True Or False for each question
True Or False for each question - 1

Answers

Answer 1
Answer:

7 true

8 false

9 false

10 false

11 false

12 true

13 true

hope this helps!


Related Questions

Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a wavelength of 3.5 cm?Express your answer in GHz and using two significant figures.f = ________GHzMicrowave signals are beamed between two mountaintops 52 km apart. How long does it take a signal to travel from one mountaintop to the other?Express your answer in ms and using two significant figures.t = ________ms
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A heavy object and a light object are dropped from the same height. If we neglect air resistance, which will hit the ground first?

The universe is filled with photons left over from the Big Bang that today have an average energy of about 2 × 10−4 eV (corresponding to a temperature of 2.7 K). As derived in lecture, the number of available energy states per unit volume for photons is ????(????)????????

Answers

Answer:

The number of available energy states per unit volume is 4.01*10^(48)

Explanation:

Given that,

Average energy  E=2*10^(-4)\ eV

Photon = 4*10^(-5)\ eV

We need to calculate the number of available energy states per unit volume

Using formula of energy

g(\epsilon)d\epsilon=(8\pi E^2dE)/((hc)^3)

Where, E = energy

h = Planck constant

c = speed of light

Put the value into the formula

g(\epsilon)d\epsilon=(8*\pi*2*10^(-4)*4*10^(-5)*1.6*10^(-19))/((6.67*10^(-34)*3*10^(8))^3)

g(\epsilon)d\epsilon=4.01*10^(48)

Hence, The number of available energy states per unit volume is 4.01*10^(48)

Statement: Baseball is not a dangerous sport unless either a bat breaks or a player is hit with a ball.Key: B = A bat breaks.
D = Baseball is a dangerous sport.
P = A player is hit with a ball.

Answers

Answer:

P

Explanation:

If the bat break it will get replaced and you will get another bat,but if u get hit with the ball it's dangerous in many ways

Final answer:

The question seeks to apply Boolean logic, specifically the 'OR' function, to the statement about baseball. In Boolean terms, 'Baseball is a dangerous sport (D)' is true if either 'A bat breaks (B)' or 'A player is hit with a ball (P)' is true.

Explanation:

The subject matter of this question is Boolean logic, which is a subfield of mathematics. In the provided statement, the sport of baseball (D) is deemed dangerous if either a bat breaks (B) or a player is hit with a ball (P). In terms of Boolean logic, this can be symbolized as 'D = B ∨ P', where ∨ signifies the logical operation 'OR'. Therefore, if either B or P is true (meaning a bat breaks or a player is hit), then D (Baseball is a dangerous sport) is consequentially deemed true.

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A 4,667 kHz AM radio station broadcasts with a power of 84 kW. How many photons does the transmitting antenna emit each second.

Answers

Answer:

Explanation:

Frequency( ν ) 4667 x 10³ Hz = 4.667 x 10⁶ Hz

Energy of one photon = hν [ h is plank's constant ]

= 6.6 x 10⁻³⁴ x 4.667 x 10⁶ = 30.8 x 10⁻²⁸ J

Power = 84 x 10³ J/s

No of photons emitted = Power / energy of one photon

= 84 x 10³ / 30.8 x 10⁻²⁸ =2.727 x 10³¹  per second .

Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit. The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm. What is the width of the slit?

Answers

Answer:

The width of the slit is 0.167 mm

Explanation:

Wavelength of light, \lambda=608\ nm=608* 10^(-9)\ m

Distance from screen to slit, D = 88.5 cm = 0.885 m

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m

We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

y=(mD\lambda)/(a)

where

a = width of the slit

a=(mD\lambda)/(y)

a=(5* 0.885\ m* 608* 10^(-9)\ m)/(0.0161\ m)

a = 0.000167 m

a=1.67* 10^(-4)\ m

a = 0.167 mm

So, the width of the slit is 0.167 mm. Hence, this is the required solution.

