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Lesechka [4]
3 years ago
6

How many kilocalories are generated when the brakes are used to bring a 1200-kg car to rest from a speed of 95 km/h ? 1 kcal = 4

186 j?
Physics
1 answer:
timofeeve [1]3 years ago
4 0
<span>Answer: 1st, identify the givens and the unknown - this will give you parameter of what concept and formula are you going to use. Given: m= 1200kg v initial = 95km/hr v final = 0 2nd, focus on the units - in most cases units speak for the concept the unit of the unknown is kcal, thus its the unit of energy or work so, W = ? 3rd, provide the appropriate formula - give formula or equation that the given and the unknown are present since W = delta K.E =delta P.E W= 0.5m( vf^2 - vi^2) ---> best formula 4th, Substitute the given to the formula since 1 Joule = 1Nm 1N = 1kgms^-2 1cal = 4.19 J we express first 95 km/hr to m/s 95km/hr x 1000m/1km x 1hr/3600sec = 26.39 m/sec W= 0.5(1200kg)[(0^2- (26.39m/sec)^2] W=600 kg(0 - 696.43m^2/s^2) W=600kg(-696.43m^2/s^2) W=417859.3Nm or 417859.3 J W = 417859.3 J x 1 cal /4.19 J W = 99,727.7 cal or 99.728 kcal</span>
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Answer:

Explanation:

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impulse = change in momentum

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average magnitude of the force exerted by floor = 1.2 x 10³ N

If R be reaction force by earth

R - mg = 1.2 x 10³

R = 1.2 x 10³ + mg

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2 years ago
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3 years ago
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Why are stars given an absolute magnitude?
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2 years ago
Most automobiles have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made o
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Answer:

There is a loss of fluid in the  container of 0.475L

Explanation:

To solve the problem it is necessary to take into account the concepts related to the change of voumen in a substance depending on the temperature.

The formula that describes this thermal expansion process is given by:

\Delta V = \beta V_0 \Delta T

Where,

\Delta V =Change in volume

V_0 =Initial Volume

\Delta T = Change in temperature

\beta = coefficient of volume expansion (Coefficient of copper and of the liquid for this case)

There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,

\Delta V_T = \Delta V_l - \Delta V_c

Where,

\Delta V_l= Change in the volume of liquid

\Delta V_c= Change in the volume of copper

Then replacing with the previous equation we have:

\Delta V = \beta_l V_0 \Delta T- \beta_c V_0 \Delta T

\Delta V = (\beta_l-\beta_c)V_0\Delta T

Our values are given as,

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\Delta T = (95\°C-10\°C)

Replacing we have that,

\Delta V = (\beta_l-\beta_c)V_0\Delta T

\Delta V = (400*10^{-6}/\°C-51*10^{-6}/\°C)(16L)(95\°C-10\°C)

\Delta V = 0.475L

Therefore there is a loss of fluid in the container of 0.475L

6 0
3 years ago
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