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mestny [16]
3 years ago
11

If 59.1 g of pure metal uses 890.9 j of energy to heat it from 26.1 c to 59.6 c, identify the metal

Physics
1 answer:
r-ruslan [8.4K]3 years ago
4 0

Answer: The metal is iron.

Explanation:

We have 59.1 grams of a given metal.

We need 890.9 joules of energy to heat the metal from 26.1°C to 59.6°C

First, we need to see the net increase in temperature, which is equal to the difference between the final temperature and the initial temperature.

ΔT = 59.6°C - 26.1°C = 33.5°C

Now we can find the specific heat, which is calculated as:

c = Ε/(ΔT*m)

Where:

E = energy given to the material = 890.9 J

m = mass of the material = 59.1 gr

ΔT = change in temperature = 33.5°C

Replacing these in the above equation we get:

c = 890.9 J/(33.5°C*59.1 gr) = 0.45 J/(gr*°C)

Now we need to look at a table of specific heats and see which metal has this specific heat.

We can see that iron is the best match, so we can assume that the metal is iron.

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Three boxes in contact rest side-by-side on a smooth, horizontal floor. Their masses are 5.0-kg, 3.0-kg, and 2.0-kg, with the 3.
Ivahew [28]

Answer:

(a)Look at the attached graphic

(b)

(b)-1 Equation 1  : m1= 5kg

       50-F1= 5 *a

(b)-2 Equation 2 : m2= 3kg

        F1-F2= 3 *a

(b)-3 Equation 3 : m3= 2kg

         F2 = 2*a  

(c) F1 =25 N

(d) F2 =10 N

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

(a) Draw the free-body diagrams for each of the boxes

Look at the attached graphic

(b) Write Newton’s equation for each mass along the horizontal direction.

Data: m1=  5.0-kg ,m2= 3.0-kg , ,m3= 2.0-kg

<em>Look</em> <em>m1 free-body diagram:</em>

∑Fx = m1*a

50-F1= 5 *a Equation 1

<em>Look</em> <em>m2 free-body diagram:</em>

∑Fx = m2*a

F1-F2= 3 *a Equation 2

<em>Look</em> <em>m3 free-body diagram:</em>

∑Fx = m3*a

F2 = 2*a     Equation 3

(c) What magnitude force does the 3.0-kg box exert on the 5.0- kg box?

<em>Look</em> <em>Free body diagram of the mass set</em>

∑Fx = m*a   m= m1+m2+m3= 5+3+2 = 10 kg

50 = 10*a

a= 50/10 = 5 m/s²

We replace a = 5 m/s² in the equation 1:

50-F1= 5 *5

50-25= F1

F1 = 25 N

<em> (d) </em><em>What magnitude force does the 3.0-kg box exert on the 2.0kg box?</em>

We replace a= 5 m/s² in the equation 3

F2 = 2*5 = 10 N

4 0
3 years ago
A projectile is fired into the air from the top of a 200-m cliff above a valley as shown below. Its initial velocity is 60 m/s a
anastassius [24]

a) y(max)  = 337.76 m

b) t₁ = 5.30 s  the time for y maximum

c)t₂ =  13.60 s  time for y = 0 time when the fly finish

d) vₓ = 30 m/s        vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα          v₀ₓ = 60*(1/2)     v₀ₓ = 30 m/s   ( constant )

v₀y = v₀ * sinα           v₀y = 60*(√3/2)     v₀y = 30*√3  m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y  =  y₀ + v₀y*t - (1/2)*g*t²      (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt  = v₀y  - g*t

dy/dt  = 0           v₀y  - g*t₁  = 0    t₁ = v₀y/g

v₀y = 60*sin60°  = 60*√3/2  = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s  the time for y maximum

And y maximum is obtained from the substitution of t₁  in equation (1)

y (max) = 200 + 30*√3 * (5.30)  - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max)  = 337.76 m

Total time of flying (t₂)  is when coordinate y = 0

y = 0 = y₀  + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂  - 4.9*t₂²            4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for  t₂

t =  [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t =  [51.96 ±√2700 + 3920]/9.8

t =  [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8         we dismiss this solution ( negative time)

t₂ =  13.60 s  time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60°       vₓ = 30 m/s  as we said before v₀ₓ is constant

vy = v₀y - g *t        vy = 30*√3  - 9.8 * (13.60)

vy = 51.96 - 133.28         vy = - 81.32 m/s

The sign minus means that vy  change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60  m

x = 408 m

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