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2 years ago

Why do balloons make noise when they pop?

2 answers:

6 0

The pressure of the air inside the balloon is higher than outside, and when you pop the balloon, the high pressure air expands quickly and makes a popping sound.

Ray Of Light [21]2 years ago

3 0

The pressure inside the balloon

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A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

$N=mg$

Therefore, frictional force, $f=\mu mg$

Now, frictional force = centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833$

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0

3 years ago

If the average velocity during the athlete's walk back

goblinko [34]

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

$T=\frac{d}{1.50}$

From the question we are told

If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,

Generally the equation Time spent is mathematically given as

T=\frac{d}{v}

Therefore

$T=\frac{d}{1.50}$

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

$T=\frac{d}{1.50}$

For more information on this visit

brainly.com/question/22271063

4 0

2 years ago

A heat engine does 200 j of work per cycle while exhausting 600 j of heat to the cold reservoir. what is the engine's thermal ef

AveGali [126]

The thermal efficiency of an engine is
$\eta= \frac{W}{Q}$
where
W is the work done by the engine
Q is the heat absorbed by the engine to do the work

In this problem, the work done by the engine is W=200 J, while the heat exhausted is Q=600 J, so the efficiency of the machine is
$\eta= \frac{W}{Q}= \frac{200 J}{600 J}=0.33 = 33 \%$

8 0

3 years ago

What is the change in internal energy if 60J of heat are released from a system and 20J of work is done on the system? Use U=Q-W

Ganezh [65]

D because 60-20= 40J

5 0

3 years ago

a college student produces about 100 kcal of heat per hour on the average what is the rate of energy production and joules

Bond [772]

Given:

Amount of heat produced = 100 kcal per hour

Let's find the rate of energy production in joules.

We know that:

1 calorie = 4.184 Joules

1 kcal = 4.184 Joules

To find the rate of energy production in Joules, we have:

$\begin{gathered} Rate=100\ast4.184 \\ \\ \text{Rate}=418.4\text{ KJ/hour} \end{gathered}$

Therefore, the rate of energy production in joules is 418.4 kJ/h which is equivalent to 418400 Joules

ANSWER:

418.4 kJ/h

6 0

10 months ago