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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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A line of dominos is knocked down. Which explanation is true?
Llana [10]

B: an increase in acceleration caused an increase in force.

This is based on the concept of force on an object.

Now, formula for force is commonly known as;

Force = mass × acceleration

Now, mass and acceleration are the input values that make the output which is the Force to either increase or decrease.

  • Now, for the line of dominoes to fall, it means that the force was so overwhelming that the dominoes couldn't resist it.

Now, this output which is the force became so much as a result of the increase in acceleration of the motion that triggered its movement.

Moreover, the force is directly proportional to the acceleration!

Thus, we can say that an increase in acceleration caused an increase in force.

Read more at; brainly.in/question/25954363

4 0
3 years ago
Read 2 more answers
In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400
liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, f_{ra} = \mu mg\\............(1)

Since frictional force is a type of force, we are safe to say f_{ra} = ma.......(2)

Equating (1) and (2)

ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}

a) Speed of A at the start of the sliding

d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s

b) Speed of B at the start of the sliding

d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s

Let the speed of car B before collision = v_{B1}

Momentum of car B before collision = m_{B} v_{B1}

Momentum after collision = m_{A} v_{A} + m_{B} v_{B2}

Applying the law of conservation of momentum:

m_{B} v_{B1}  = m_{A} v_{A} +m_{B} v_{B2}

m_{A} = 1100 kg\\m_{B} = 1400 kg

(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s

3 0
4 years ago
Read 2 more answers
A 43.9-g piece of copper (CCu= 0.385 J/g°C) at 135.0°C is plunged into 254 g of water at 39.0°C. Assuming that no heat is lost t
Semmy [17]

Answer:

T = 40.501\,^{\textdegree}C

Explanation:

The interaction of the piece of copper and water means that the first one need to transfer heat in order to reach a thermal equilibrium with water. Then:

-Q_{out,Cu} = Q_{in,H_{2}O}

After a quick substitution, the expanded expression is:

-(43.9\,g)\cdot (0.385\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-135^{\textdegree}C) = (254\,g)\cdot (4.187\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-39\,^{\textdegree}C)

-16.902\,\frac{J}{^{\textdegree}C}\cdot (T-135^{\textdegree}C) = 1063.498\,\frac{J}{^{\textdegree}C} \cdot  (T-39^{\textdegree}C)

43758,192\,J = 1080.4\,\frac{J}{^{\textdegree}C}\cdot T

The final temperature of the system is:

T = 40.501\,^{\textdegree}C

8 0
3 years ago
Read 2 more answers
A billiard ball of mass m moving with speed v1=1 m/s collides head-on with a second ball of mass 2m which is at rest. What is th
Soloha48 [4]

Answer:

The velocity of mass 2m is  v_B = 0.67 m/s

Explanation:

From the question w are told that

     The mass of the billiard ball A is =m

     The initial speed  of the billiard ball A = v_1 =1 m/s

    The mass of the billiard ball B is = 2 m

    The initial speed  of the billiard ball  B = 0

Let the final speed  of the billiard ball A  = v_A

Let The finial speed  of the billiard ball  B = v_B

      According to the law of conservation of Energy

                 \frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2

              Substituting values  

                \frac{1}{2} m (1)^2  = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2

Multiplying through by \frac{1}{2}m

                1 =v_A^2 + 2 v_B ^2 ---(1)

    According to the law of conservation of Momentum

            mv_1 + 2m(0) = mv_A + 2m v_B

    Substituting values

            m(1)  = mv_A + 2mv_B

Multiplying through by m

           1 = v_A + 2v_B ---(2)

making v_A subject of the equation 2

            v_A = 1 - 2v_B

Substituting this into equation 1

         (1 -2v_B)^2 + 2v_B^2 = 1

         1 - 4v_B + 4v_B^2 + 2v_B^2 =1

          6v_B^2  -4v_B +1 =1

          6v_B^2 -4v_B =0

Multiplying through by \frac{1}{v_B}

          6v_B -4 = 0

            v_B = \frac{4}{6}

            v_B = 0.67 m/s

4 0
3 years ago
The drawing shows a hydraulic chamber with a spring (spring constant = 1600 N/m) attached to the input piston and a rock of mass
Triss [41]

Answer:

\Delta x=245\ mm

Explanation:

Given:

  • spring constant of the spring attached to the input piston, k=1600\ N.m^{-1}
  • mass subjected to the output plunger, m=40\ kg

<u>Now, the force due to the mass:</u>

F=m.g

F=40\times 9.8

F=392\ N

<u>Compression in Spring:</u>

\Delta x=\frac{F}{k}

\Delta x=\frac{392}{1600}

\Delta x=0.245\ m

or

\Delta x=245\ mm

8 0
3 years ago
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