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motikmotik
2 years ago
11

Which of the following is the BEST example of increasing the intensity of a workout? A. running one mile further than normal B.

running one mile faster than normal C. running one mile more every other workout D. jogging for three miles Please select the best answer from the choices provided. A B C D
Physics
2 answers:
stepladder [879]2 years ago
4 0

Answer:

B. running one mile faster than normal

Explanation:

OverLord2011 [107]2 years ago
4 0

Answer:

B. Running one mile faster than normal

Explanation:

Just took the quiz, good luck ❤

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The equation for photosynthesis is 6H2O (water) + 6CO2 (carbon dioxide) + Light Energy → C6H12O6 (glucose) + 6O2 (oxygen). When
aleksklad [387]

Answer:

b

Explanation:

bc

8 0
2 years ago
If the humidity in a room of volume 450 m3 at 30 ∘C is 75%, what mass of water can still evaporate from an open pan?
Yuliya22 [10]
From tables,

SVP at 30°C = 4.24 kPa

From ideal gas expressions;
n = PV/RT = (4.24*1000*450)/(8.314*303) = 757.4 moles

Now, 75% of 757.4 moles will evaporate leaving 20%. Then, 25% of 757.5 moles...
25% of 757.4 moles = 25/100*757.4 = 189.35 moles
Mass of 189.35 moles = 189.35 moles*18 g/mol = 3408.3 g ≈ 3.4 kg
5 0
2 years ago
Q 32.35: The isotope 235U decays by alpha emission with a half-life of 7.0 x 108 y. It also decays (rarely) by spontaneous fissi
Julli [10]

Answer:

rate of fission =5.89*10^3 1\Year

Explanation:

we know that

rate of fission is given asrate of fission = \frac{0.69}{(T_{1/2})_{fission}} *\frac{ Mass* Avogardo\ number}{Molar\ mass}

(T_{1/2})_{fission} = 3*10^{17} y

mass = 1.0 g

avogardo number = 6.02*10^23

molar mass of isotopes 235U =235

Putting all value to get rate of emission

rate of fission = \frac{0.69}{3*10^{17}} *\frac{1.0*6.02*10^{23}}{235}

rate of fission =5.89*10^3 1\Year

5 0
2 years ago
A point charge q1 = 4.50 nC is located on the x-axis at x = 1.95 m , and a second point charge q2 = -6.80 nC is on the y-axis at
Vinvika [58]

Answer:

Explanation:

One charge is situated at x = 1.95 m . Second charge is situated at y = 1.00 m

These two charges are situated outside sphere as it has radius of .365 m with center at origin. So charge inside sphere = zero.

Applying Gauss's theorem

Flux through spherical surface = charge inside sphere / ε₀

= 0 / ε₀

= 0 Ans .

3 0
2 years ago
Kara Less was applying her makeup when she drove into South's busy parking lot last Friday morning. Unaware that Lisa Ford was s
exis [7]

Answer

given,

Mass of Kara's car = 1300 Kg

moving with speed = 11 m/s

time taken to stop = 0.14 s

final velocity = 0 m/s

distance between Lisa ford and Kara's car = 30 m

a) change in momentum of Kara's car

  Δ P = m Δ v                  

  \Delta P = m (v_f-v_i)

  \Delta P = 1300 (0 - 11)

  Δ P = - 1.43 x 10⁴ kg.m/s

b) impulse is equal to change in momentum of the car

    I = - 1.43 x 10⁴ kg.m/s

c) magnitude of force experienced by Kara

  I = F x t

 I is impulse acting on the car

 t is time

  - 1.43 x 10⁴= F x 0.14

    F = -1.021 x 10⁵ N

negative sign represents the direction of force

8 0
2 years ago
Read 2 more answers
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