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taurus [48]
1 year ago
8

Which one of the following devices converts radioactive emissions to light for detection?

Physics
1 answer:
Finger [1]1 year ago
6 0

the radiotracer converts radioactive emissions to light for detection. the answer is D.

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A 10.0 cm3 sample of copper has a mass of 89.6
Romashka-Z-Leto [24]
Density is mass divides by volume, so
89.6g / 10cm^3 =8.96g /cm^3

*cm^3 is a standard unit of volume*
4 0
2 years ago
A sprinter accelerates from rest to a velocity of 12m/s in the first 6 seconds of the 100 meter dash .
GREYUIT [131]

Answer:

a) 36 m

b) 64 m

Explanation:

Given:

v₀ = 0 m/2

v = 12 m/s

t = 6 s

Find: Δx

Δx = ½ (v + v₀) t

Δx = ½ (12 m/s + 0 m/s) (6 s)

Δx = 36 m

The track is 100 m, so the sprinter still has to run another 64 m.

5 0
2 years ago
After an incandescent lamp is turned on, the temperature of its filament rapidly increases from room temperature to its operatin
IrinaVladis [17]

Answer: (1) The resistance increases and the current decreases.

Explanation:  

When the temperature of the filament increases, the vibrational energy of the constituent atoms increases which leads to increase in inter-atomic collision. Thus, the resistance would increase. The increases in resistance would obstruct the flow of charges more leading to decrease in the value of the current.

Hence, when the temperature of the filament increase, the resistance increases and current decreases.

6 0
1 year ago
Read 2 more answers
A rigid vessel of 0.06 m3 volume contains an ideal gas , CV =2.5R, at 500K and 1 bar.a). if 15000J heat is transferred to the ga
andreev551 [17]

Answer:

Given that

V= 0.06 m³

Cv= 2.5 R= 5/2 R

T₁=500 K

P₁=1 bar

Heat addition = 15000 J

We know that heat addition at constant volume process ( rigid vessel ) given as

Q = n Cv ΔT

We know that

P V = n R T

n=PV/RT

n= (100 x 0.06)(500 x 8.314)

n=1.443 mol

So

Q = n Cv ΔT

15000 = 1.433 x 2.5 x 8.314 ( T₂-500)

T₂=1000.12 K

We know that at constant volume process

P₂/P₁=T₂/T₁

P₂/1 = 1000.21/500

P₂= 2 bar

Entropy change given as

\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}

Cp-Cv= R

Cp=7/2 R

Now by putting the values

\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}

\Delta S=1.443\times 3.5\times 8.314\ln \dfrac{1000.21}{500}-1.443\times 8.314\ln \dfrac{2}{1}

a)ΔS= 20.79 J/K

b)

If the process is adiabatic it means that heat transfer is zero.

So

ΔS= 20.79 J/K

We know that

\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}

Process is adiabatic

\Delta S_{surr}=0

\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}

\Delta S_{univ}= 20.79 +0

\Delta S_{univ}= 20.79

3 0
1 year ago
A good quarterback can throw a football at 27 m/s (about 60 mph). If we assume that the ball is caught at the same height from w
bezimeni [28]

Answer:

The ball was in air for 3.896 s

Explanation:

given,

g = 9.8 m/s², acceleration due to gravity,

If the launch angle is 45°, the horizontal range will be maximum.

The horizontal and vertical launch velocities are equal, and each is equal to

v_h  =  v cos θ

v_h  =  27 × cos 45°

         = 19.09 m/s.

The time to attain maximum height is one half of the time of flight.

v = u + at                     ∵ v = 0 (max. height)

19.09 - 9.8 t₁ = 0

t₁ = 1.948 s

The time of flight is twice of the maximum height time

2 t₁ = 3.896 s

The horizontal distance traveled is

D = v × t

D = 3.896×19.09

   = 74.375 m

The ball was in air for 3.896 s

8 0
1 year ago
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