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love history [14]
4 years ago
15

What is the specific heat (J/g⋅∘C) of a metal object whose temperature increases by 3.0∘C when 17.5 g of metal was heated with 3

8.5 J?
Physics
1 answer:
lesya692 [45]4 years ago
6 0

Answer:

0.733J/g°C

Explanation:

Using the formula

Q=mcΔθ

Q=38.5J, c=? , m=17.5g , Δθ=3°C

c= Q/(mΔθ)

c=\frac{38.5}{17.5*3}

c=0.733J/g°C

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Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block
Oksana_A [137]

Answer:

av=0.333m/s, U=3.3466J

b.

v_{A2}=-1.333m/s,\\ v_{B2}=0.667m/s

Explanation:

a. let m_A be the mass of block A, andm_B=10.0kg be the mass of block B. The initial velocity of A,\rightarrow v_A_1=2.0m/s

-The initial momentum =Final momentum since there's no external net forces.

pA_1+pB_1=pA_2+pB_2\\\\P=mv\\\\\therefore m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

v_A_1-v_B_1=v_{B2}-v_{A2}

-Applying the conservation of momentum. The blocks have the same velocity after collision:

v_{B2}=v_{A2}=v_2\\\\2\times 2+10\times 0=2v_2+10v_2\\\\v_2=0.3333m/s

#Total Mechanical energy before and after the elastic collision is equal:

K_1+U_{el,1}=K_2+U_{el,2}\\\\#Springs \ in \ equilibrium \ before \ collision\\\\U_{el,2}=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\\\\U_{el,2}=0.5\times 2\times 2^2-0.5(2+10)(0.333)^2\\\\U_{el,2}=3.3466J

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0

We plug these values in the equation:

m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}

2\times2+10\times0=2v_A_2+10v_B_2\\\\2=v_A_2+5v_B_2\\\\#Eqtn 2:\\v_A_1-v_B_1=v_{B2}-v_{A2}\\\\2-0=v_{B2}-v_{A2}\\\\2=v_{B2}-v_{A2}\\\\#Solve \ to \ eliminate \ v_{A2}\\\\6v_{B2}=2.0\\\\v_{B2}==0.667m/s\\\\#Substitute \ to \ get \ v_{A2}\\\\v_{A2}=\frac{4}{6}-2=1.333m/s

7 0
4 years ago
What average net force is required to accelerate a 9.5 g
Nesterboy [21]

Answer:

2361 Newtons

Explanation:

From the second Newton's law of motion;

F = ma

In this case;

we are given;

Mass as 9.5 g

Initial speed as 0 m/s

Final velocity as 650 m/s

Distance is 0.85 m

Using the equation;

V² = U² + 2as

But u = 0

v² = 2as

Therefore;

a = v² ÷ 2s

  = 650² ÷ 2(0.85)

  = 248,529.40 m/s²

But;

F = ma

   = 0.0095 kg × 248,529.40 m/s²

   = 2361 Newtons

Therefore;

The average net force required to accelerate the bullet is 2361 Newtons.

6 0
3 years ago
A cylindrical storage tank has a radius of 1.35 m. When filled to a height of 3.45 m, it holds 14,014 kg of a liquid industrial
Kruka [31]

Answer:

709.93 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of a body to the volume of that body. The S.I unit of density is kg/m³

From the question above,

D = m/v.............................. Equation 1

Where D = density, m = mass, v = volume.

Note: The volume of the liquid is equal to the volume of the height occupied by the liquid in the container

Since the tank is cylindrical,

v = πr²h........................ Equation 2

Where r = radius of the the tank, h = height of the liquid in the tank

Substitute equation 2 into equation 1

D = m/(πr²h)............... Equation 3

Given: m = 14014 kg, r = 1.35 m, h = 3.45 m, π = 3.14

Substitute into equation 3

D = 14014/(3.14×1.35²×3.45)

D = 14014/19.74

D = 709.93 kg/m³

6 0
3 years ago
What's the general relationship between mass and gravitational force ?
Hitman42 [59]

Answer:

Gravity is the attraction between two objects that have mass. The amount of gravity is directly proportional to the amount of mass of the objects and inversely proportional to the square of the distance between the objects. Gravity is a force that increases the velocity of falling objects - they accelerate.

Explanation:

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3 years ago
Which best describes emerging scientific ideas?
natka813 [3]
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3 years ago
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