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3 years ago

How does the area of 0.3 m2 of iron change when heated from 20 ° C to 60 ° C?

Physics

1 answer:

DerKrebs [107]3 years ago

6 0

Answer:

The quantitative relationship between heat transfer and temperature change contains all three factors: Q = mcΔT, where Q is the symbol for heat transfer, m is the mass of the substance, and ΔT is the change in temperature. The symbol c stands for specific heat and depends on the material and phase. The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00ºC. The specific heat c is a property of the substance; its SI unit is J/(kg ⋅ K) or J/(kg ⋅ ºC). Recall that the temperature change (ΔT) is the same in units of kelvin and degrees Celsius. If heat transfer is measured in kilocalories, then the unit of specific heat is kcal/(kg ⋅ ºC).

Explanation:

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A horse of mass 242 kg pulls a cart of mass 224 kg. The acceleration of gravity is 9.8 m/s 2 . What is the largest acceleration

Rufina [12.5K]

To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as

F = ma

Where,

m= Mass

a = Acceleration

At the same time the frictional force can be defined as,

$F_f = \mu N$

Where,

$\mu =$ Frictional coefficient

N = Normal force (mass*gravity)

Our values are given as,

$m_h = 242 kg\\m_c = 224 kg\\\mu = 0.894\\$

By condition of Balance the friction force must be equal to the total net force, that is to say

$F_{net} = F_f$

$m_{total}a = \mu m_hg$

$(m_h+m_c)a = \mu*m_h*g$

Re-arrange to find acceleration,

$a= \frac{\mu*m_h*g}{(m_h+m_c)}$

$a = \frac{0.894*242*9.8}{(242+224)}$

$a = 4.54 m/s^2$

Therefore the acceleration the horse can give is $4.54m/s^2$

3 0

3 years ago

A tank contains 350 liters of fluid in which 10 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

aleksandr82 [10.1K]

Answer:

$A(t) = -340e^{-t/70} + 350$

Explanation:

Since fluid is pumping in and out at the same rate (5L/min), the total fluid volume in the tank stays constant at 350L. Only the amount of salt and its concentration changed overtime.

Let A(t) be the amount of salt (g) at time t and C(t) (g/L) be the concentration at time t

A(0) = 10 g

Brine with concentration of 1g/L is pouring in at the rate of 5L/min so the salt income rate is 5 g/min

The well-mixed solution is pouring out at the rate of 5L/min at concentration C(t) so the salt outcome rate is 5C g/min

But the concentration is total amount of salt over 350L constant volume

C = A / 350

Therefore our rate of change for salt A' is

A' = 5 - 5A/350 = 5 - A/70

This is a first-order linear ordinary differential equation and it has the form of y' = a + by. The solution of this is

$y = ce^{bt} + \frac{a}{b}$

So $A = ce^{\frac{-t}{70}} + \frac{5}{1/70} = ce^{-t/70} + 350$

with A(0) = 10

c + 350 = 10

c = 10 - 350 = -340

$A(t) = -340e^{-t/70} + 350$

4 0

3 years ago

Q = cmAT

castortr0y [4]

Answer: $Q=3000 cal$

Explanation:

We are given the following formula:

$Q=m. c. \Delta T$ (1)

Where:

$Q=3000 cal$ is the amount of heat

$m=300g$ is the mass of water

$c=1 cal/g \°C$ is the specific heat of water

$\Delta T$ is the variation in temperature, which in this case is $\Delta T=30\°C-20\°C=10\°C$

Rewriting equation (1) with the known values at the right side, we will prove the result is $3000 cal$ :

$Q=(300g)(1 cal/g \°C)(10\°C)$ (2)

$Q=3000 cal$ This is the result

8 0

3 years ago

How do I solve this arithmetic sequence?

Serjik [45]

Answer:

The 16ᵗʰ term of this sequence is 82

Step-by-step explanation:

Here,

First Term = a₁ = 9

Common Difference = (d) = 2

Now, For 16ᵗʰ term, n = 16

a₁₆ = 7 + (16 - 1) × 2

a₁₆ = 7 + 15 × 5

a₁₆ = 7 + 75

a₁₆ = 82

Thus, The 16ᵗʰ term of this sequence is 82

<u>-TheUnknownScientist</u>

4 0

2 years ago

The upward velocity of a 2540kg rocket is v(t)=At + Bt2. At t=0 a=1.50m/s2. The rocket takes off and one second afterwards v=2.0

Alexeev081 [22]

Answer:

The value of A is 1.5m/s^2 and B is 0.5m/s^³

Explanation:

The mass of the rocket = 2540 kg.

Given velocity, v(t)=At + Bt^2

Given t =0

a= 1.50 m/s^2

Now, velocity V(t) = A*t + B*t²

If, V(0) = 0, V(1) = 2

a(t) = dV/dt = A+2B × t

a(0) = 1.5m/s^²

1.5m/s^² = A + 2B × 0

A = 1.5m/s^2

now,

V(1) = 2 = A× 1 + B× 1^²

1.5× 1 +B× 1 = 2m/s

B = 2-1.5

B = 0.5m/s^³

Now Check V(t) = A× t + B × t^²

So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² × 1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s

Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)

7 0

3 years ago

How does the area of ​​0.3 m2 of iron change when heated from 20 ° C to 60 ° C?

How does the area of 0.3 m2 of iron change when heated from 20 ° C to 60 ° C?