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Rufina [12.5K]
4 years ago
5

WheN a car is driven some distance the air pressure in the tyre increases,why?

Physics
1 answer:
KatRina [158]4 years ago
8 0
Over the course of a long drive, as a result of alternating between braking and acceleration and of course steering, tires get heated up. When this happens, the air within the tires gets heated up, causing the air within the tire to expand. This expanding air exerts force on the tire from inside thus increasing tire pressure.
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What is the velocity in an object
sladkih [1.3K]
How far it moves over a certain period of time, but unlike the term "speed" velocity is a vector quantity and tells you the direction of an object (speed is also distance over time just without a direction). This means an object can have negative velocity, but it can't have negative speed.
3 0
3 years ago
A car leaves an intersection traveling west. Its position 4 sec later is 21 ft from the intersection. At the same time, another
Alex_Xolod [135]

Answer:

15.8 ft/s

Explanation:

\frac{da}{dt} = Velocity of car A = 9 ft/s

a = Distance car A travels = 21 ft

\frac{db}{dt} = Velocity of car B = 13 ft/s

b = Distance car B travels = ft

c = Distance between A and B after 4 seconds = √(a²+b²) = √(21²+28²) = √1225 ft

From Pythagoras theorem

a²+b² = c²

Now, differentiating with respect to time

2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}\\\Rightarrow a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}\\\Rightarrow \frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}\\\Rightarrow \frac{dc}{dt}=\frac{21\times 9+28\times 13}{\sqrt{1225}}\\\Rightarrow \frac{dc}{dt}=15.8\ ft/s

∴ Rate at which distance between the cars is increasing three hours later is 15.8 ft/s

6 0
4 years ago
How many newtons does a 5-kg backpack weigh on Earth?
zvonat [6]
A 5kg backpack will weigh 49 newtons on earth
7 0
3 years ago
Joey, whose mass is m = 36 kg, stands at rest at the outer edge of a frictionless merry-go-round with the mass M = 300 kg and th
Evgen [1.6K]

Answer:

\omega=0.24\ rad.s^{-1}

Explanation:

Given:

mass of person, m=36\ kg

mass of merry go-round, M=300\ kg

radius of merry go-round, R=2\ m

velocity of the person running, v=4\ m.s^{-1}

<u>We consider merry go-round as a ring:</u>

Now the moment of inertial of the ring is given as,

I=M.R^2

I=300\times 2^2

I=1200\ kg.m^{-2}

<u>Moment of inertia of the person considering as a point mass:</u>

I_p=m.R^2

I_p=36\times 2^2

I_p=144\ kg.m^2

<u>Now according to the conservation of angular momentum:</u>

I.\omega=I_p.\omega_p

where:

\omega = angular velocity of the merry-go-round

\omega_p= angular velocity of the person running

1200\times \omega=144\times \frac{v}{R}

\omega=\frac{144}{1200} \times \frac{4}{2}

\omega=0.24\ rad.s^{-1}

4 0
4 years ago
Read 2 more answers
You illuminate a slit with a width of 70.3 μm with a light of wavelength 719 nm and observe the resulting diffraction pattern on
Zielflug [23.3K]

Answer:

4.3 cm

Explanation:

We are given that

Width,d=70.3\mu m=70.3\times10^{-6} m

1\mu m=10^{-6} m

Wavelength,\lambda=719 nm=719\times 10^{-9} m

1nm=10^{-9} m

r=2.11 m

We have to find the width in cm of the pattern.

The angle for the first minimum m=1

sin\theta_{min}\approx \theta=\frac{\lambda}{d}=\frac{719\times 10^{-9}}{70.3\times 10^{-6}}=0.0102 rad

y=r\theta=2.11\times 0.0102=0.0215 m

The width of the pattern=2y=2\times 0.0215=0.043 m=0.043\times 100=4.3 cm

8 0
3 years ago
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