SCIENCE HELP! PLEASE HELP URGENT!

the specific heat of water is 4.2 j/c. if it takes 31,500 joules to heat to warm 750 g of water, what was the temperature change

defon

The amount of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q=mC_s \Delta T$
where
m is the mass of the substance
$C_s$ the specific heat capacity
$\Delta T$ the increase in temperature

In our problem, the mass of the water is m=750 g, the specific heat is $C_s = 4.2 J/g ^{\circ}C$ and the amount of heat supplied is $Q=31500 J$ , so if we re-arrange the previous formula we find the increase in temperature of the water:
$\Delta T= \frac{Q}{m C_s}= \frac{31500 J}{(750 g)(4.20 J/g^{\circ} C)}=10^{\circ}C$

7 0

2 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

2 years ago

Think about the way optical fibers are designed. why are the ends of the fibers glowing very brightly while the sides of the fib

Effectus [21]

Answer:

D. because the light is reflected back into the fiber along its sides

Explanation:

The fiber is constructed in a way that the light is bent/reflected/refracted toward the center core of glass. So, from the center core, there is a layer above it that has a different propagation than the core, and above that the same thing. To give you a real world visual example, if you look down in a pool of water, then stick a straight stick into it, you see that the straight stick appears to bend. That is what is happening to the light as it travels through a different medium (air to water). This same effect is incorporated in the fiber optic cable construction.

6 0

2 years ago

A 2:2 kg toy train is con ned to roll along a straight, frictionless track parallel to the x-axis. The train starts at the origi

Liula [17]

Answer:

a) 10.51 J

b) 3.48 m/s

Explanation:

Given data :

mass of train ( M ) = 2.2 kg

Given initial velocity ( u ) = 1.6 m/s

<u>a) calculating work done by the force over the journey of the train</u>

F = mx + b ------ ( 1 )

m = slope = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m

x = distance travelled on the x axis by the train = 7.5 m

F = force experienced by the train = 2.8 N

x = 0

∴ b = 2.8

hence equation 1 can be written as

F = ( -0.373) x + 2.8 ----- ( 2 )

hence to determine the work done by the force

W = $\int\limits^7_0 { ( -0.373) x + 2.8 )} \, dx$ Note: the limits are actually 7.5 and 0

∴ W ( work done ) = -10.49 + 21 = 10.51 J

<u>b) calculate the speed of the train at the end of its journey</u>

we will apply the work energy theorem

W = 1/2 m*v^2 - 1/2 m*u^2

∴ V^2 = 2 / M ( W + 1/2 M*u^2 ) ( input values into equation )

V^2 = 12.11

hence V = 3.48 m/s

6 0

3 years ago

What practical function is provided by the ionosphere? Produces auroras charges atoms heats oxygen and nitrogen permits worldwid

Karolina [17]

Answer:

Worldwide Radio Communication

Explanation:

The ionosphere is important because it is through the ionosphere that world wide radio communication is possible.

6 0

3 years ago