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AfilCa [17]
3 years ago
10

What mass of water will change its temperature by 20.0 degrees Celsius when 515 J of heat is added to it? the specific heat of w

ater is 4.18J/g*C.
Physics
1 answer:
andrey2020 [161]3 years ago
4 0
The problem can be solve using the formula:
q = m Cp T
where q is the heat added to the substance
m is the mass of the substance
Cp is the specific of the substance
T is the change in temperature of the subtance

q = m Cp T
515 = m ( 4.18 ) ( 20)
m = 515 / ( 4.18) ( 20 )
m = 6.16 g of water

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Read 2 more answers
To test the performance of its tires, a car
velikii [3]

<u>Answer</u>:

The coefficient of  static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r =  516 m

Tangential Acceleration a_r=  3.89 m/s^2

Speed,v =  32.8 m/s

<u>To Find:</u>

The coefficient of  static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

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

a_{R = \frac{(32.8)^2}{516}

a_{R = \frac{(1075.84)}{516}

a_{R = 2.085 m/s^2

Now the total acceleration is

\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2

=>= \sqrt{ (a_r)^2+(a_R)^2}

=>\sqrt{ (3.89 )^2+( 2.085)^2}

=>\sqrt{ (15.1321)+(4.347)^2}

=>19.4791 m/s^2

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of  static friction is

\mu =\frac{f}{W}

From (1) and (2)

\mu =\frac{ma}{mg}

\mu =\frac{a}{g}

Substituting the values, we get

\mu =\frac{19.4791}{9.8}

\mu =1.987

8 0
3 years ago
(15 points) :^|
mihalych1998 [28]
That is FALSE. The equation to calculate the charges has a distance component that is in the denominator which means that it is inversely proportional (as the distance os greater the force is smaller)
7 0
3 years ago
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