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3 years ago

Which value is represented by the slope of the line?

Physics

1 answer:

netineya [11]3 years ago

5 0

The slope of the line is

(change in ' y ' between the ends) / (change in ' x ' between the ends)

Slope = (630g - 0) / (70 cm^3 - 0)

Slope = (630 / 70) g/cm^3

<em>Slope = 9.0 g/cm^3</em>

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i need help please. this is for physics but everything i search for related to this comes up as chemistry

Annette [7]

The car tyre contains air initially at a pressure of 195 kPa after travelling several km the temperature of the air inside a car tyre rises from 30 to 70°C if the tyre is rigid and does not expand then the new pressure inside the tyre would be 220.74 kPa.

<h3>What is pressure?</h3>

The total applied force per unit of area is known as the pressure.

The pressure depends both on externally applied force as well the area on which it is applied.

The mathematical expression for the pressure

Pressure = Force /Area

the pressure is expressed by the unit pascal or N /m²

By using the Charles law for gases which states that the volume of the gas remains constant then the pressure of the gas is directly proportional to the temperature.

As given in the problem the tyre is rigid and does not expand this means the volume of the tyre remains constant.

The mathematical expression for Charles's law is as follows

P₁/P₂ = T₁/T₂

First, we have to change the temperature from degree Celcius to the kelvin temperature scale

K = 273 + C

where k is the temperature in kelvin and the C is degrees of Celcius

Initially, the temperature was 30° C

T₁ = 273 + 30

T₁ = 303 K

Then after travelling the temperature of the air inside a car tyre rises from 30 to 70°C

T₂= 273+ 70

T₂ =343 K

The car tyre contains air initially at a pressure of 195 KPa

P₁ = 195 kPa

Lets us take the final pressure of the air would be P₂

By substituting the values in the formula

P₁/P₂ = T₁/T₂

195/P₂ = 303/343

P₂ = 220.74 kPa

Thus, the new pressure inside the tyre would be 220.74 kPa.

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7 0

2 years ago

A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T

Nady [450]

The first collision because a greater amount of momentum must be taken and used in order to push the cart back, giving it a greater mass and impulse

6 0

4 years ago

A guitar string produces 3 beats/s when sounded with a 352-hz tuning fork and 8 beats/s when sounded with a 357-hz tuning fork.

MA_775_DIABLO [31]

Explanation:

The fluctuating sound heard when two objects vibrate with different frequencies is called beats. It is given that guitar string produces 3 beats/s when sounded with a 352 Hz tuning fork and 8 beats/s when sounded with a 357 Hz tuning fork.

It is assumed to find the vibrational frequency of the string.

For 3 beats/s, beat frequency can be :

352 - 3 or 352 + 3 = 349 Hz or 355 Hz

For 8 beats/s, beat frequency can be :

357 - 8 or 357 + 8 = 349 Hz or 365 Hz

It means that the vibrational frequency is 349 Hz.

5 0

4 years ago

If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical

alexdok [17]

Answer:

The critical stress required for the propagation of an initial crack $\sigma_{c}$ = 21.84 M pa

Explanation:

Given data

Modulus of elasticity E = 225 × $10^{9}$ $\frac{N}{m^{2} }$

Specific surface energy for magnesium oxide is $\gamma_{s}$ = 1 $\frac{J}{m^{2} }$

Crack length (a) = 0.3 mm = 0.0003 m

Critical stress is given by $\sigma_{c}^{2} }$ = $\frac{2 E \gamma}{\pi a}$ -------- (1)

⇒ 2 E $\gamma_{s}$ = 2 × 225 × $10^{9}$ × 1 = 450 × $10^{9}$

⇒ $\pi$ a = 3.14 × 0.0003 = 0.000942

⇒ Put these values in equation 1 we get

⇒ $\sigma_{c}^{2} }$ = $\frac{450 }{0.000942} 10^{9}$

⇒ $\sigma_{c}^{2} }$ = 4.77 × $10^{14}$

⇒ $\sigma_{c}$ = 2.184 × $10^{7}$ $\frac{N}{m^{2} }$

⇒ $\sigma_{c}$ = 21.84 $\frac{N}{mm^{2} }$

⇒ $\sigma_{c}$ = 21.84 M pa

This is the critical stress required for the propagation of an initial crack.

4 0

3 years ago

The inventor of a new machine claims that its actual mechanical advantage is 4. Literature on the machine reports its input dist

Andreas93 [3]

To solve this we are going to use the formula for ideal mechanical advantage: $IMA= \frac{D_{I} }{D_{O}}$
where
$IMA$ is the machine mechanical advantage
$D_{I}$ is the input distance
$D_{O}$ is the output distance

We know for our problem that $D_{I}=81$ and $D_{O}=27$ . Lets replace those values in our formula to find $IMA$ :
$IMA= \frac{D_{I} }{D_{O}}$
$IMA= \frac{81}{27}$
$IMA=3$

The ideal machine advantage of the machine is 3. The inventor is claiming that the actual mechanical advantage of the machine is 4. Since the actual mechanical advantage takes into account energy losses, it is always less than the ideal mechanical advantage. We can conclude that the developer's claim is false.

8 0

3 years ago