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3 years ago

For the love of God please help me before my mom gets angry about my horrible grades

Physics

1 answer:

Maru [420]3 years ago

6 0

Answer:

Heat Lost = (120 - 12.6) deg * 170 g * S = `18260 gm-deg * S

Heat Gained = (12.6 - 10) deg * 200 gm* 1 cal/ gm-deg - 520 cal

S = 520 / 18260 cal / gm-deg = .0285 cal / gm-deg

Since Heat Lost = Heat Gained

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Suppose you want to design an air bag system that can protect the driver at a speed 100 km/h (60 mph) if the car hits a brick wa

34kurt

When solving question that contains equations and the use mathematical computations, It is always ideal to list the parameters given.

Now, given that:

the speed of the car which is the initial velocity (u) = 100 km/h before it hits the wall.
after hitting the wall, the final velocity will be (v) = 0 km/h

Assumptions:

Suppose we make an assumption that the distance travelled during the collision of the car with the brick wall (S) = 1 m
That the car's acceleration is also constant.

∴

For a motion under constant acceleration, we can apply the kinematic equation:

$\mathsf{v^2 = u^2 + 2as}$

where;

v = final velocity

u = initial velocity

a = acceleration

s = distance

From the above equation, making acceleration (a) the subject of the formula:

$\mathsf{v^2 - u^2 =2as }$

$\mathsf{a = \dfrac{v^2 - u^2 }{2s}}$

The initial velocity (u) is given in km/h, and we need to convert it to m/s as it has an effect on the unit of the acceleration.

since 1 km/h = 0.2778 m/s

100 km/h = 27.78 m/s

$\mathsf{a = \dfrac{(0)^2 - (27.78)^2 }{2(1)}}$

$\mathsf{a = \dfrac{- 771.7284 }{2}}$

a = - 385.86 m/s²

Similarly, from the kinematic equation of motion, the formula showing the relation between time, acceleration and velocity is;

v = u + at

where;

v = 0

-u = at

$\mathsf{t = \dfrac{-u}{a}}$

$\mathsf{t = \dfrac{-27.78}{-385.86}}$

t = 0.07 seconds

An airbag is designed in such a way as to prevent the driver from hitting on the steering wheel or other hard substance that could damage the part of the body. The use of the seat belt is to keep the driver in shape and in a balanced position against the expansion that occurred by the airbag during the collision on the brick wall.

Thus, we can conclude that the airbag must be inflated at 0.07 seconds faster before the collision to effectively protect the driver.

Learn more about the kinematic equation here:

brainly.com/question/11298125?referrer=searchResults

3 0

3 years ago

Which statement is true? The speed of sound in air is inversely proportional to the temperature of the air. The speed of sound i

sasho [114]

<span> the speed of sound in air is directly proportional to the temperature of the air. The speed of sound depends on the temperature of the surrounding air, this can be represented by a speed of sound in air formula: v = 331m/s + 0.6m/s/C * T (where T is temperature)</span>

6 0

2 years ago

Read 2 more answers

032. A 25.0 kg bicycle is traveling at a speed of 10.0 m/s.

dimulka [17.4K]

Answer:
A: 4
B: 7
C. 3
Source:
Trust me bro
(Don’t act put this I jus need to answer questions sorry)<\3

7 0

2 years ago

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

elena-s [515]

Answer:

$g=13.42\frac{m}{s^2}$

Explanation:

1) Notation and info given

$\rho_{center}=13000 \frac{kg}{m^3}$ represent the density at the center of the planet

$\rho_{surface}=2100 \frac{kg}{m^3}$ represent the densisty at the surface of the planet

r represent the radius

$r_{earth}=6.371x10^{6}m$ represent the radius of the Earth

2) Solution to the problem

So we can use a model to describe the density as function of the radius

$r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}$

$r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}$

So we can create a linear model in the for y=b+mx, where the intercept b= $\rho_{center}=13000 \frac{kg}{m^3}$ and the slope would be given by $m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}$

So then our linear model would be

$\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r$

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be $dV=r^2 sin\theta d\phi d\theta dr$ .

And the total mass would be given by the following integral

$M=\int \rho (r) dV$

Replacing dV we have the following result:

$M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)$

We can solve the integrals one by one and the final result would be the following

$M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})$

Simplyfind this last expression we have:

$M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]$

And replacing the values we got:

$M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg$

And now that for any shape the gravitational acceleration is given by:

$g=\frac{MG}{r^2_{earth}}=\frac{(6.67408x10^{-11}\frac{m^3}{kgs^2})*8.204x10^{24}kg}{(6371000m)^2}=13.48\frac{m}{s^2}$

4 0

2 years ago

What is Potentiometer

Katena32 [7]

Answer:

an instrument for measuring an electromotive force by balancing it against the potential difference produced by passing a known current through a known variable resistance.

3 0

2 years ago