A ball of mass 0.7 kg flies through the air at low speed, so that air resistance is negligible. (a) What is the net force acting on the ball while it is in motionWhich components of the ball's momentum will be changed by this force? What happens to the x component of the ball's momentum during its flight? What happens to the y component of the ball's momentum during its flight? It decreases. What happens to the z component of the ball's momentum during its flight?

Answers

A) Net Force is -6.86N

B) The y component of momentum.

C) The x component of momentum should remain the same.

D)The y component of momentum decreases.

E)The z component of momentum should remain constant.

The following information should be considered:

(A)

The net force should be

= -9.8 (0.7)

= -6.86N

(B)

Due to the net force is on the y-axis, so only the vertical component of the momentum should be changed because to the force.

(C)

Because there is no resistance of air, the ball should be in projectilemotion problems, this represents hat the x component of the velocity remains constant, also does the mass.

D)

The y component of momentum reduced, this is due to gravity reduced the y component of the velocity.

E)Because there is no z component of the force there is no change in the z component of the momentum.

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Final answer:

With negligible air resistance and low speed, the only significant net force on a 0.7 kg ball is gravity, affecting the ball's y component of momentum. The x component remains constant, and z component changes are not discussed without additional forces.

Explanation:

When a ball of mass 0.7 kg flies through the air at low speed with air resistance negligible, the net force acting on the ball while it is in motion is primarily due to gravity, which will be impacting the y component of the ball's momentum. The x component of the ball's momentum remains unchanged because no horizontal force is applied, while the y component changes due to gravity, and the z component would only change if there were forces acting in a direction out of the horizontal plane, which are not mentioned in the scenario. As for the Earth-ball system, momentum is conserved in the vertical direction because the system experiences no net external vertical force.

A concave mirror is to form an image of the filament of a headlight lamp on a screen 8.50 m from the mirror. The filament is 6.00 mm tall, and the image is to be 37.5 cm tall. Part A: How far in front of the vertex of the mirror should the filament be placed?Part B: to what radius of curvature should you grind the mirror?

Answers

Answer:13.6 cm

Explanation:

Given

v(image distance)=-8.5 m

height of object(h_1)=6 mm

height of image (h_2)=37.5 cm

and magnification of concave mirror is given by m=(-v)/(u)=(-h_2)/(h_1)

m=(-37.5* 10)/(6)=-62.5

-62.5=(8.5* 100)/(u)

u=13.6 cm

so object is at a distance of 13.6 cm from mirror.

for focal length

(1)/(f)=(1)/(v)+(1)/(u)

(1)/(f)=(-1)/(850)+(-1)/(13.6)

(1)/(f)=-0.00117-0.0735

f=-13.4 cm

thus radius of curvature of mirror is R=2f=26.8 cm

Final answer:

The filament of the headlight lamp should be placed about 0.85 m in front of the vertex of the mirror. The radius of curvature for the concave mirror should be approximately 0.85 m.

Explanation:

To determine how far in front of the vertex of the mirror the filament should be placed, we can use the mirror equation:

1/f = 1/do + 1/di

Where f is the focal length of the concave mirror, do is the object distance, and di is the image distance.

With the given information, we have:

do = ?

di = 8.50 m

Using the magnification formula:

magnification = -di/do

By substituting the values we know, we can solve for do:

37.5 cm / 6.00 mm = -8.50 m / do

Solving for do, we find that do ≈ - 0.85 m.

Since the object distance cannot be negative, we conclude that the filament of the headlight lamp should be placed about 0.85 m in front of the vertex of the mirror.

To find the radius of curvature for the concave mirror, we use the mirror formula:

1/f = 1/do + 1/di

With do = -0.85 m and di = 8.50 m, we can rearrange the formula to solve for f:

1/f = 1/-0.85 + 1/8.50

1/f ≈ -1.1765

Solving for f, we find that the focal length is approximately 0.85 m.

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