]
Density = Mass / Volume
Volume = Area x Length = (4.50 x 5.20) x 6.00 = 140.4
Density = 1587 x 140.4 = 222,814.8 g/cm3 (cubed)
222,814.8 / 1000 = 222.8148 kg/m3 (cubed)
4 0
3 years ago
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