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AVprozaik [17]
3 years ago
13

A car of 900 kg mass is moving at the velocity of 60 km/hr. It is brought into rest at 50 meter distance by applying a brake. No

w, calculate the force required to stop the car.
Physics
1 answer:
kozerog [31]3 years ago
8 0

Answer: -2502N

Explanation:

(V_2)^2=(V_1)^2+2ad

where;

V_2 = final velocity = 0

V_1 = initial velocity = 60 km/h = 16.67 m/s

a = acceleration

d = distance

First all of, because acceleration is given in m/s and not km/h, you need to convert 60km/h to m/s. Our conversion factors here are 1km = 1000m and 1h = 3600s

60km/h(\frac{1000m}{1km} )(\frac{1h}{3600s} )=16.67m/s

Solve for a;

(V_2)^2=(V_1)^2+2ad

Begin by subtracting (V_1)^2

(V_2)^2-(V_1)^2=2ad

Divide by 2d

\frac{(V_2)^2-(V_1)^2}{2d} =a

Now plug in your values:

a=\frac{(0)^2-(16.67 m/s)^2}{2(50m)}

a=\frac{0-277.89m^2/s^2}{100m}

a=-2.78m/s

If you're wondering why I calculated acceleration first is because in order to find force, we need 2 things: mass and acceleration.

F=ma

m = mass = 900kg

a = acceleration = -2.78m/s

F=(900kg)(-2.78m/s)\\F=-2502N

It's negative because the force has to be applied in the opposite direction that the car is moving.

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Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

  • Length of the massless beam;L = 4.00m
  • Distance of support from the left end; x = 3.00m
  • First mass; m1 = 31.3 kg
  • Distance of beam from  the left end( m₁ is attached to ); x_1 = ?
  • Second mass; m_2 = 61.7 kg
  • Distance of beam from  the right of the support( m₂ is attached to ); x_1 = 0.273m

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, m_1g( x-x_1) = m_2gx_2

we divide both sides by g

m_1( x-x_1) = m_2x_2

Next, we make x_1, the subject of the formula

x_1 = x - [ \frac{m_2x_2}{m_1} ]

We substitute in our given values

x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]

x_1 = 3.00m - 0.538m

x_1 = 2.46m

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

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3 years ago
What is the kinetic energy in Joules of a 1,500kg car traveling at 75mph?What is the kinetic energy in Joules of a 1,500kg car t
Sergio039 [100]

Answer: 2812500 joules

Explanation:

Mass of car = 1500kg

Velocity of car = 75mph

Kinetic energy = ?

Recall that kinetic energy is the energy possessed by a moving object, and it depends on its mass M and velocity, V

Thus, Kinetic energy = 1/2 x mv^2

= 1/2 x 1000kg x (75mph)^2

= 0.5 x 1000kg x (75mph)^2

= 500 x 5625

= 2812500 joules

Thus, the car travels with a kinetic energy of 2812500 joules

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If the force of attraction (gravity) on the moon is 1/6 that of the force on Earth, what would
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I believe that you would weigh around 68 or 69 N, or 7 kilograms.

4 0
2 years ago
When a pendulum is at the midpoint of its oscillation, hanging straight down, which statement is true?
Svetach [21]
When a pendulum is at the midpoint of its oscillation, hanging straight down ...

-- that's the fastest it's going to swing, so its kinetic energy is maximum;
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3 years ago
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Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C.
salantis [7]

Explanation:

Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

u_{1} = u_{g} = 2553.6 kJ/kg

v_{1} = v_{g} = 0.4625 m^{3}/kg

At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, v_{2} = v_{g} = 0.4625 m^{3}/kg

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at v_{2} = v_{g} = 0.4625 m^{3}/kg and temperature T_{2} = 360^{o}C so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

 u_{2} = u_{\text{at 5 bar, 400^{o}C}} + (\frac{v_{2} - v_{\text{at 5 bar, 400^{o}C}}}{v_{\text{at 7 bar, 400^{o}C - v_{at 5 bar, 400^{o}C}}}})(u_{at 7 bar, 400^{o}C - u_{at 5 bar, 400^{o}C}})

     u_{2} = 2963.2 + (\frac{0.4625 - 0.6173}{0.4397 - 0.6173})(2960.9 - 2963.2)

                   = 2963.2 - 2.005

                   = 2961.195 kJ/kg

Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

      Q - W = \Delta U + \Delta K.E + \Delta P.E ......... (1)

Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

            \Delta K.E = \Delta P.E = 0

Now, equation will be as follows.

           Q - W = \Delta U + \Delta K.E + \Delta P.E

           Q - 0 = \Delta U + 0 + 0

           Q = \Delta U

Now, we will obtain the heat transfer per unit mass as follows.

          \frac{Q}{m} = \Delta u

         \frac{Q}{m} = u_{2} - u_{1}

                      = (2961.195 - 2553.6)

                      = 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.

4 0
3 years ago
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