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ollegr [7]
3 years ago
15

Someone please help me, I’m so lost my brain hurts

Physics
2 answers:
tiny-mole [99]3 years ago
7 0

-- It takes 100 calories of heat to make 10 grams of the stuff 20° warmer.

How much of the heat warms each gram ?

-- It takes 10 calories of heat to make each gram of the stuff 20° warmer.

How much of the heat warms that gram each degree ?

-- It takes 1/2 calorie of heat to make each gram of the stuff 1° warmer.

The specific heat of that stuff is

(1/2 calorie) per gram per °C.

That's choice-3 .

maw [93]3 years ago
5 0

Explanation:

The answer is No. 3

0.5 cal/g°C

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A 13561 N car traveling at 51.1 km/h rounds
Minchanka [31]

Answer:

a) The centripetal acceleration of the car is 0.68 m/s²

b) The force that maintains circular motion is 940.03 N.

c) The minimum coefficient of static friction between the tires and the road is 0.069.

Explanation:

a) The centripetal acceleration of the car can be found using the following equation:

a_{c} = \frac{v^{2}}{r}

Where:

v: is the velocity of the car = 51.1 km/h

r: is the radius = 2.95x10² m

a_{c} = \frac{(51.1 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2.95 \cdot 10^{2} m} = 0.68 m/s^{2}

Hence, the centripetal acceleration of the car is 0.68 m/s².

b) The force that maintains circular motion is the centripetal force:

F_{c} = ma_{c}

Where:

m: is the mass of the car

The mass is given by:

P = m*g

Where P is the weight of the car = 13561 N

m = \frac{P}{g} = \frac{13561 N}{9.81 m/s^{2}} = 1382.4 kg

Now, the centripetal force is:

F_{c} = ma_{c} = 1382.4 kg*0.68 m/s^{2} = 940.03 N

Then, the force that maintains circular motion is 940.03 N.

c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:

F_{c} = F_{\mu}

F_{c} = \mu N = \mu P

\mu = \frac{F_{c}}{P} = \frac{940.03 N}{13561 N} = 0.069

Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.

I hope it helps you!                

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