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Katarina [22]
2 years ago
5

When the mass of the bottle is 0.125 kg, the KE is______ kg m2/s2.

Physics
1 answer:
tensa zangetsu [6.8K]2 years ago
5 0
Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2


Explanation:

(1) Given mass = 0.125 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2

(2) Given mass = 0.250 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2

(3) Given mass = 0.375 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2

(4) Given mass = 0.500 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
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You place a 500 g block of an unknown substance in an insulated container filled 2 kg of water. The block has an initial tempera
Nina [5.8K]

Answer:

3349J/kgC

Explanation:

Questions like these are properly handled having this fact in mind;

  • Heat loss = Heat gained

Quantity of heat = mcΔ∅

m = mass of subatance

c = specific heat capacity

Δ∅ = change in temperature

m₁c₁(∅₂-∅₁) = m₂c₂(∅₁-∅₃)

m₁ = mass of block = 500g = 0.5kg

c₁  = specific heat capacity of unknown substance

∅₂ = block initial temperature = 50oC

∅₁ = equilibrium temperature of block and water after mix= 25oC

m₂= mass of water = 2kg

c₂ = specific heat capacity of water = 4186J/kg C

∅₃ = intial temperature of water = 20oC

0.5c₁(50-25) = 2 x 4186(25-20)

And we can find c₁ which is the unknown specific heat capacity

c₁ = \frac{2*4186*5}{0.5*25}= 3348.8J/kg C≅ 3349J/kg C

4 0
2 years ago
8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres
Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T

hence, B_o = 1.013μT

6 0
2 years ago
A car experiences a centripetal acceleration of 4.4 m/s ^2 as ur rounds a corner with a speed of 15 m/s. What is the radius of t
damaskus [11]
The calculation of the centripetal acceleration of an object following a circular path is based on the equation,

                  a = v² / r

where a is the acceleration, v is the velocity, and r is the radius.

Substituting the known values from the given above,

             4.4 m/s² = (15 m/s)² / r

The value of r from the equation is 51.14 m.

Answer: 51.14 m
3 0
2 years ago
An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. in
Yakvenalex [24]
The charges are the same in absolute value, so the change of potential energy is the same. That means that the change in kinetic energy is also the same. Then:

1 = Ke/Kp = m_e *v_e^2 / m_p * v_p^2, or

v_e/v_p = sqrt( m_p/m_e),

So the speed of the electron will be sqrt( m_p/m_e) times greater than the speed of the proton
4 0
2 years ago
16. A 7500 kg 18-wheeler traveling at 20 m/s exits onto the runaway truck ramp on the freeway.
miskamm [114]

Answer:

<em>765,000 Joule</em>

Explanation:

<u>Principle of Conservation of Energy </u>

The total energy in an isolated system cannot be created or destroyed, but transformed. Moving objects have kinetic energy, objects placed in some height above a reference level have gravitational potential energy. When they change their motion variables, one energy converts into the other, but if the numbers don't fit, we know there was some other type of energy acting into the system. The most common reason for energy 'losses' is the thermal energy, produced when objects move in rough surfaces or take friction from the air.

The 7,500 kg truck is originally traveling at 20 m/s to a certain height we'll set to 0. Thus, its total energy is  

\displaystyle E_1=\frac{mv^2}{2}

\displaystyle E_1=\frac{7,500\ 20^2}{2}

E_1=1,500,000\ Joule

When it comes to a stop, its speed is 0 and its height is 10 m higher than before. It means all the kinetic energy was transformed into other types of energy. The gravitational potential energy is

U=mgh=(7,500)(9.8)(10)=735,000\ Joule

Since this number is not equal to the previous value of the energy, the difference is due to thermal energy dissipated by friction

E_t=1,500,000\ Joule-735,000\ Joule=765,000\ Joule

7 0
2 years ago
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