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3 years ago

10

A blacksmith heats a 35 g lump of iron from room temperature until it glows (2425 degrees C) to work it. If the specific heat of

iron is 0.450 j/g degrees C, how much thermal energy has it gained?

1 answer:

Drupady [299]3 years ago

6 0

Answer:

37.9 kJ

Explanation:

We can calculate the thermal energy gained by the iron using the formula:

$Q=m C_s \Delta T$

where

m = 35 g is the mass of the iron

Cs = 0.450 j/g is the iron's specific heat capacity

$\Delta T= 2425 C - 20 C = 2405 C$ is the change in temperature of the iron (assuming that the room's temperature is 20 C degrees)

Substituting numbers into the formula, we find

$Q=(35 g)(0.450 J/g)(2405 C)=3.79\cdot 10^4 J=37.9 kJ$

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You could use newton’s second law to calculate the force applied to an object if you knew the objects mass and its _____.

olga2289 [7]

Answer:

You could use newton’s second law to calculate the force applied to an object if you knew the objects mass and its <u>acceleration.</u>

Explanation:

By, Newtons second law, the force applied on an object directly varies with the acceleration caused and the mass of the object.

This is given by :

$F=m\ a$

Where $F$ represents force applied on the object , $m$ represents mass of the object and $a$ represents the acceleration.

In order to calculate force applied on object we require the mass of the object and its acceleration. The force can be calculated by finding the product of mass and acceleration of the object.

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2.Cars were previously manufactured to be as sturdy as possible, whereas today's cars

Marina CMI [18]

Answer:

Crumple zones are designed to absorb and redistribute the force of a collision. ... Also known as a crush zone, crumple zones are areas of a vehicle that are designed to deform and crumple in a collision. This absorbs some of the energy of the impact, preventing it from being transmitted to the occupants.

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3 years ago

A nickel (5 cent coin) has a mass of 5.0 g. How many nickels are there in a stack of nickels with a mass of 10.0 kg?

Marrrta [24]

Answer:

2000 nickels

Explanation:

One way to solve proportionality problems, direct and inverse: the simple 3 rule.

If the relationship between the magnitudes is direct (when one magnitude increases so does the other), the simple direct rule of three must be applied.

On the contrary, if the relationship between the magnitudes is inverse (when one magnitude increases the other decreases) the rule of three simple inverse applies.

The simple 3 rule is an operation that helps us quickly solve proportionality problems, both direct and inverse.

To make a simple rule of three we need 3 data: two magnitudes proportional to each other, and a third magnitude. From these, we will find out the fourth term of proportionality.

In the simple three rule, therefore, the proportionality relationship between two known values A and B is established, and knowing a third value C, a fourth value D is calculated.

A -> B

C -> D

Calculation

1 nickel --> 5 g

X? nickel --> 10000g

X = (10000 g * 1 nickel) / 5 g

X = 2000 nickels

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3 years ago

A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas

Radda [10]

Answer:

change in momentum, $\Delta p=7.65 \,kg.m.s^{-1}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

Average Force, $F=144.3396\,N$

$F_x=125.0018\,N$

$F_y=72.1698\,N$

Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$ .....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$ .........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

$F_x=125.0018\,N$

&

$F_y=144.3396\times sin 30^{\circ}$

$F_y=72.1698\,N$

6 0

3 years ago

Given a second class lever with a distance of 5.00 feet from the fulcrum to the effort and a distance of 33.0 inches from the re

leva [86]

Answer:

The correct answer is C. 45.5 lbs.

Explanation:

In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.

The formula for any problem involving a lever is:

$F_ed_e=F_ld_l$

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.

The parameter of the formula that you need is F_l:

$F_l=\frac{F_ed_e}{d_l}$

The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.

$F_l=\frac{25*60}{33}$

F_l=45.5 lbs

7 0

4 years ago