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3 years ago

how much heat must be added to 30.0 g of solid iron to melt it, assuming the iron is already at it's melting point?

Physics

1 answer:

Dmitry [639]3 years ago

6 0

Answer:

2,800°F (1,538°C)

Explanation:

30g of iron will melt at the same temperature as 1g of iron.

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Find the resistance of wire of 0.65m Radius 0.25 and resistivity 3x10-6 OHM

densk [106]

Complete Question:

Find the resistance of a wire of length 0.65 m, radius 0.25 mm and resistivity 3 * 10^{-6} ohm-metre.

Answer:

Resistance = 9.95 Ohms

Explanation:

Given the following data;

Length = 0.65 m

Radius = 0.25 mm to meters = 0.00025 m

Resistivity = 3 * 10^{-6} ohm-metre.

To find the resistance of the wire;

Mathematically, resistance is given by the formula;

$Resistance = P \frac {L}{A}$

Where;

P is the resistivity of the material.
L is the length of the material.
A is the cross-sectional area of the material.

First of all, we would find the cross-sectional area of the wire.

Area of circle = πr²

Substituting into the equation, we have;

Area = 3.142 * (0.00025)²

Area = 3.142 * 6.25 * 10^{-8}

Area = 1.96 * 10^{-7} m²

Now, to find the resistance of the wire;

$Resistance = 3 * 10^{-6} * \frac {0.65}{1.96 * 10^{-7}}$

$Resistance = 3 * 10^{-6} * 3316326.531$

Resistance = 9.95 Ohms

5 0

3 years ago

Help please!!!

exis [7]

Answer:

5 years worth of work (aka all of the homework i currently have)

3 0

3 years ago

008 (part 3 of 4) 3.0 points

motikmotik

Car A take a time of 2.55hr and car B take a time of 2.14 hr

We know that distance divide by time is speed

here it is given that car A to reach a gas station a distance 189 km from the school traveling at a speed of 74 km/hr

so speed=distance/time

s=d/t

t=d/s

=189/74

=2.55hr

In case of car B it is given that The distance from the is 199.8km, car b is traveling at a speed of 93 km/hr

s=d/t

t=d/s

=199.8/93

=2.14hr

so from the above given data and the formula we solved and found out the time taken by car A is 2.55h and car B is 2.14h

learn more about Speed here brainly.com/question/13943409

#SPJ9

5 0

1 year ago

Two objects are dropped from rest from the same height. Object A falls through a distance during a time t, and object B falls th

UNO [17]

Answer:

Distance covered by B is 4 times distance covered by A

Explanation:

For an object in free fall starting from rest, the distance covered by the object in a time t is

$s=\frac{1}{2}gt^2$

where

s is the distance covered

g is the acceleration due to gravity

t is the time elapsed

In this problem:

- Object A falls through a distance $s_A$ during a time t, so the distance covered by object A is

$s_A=\frac{1}{2}gt^2$

- Object B falls through a distance $s_B$ during a time 2t, so the distance covered by object B is

$s_B=\frac{1}{2}g(2t)^2 = 4(\frac{1}{2}gt^2)=4s_A$

So, the distance covered by object B is 4 times the distance covered by object A.

5 0

2 years ago

The suspension system of a 2100 kg automobile "sags" 8.5 cm when the chassis is placed on it. Also, the oscillation amplitude de

Gennadij [26K]

Answer:

Part a)

$k = 6.06 \times 10^4 N/m$

Part b)

$b = 1795.4 kg/s$

Explanation:

Part a)

as the mass of the suspension system is given as

$m = 2100 kg$

also we have

$x = 8.5 cm$

so now for force balance we have

$mg = kx$

$(525)(9.81) = k(0.085)$

$k = 6.06 \times 10^4 N/m$

Part b)

Now we know that amplitude decreases by 63% in each cycle

so after one cycle the amplitude will become 37% of initial amplitude

so it is given as

$A = 0.37 A_o$

also we know

$A = A_o e^{-bt/2m}$

$0.37 A_o = A_o e^{-bt/2m}$

$\frac{bt}{2m} = 1$

$b = \frac{2m}{t}$

here t = time period of one oscillation

so it is

$t = 2\pi\sqrt{\frac{m}{k}}$

$t = 2\pi\sqrt{\frac{525}{6.06 \times 10^4}}$

$t = 0.58 s$

now damping constant is

$b = \frac{2(525)}{0.58}$

$b = 1795.4 kg/s$

7 0

3 years ago

how much heat must be added to 30.0 g of solid iron to melt it, assuming the iron is already at it's melting point?​

how much heat must be added to 30.0 g of solid iron to melt it, assuming the iron is already at it's melting point?