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Usimov [2.4K]
3 years ago
12

Which of these experiments would make use of quantitative data

Physics
2 answers:
Sever21 [200]3 years ago
8 0

the answer is (B). The answers (A), (C), and (D) are not qualitative.

Hope this helps!

kumpel [21]3 years ago
7 0
Study on the effects of sunlight on the growth rate of plants
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The core of a certain reflected reactor consist of a cylinder 10 ft high 10 ft in diameter The measured maximum-to-average flux
Westkost [7]

Answer:

The maximum power density in the reactor is 37.562 KW/L.

Explanation:

Given that,

Height = 10 ft = 3.048 m

Diameter = 10 ft = 3.048 m

Flux = 1.5

Power = 835 MW

We need to calculate the volume of cylinder

Using formula of volume

V =\pi r^2 h

Put the value into the formula

V=\pi\times(1.524)^2\times 3.048

V= 22.23\m^3

V = 22.23\times10^{3}\ Liter

We need to calculate the maximum power density in the reactor

Using formula of power density

P=\dfrac{E}{V}

Where, P = power density

E = energy

V = volume

Put the value into the formula

P=\dfrac{835\times10^{6}}{22.23\times10^{3}}

P=37561.85 = 37.562\times KW/L

Hence, The maximum power density in the reactor is 37.562 KW/L.

6 0
3 years ago
A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
4 years ago
An artist working on a piece of metal in his forging studio plunges the hot metal into oil in order to harden it. The metal piec
lutik1710 [3]

Answer:

The temperature of the metal is  T_m  =  376.8 ^o C

Explanation:

From the question we are told that

     The mass of the metal is  M =  60 \ kg

     The specific heat of the metal is  c_p  =  0.1027 kcal/(kg \cdot ^oC)

       The mass of the oil is M_o  =  810 \ kg

       The temperature of the oil is  T_o  =  35^oC

       The specific heat of oil is  c_o  =  0.7167 kcal/(kg \cdot ^oC )

       The equilibrium temperature is T_e  =  39 ^oC

According to the law of energy conservation

     Heat lost by metal  =  heat gained by the oil

So  

   The quantity  of heat lost by the metal is mathematically represented as

               Q =  - Mc_p \Delta T

=>            Q =  -Mc_p (T_m  -  T_c)

Where T_ m  the temperature of metal before immersion

The negative sign show heat lost

The quantity  of gained t by the metal is mathematically represented as      

           Q =  M_o c_o \Delta T

=>        Q =  M_o c_o (T_c - T_o)

So  

         Mc_p (T_m  -  T_c)   =   M_o c_o (T_c - T_o)

substituting values

          - 60 * 0.1027 (T_m  - 39)   =   810 * 0.7167 *  (39 - 35)

=>       T_m  =  376.8 ^o C

         

6 0
3 years ago
A spherical shell of radius 9.0 cm carries a uniform surface charge density σ= 9.0 nC/m2. The electric field at r= 9.1 cm is app
maria [59]

Answer:

995.12 N/C

Explanation:

R = 9 cm = 0.09 m

σ = 9 nC/m^2 = 9 x 10^-9 C/m^2

r = 9.1 cm = 0.091 m

q = σ x 4π R² = 9 x 10^-9 x 4 x 3.14 x 0.09 x 0.09 = 9.156 x 10^-10 C

E = kq / r^2

E = ( 9 x 10^9 x 9.156 x 10^-10) / (0.091 x 0.091)

E = 995.12 N/C

8 0
3 years ago
At a place where an object is thrown vertically downward with a speed of while a different object is thrown vertically upward wi
Bumek [7]

Answer:

Both objects will undergo the same change in velocity

Explanation:

m = Mass of the Earth =  5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

m = Mass of object

Any object which is falling has only the acceleration due to gravity.

ma=\dfrac{GMm}{r^2}\\\Rightarrow a=\dfrac{GM}{r^2}\\\Rightarrow a=\dfrac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{(6.371\times 10^6)^2}\\\Rightarrow a=9.81364\ m/s^2

The acceleration due to gravity on Earth is 9.81364 m/s²

So, the speeds of the objects will change at an equal rate of 9.81364 m/s² but the change will be negative when an object is thrown up.

Hence, both objects will undergo the same change in velocity.

4 0
3 years ago
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