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2 years ago

The gauge pressure in a car tyre is 300 kPa at 20°C. After moving at high speed, the tyre has

Physics

2 answers:

Makovka662 [10]2 years ago

7 0

Answer:

49.3°C

Explanation:

Since the tire was inflated with air after been gauged. It means the volume of air in the tire stays the same.

At constant volume, Charles law applies.

Charles law states pressure is proportional to temperature.

Hence P1/T1=P2/T2

The formula uses Kelvin for temperature and to convert temperature from degree Celsius to Kelvin we add 273k

T2 = P2 ×T1/P1. =330 × 293/300= 322.3K

322.3-273= 49.3°C

Leno4ka [110]2 years ago

4 0

Answer:

229kPa

Explanation:

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Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the ball's final speed, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

2 years ago

A string has its 4th harmonic at 31.5 Hz. What is the fundamental frequency?

seropon [69]

Given data

*The given 4th harmonic frequency is 31.5 Hz

The fundamental frequency is calculated as

$\begin{gathered} f_n=\frac{31.5}{4} \\ =7.875\text{ Hz} \end{gathered}$

Hence, the fundamental frequency is 7.875 Hz

5 0

9 months ago

Alli was in the park playing on the equipment. she noticed that on the highest slide she slides down

LUCKY_DIMON [66]

Answer:

Is this answer complete????

3 0

2 years ago

Read 2 more answers

A double insulated drill ____. A. has two ground wires to insure proper grounding B. generally has a plastic case with the elect

Len [333]

Answer:

Option B is correct.

A double insulated drill generally has a plastic case with the electrical connections and motor insulated within the tool

Explanation:

A double insulated drill's design in real life has plastic casing as the insulator & it wouldn't be a drill without its motor now, would it!?

8 0

2 years ago

Three point charges are arranged along the x-axis. Charge q1 = +3.00 uC is at the origin, and charge q2= -5.00 uC is at x= 0.200

artcher [175]

We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2

F(1) = F (2on1) + F (3on1)

F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N

Since F1 is 7N,

F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)

Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N

F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m

So it's located in -.144m

Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.

6 0

2 years ago

Read 2 more answers