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s344n2d4d5 [400]
4 years ago
8

Which metal would cause the greatest increase in the temperature of the water in the calorimeter: the one with the highest speci

fic heat, or the one with the lower specific heat?
Physics
1 answer:
masha68 [24]4 years ago
4 0
<h3><u>Answer;</u></h3>

the one with the highest specific heat

<h3><u>Explanation;</u></h3>
  • The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.
  • The specific heat of water is 1 calorie/gram °C or 4.186 joule/gram °C which is higher than any other common substance.
  • <em><u>A metal with the highest specific heat will cause the greatest increase in temperature of water in a calorimeter because the metal would hold more heat, and then transfer the greater quantity of heat to the water.</u></em>
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Water is entering the prism at a rate of A m^3/hr. The prism is empty at time 0. Express the depth d of the water in meters in t
zavuch27 [327]

This question is incomplete, the complete question is;

The picture shows a triangular prism. The end of prism are equilateral triangles with x meters. the other dimension of the prism is L meters

a) Find the volume V in terms of x and L

b) Water is entering the prism at a rate of A m³/hr. The prism is empty at time 0. Express the depth d of the water in meters in terms of A, the length of time t the water has been entering the trough, and the length L of the prism.  

Answer:

a) the volume V in terms of x and L is  ((√3/4)x²L) m³

b) required expression is (2/(3)^(1/u))√(At/L)

Explanation:

Given that;

form the question and image below;

triangular prism ends are equilateral triangle

side length = x meter

Dimension of the prism = L meter

Area of the equilateral triangle = √3/4 (side)² = √3/4 (x)² meter

Volume of the triangular prism = Area × height

= √3/4 (x)² × L

V = ((√3/4)x²L) m³

Therefore, the volume V in terms of x and L is  ((√3/4)x²L) m³

b)

Rate of water entering = A m³/hr

Depth of water tank = d meter

Time = t

Length of prism = L

now Rate of water entering is A m³/hr

dv/d = A                             [  V = ((√3/4)x²L) m³ ]

and

dv/dt = √3/4 [2x dx/dt ] L                   { L is constant }

so

A = √3/4 [2x dx/dt ] L  

∫A dt = √3/2 [ Lx dx ]                   { Integrate both sides}

At = √3/2 × Lx × x²/2

x² = uAt / √3L                              { we find square root of both sides}

x = √( uAt / √3L )

x = (2/(3)^(1/u))√(At/L)

Therefore; required expression is (2/(3)^(1/u))√(At/L)

8 0
3 years ago
Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend
Rasek [7]

Answer:

Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend them into a circular path of radiuse c = 0.25m

Explanation:

Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend them into a circular path of radiuse c = 0.25m ok

3 0
4 years ago
The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the
xz_007 [3.2K]

Answer:

a) -1.14 rev/min²

b) 9900 rev

c) -9.92×10⁻⁴ m/s²

d) 30.8 m/s²

Explanation:

First, convert hours to minutes:

2.2 h × 60 min/h = 132 min

a) Angular acceleration is change in angular velocity over change in time.

α = (ω − ω₀) / t

α = (0 rev/min − 150 rev/min) / 132 min

α = -1.14 rev/min²

b) θ = θ₀ + ω₀ t + ½ αt²

θ = 0 rev + (150 rev/min) (132 min) + ½ (-1.14 rev/min²) (132 min)²

θ = 9900 rev

c) The tangential component of linear acceleration is:

a_t = αr

First,  convert α from rev/min² to rad/s²:

-1.14 rev/min² × (2π rad/rev) × (1 min / 60 s)² = -1.98×10⁻³ rad/s²

Therefore:

a_t = (-1.98×10⁻³ rad/s²) (0.50 m)

a_t = -9.92×10⁻⁴ m/s²

d) The magnitude of the net linear acceleration can be found from the tangential component and the radial component:

a² = (a_t)² + (a_r)²

The radial component is the centripetal acceleration:

a_r = v² / r

a_r = ω² r

First, convert 75 rev/min to rad/s:

75 rev/min × (2π rad/rev) × (1 min / 60 s) = 7.85 rad/s

Find the radial component:

a_r = (7.85 rad/s)² (0.50 m)

a_r = 30.8 m/s²

Now find the net linear acceleration:

a² = (-9.92×10⁻⁴ m/s²² + (30.8 m/s²)²

a = 30.8 m/s²

5 0
3 years ago
Q1: We drop this 0.1 kg apple 1 m. What speed does it hit the ground
77julia77 [94]
Potential energy is measured using formula Ep=mgh
m=mass (kg)
g= acceleration due to gravity (which is 9.8 on earth)
h= height in metres above ground

For this question
m=0.1
g=9.8
h=1
So Ep=0.1(9.8)(1)
Ep=0.98 Joules

When it is dropped all of this potential energy is converted into kinetic energy which can be measured using formula
Ek=1/2m(v^2) (v=final velocity)

Since all potential energy in this q is converted to kinetic we know Ek=0.98Joules and our mass is the same (0.1kg)

So when we sub everything in we get
0.98=1/2(0.1)(v^2)
0.98=0.05(v^2)||divide both side by 0.05
19.6=v^2 ||square root both sides
v=4.4 m/s
5 0
3 years ago
According to Dr. paul Narguizian professor of Biology and Science Education at California State University, ______ are generaliz
Mazyrski [523]

Answer:

I believe the correct answer would be A :)

Explanation:

3 0
3 years ago
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