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Blababa [14]
4 years ago
10

A 2,537-kg truck moving at 14 m/s strikes a car waiting at a traffic light, hooking bumpers. The two continue to move together a

t 8 m/s. What was the mass (in kg) of the struck car
Physics
1 answer:
slavikrds [6]4 years ago
5 0

Answer:

1902.75 kg

Explanation:

From Law of conservation of momentum,

m₁u₁ + m₂u₂ = V (m₁ + m₂).................... Equation 1

make m₂ the subject of the equation,

m₂ = (m₁V - m₁u₁)/(u₂-V)..................... Equation 2

Where m₁ = mass of the truck, m₂ = mass of the car, u₁ initial velocity of the truck, u₂ = initial velocity of the car V = common velocity

Given: m₁ = 2537 kg, u₁ = 14, V= 8 m/s, u₂ = 0 m/s ( as the car was at rest waiting at a traffic light)

Substituting into equation 2.

m₂ =[2537(8) - 2537(14)]/(0-8)

m₂ = (20296-35518)/-8

m₂ = -15222/-8

m₂ = 1902.75 kg.

Thus the mass of the car = 1902.75 kg

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Two planets both have the same mass, but different sizes. Planet Bluper has 3 times the radius as planet Ruper. If your weight o
Kay [80]

Answer:

72.8

Explanation:

8 0
3 years ago
Read 2 more answers
Compare and contrast the electric force with the gravitational force
Olenka [21]

Similarities:

- The gravitational force and the electric force have a similar form.

In fact, gravitational force is given by:

F=G\frac{m_1 m_2}{r^2}

where G is the gravitational constant, m1 and m2 are the masses of the two objects, r is the separation between the two objects.

The electric force is given by:

F=k\frac{q_1 q_2}{r^2}

where k is the Coulomb constant, q1 and q2 are the charges of the two objects, r is the separation between the two objects.

As we see, the two forces have same dependency on the distance (inversely proportional to the square of the distance)

- Both forces are non-contact forces: this means that in both cases, the two objects do not to be in contact with each other, because the force is still acting from a distance also.

Differencies:

- The gravitational force is attractive only; on the contrary, the electric force can be either attractive or repulsive, depending on the relative sign of the two charges. In fact, when the two charges have same sign, the force is repulsive; when the two charges have opposite sign, the force is attractive.

- The gravitational force is much weaker than the electric force. This can be seen by comparing the value of the constant in the two formulas; we have:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} for the gravitational force

k=8.99 \cdot 10^9 Nm^2C^{-2} for the electric force

As we see, the electric force is much stronger than the gravitational force.

6 0
3 years ago
A system of mass 13 kg undergoes a process during which there is no work, the elevation decreases by 50 m, and the velocity incr
marin [14]

Answer:

The change in kinetic energy is 4.3875 kJ

The amount of energy transferred by heat for the process is -66.98 kJ

Explanation:

Given;

mass of the system, m = 13 kg

change in height, Δh = -50 m

initial velocity, u = 15 m/s

final velocity, v = 30 m/s

change in internal energy per mass, ΔU = -5 kJ/kg

The change in kinetic energy is given by;

ΔK.E = K.E₂ - K.E₁

ΔK.E = ¹/₂mv² - ¹/₂mu²

ΔK.E = ¹/₂m(v² - u²)

ΔK.E = ¹/₂ ₓ 13 (30² - 15²)

ΔK.E = 4387.5 J

ΔK.E = 4.3875 kJ

The amount of energy transferred by heat for the process;

Q = W + ΔP.E + ΔK.E + ΔU

Where;

ΔP.E = mgΔh

ΔP.E = 13 x 9.8 x (-50)

ΔP.E = -6370 J = -6.37 kJ

W = 0

ΔU = -5kJ/kg x 13kg

ΔU = -65 kJ

Q = W + ΔP.E + ΔK.E + ΔU

Q = 0 + (-6.37) + (4.3875) + (-65)

Q = -66.98 kJ

7 0
4 years ago
A thin, uniformly charged spherical shell has a potential of 832 V on its surface. Outside the sphere, at a radial distance of 2
Zolol [24]

Answer:

a

The radius is   r_1 = 0.315m            

b

The total  charge is   Q= 2.912*10^{-8}C

c

The electric potential inside sphere is the same with that outside the sphere which implies that the electric potential is 832 V

d

The magnitude of the electric field is  E= 2641.3 V/m

e

The velocity is   v= 1.76 *10^{14} m/s

Explanation:

From the question we are told that

     The potential is V_1 = 832 V

       The radial distance from the sphere is d = 21.0cm = \frac{21}{100} = 0.21m

       The potential at the radial distance is V_2 = 499V

The potential at the surface of the sphere is mathematically represented as

                     V = \frac{kQ}{r}

                    Vr = kQ

Where kQ is  a constant what this means that the the charge Q and the coulomb constant do  not change

  This means that

              V_1 r_1 = V_2 r_2

Where r_1 is the radius of the sphere

     and r_2 is the distance  from that point where the second potential was measured to the center of the sphere which is mathematically represented as

             r_2 = r_1 + d

Substituting  this into the equation

                      v_1 r_1 = V_2 (r_1 +d)

 Now substituting value

                   832 * r_1 = 499 * (r_1 + 0.21)

                   832r_1 - 499r_1 = 104.79

                   333r_1 = 104.79

                       r_1 = \frac{104.79}{333}

                           r_1 = 0.315m              

From the equation above

          V = \frac{kQ}{r_1}

making Q the subject

        Q = \frac{V r_1 }{k}

k has a values of k = 9*10^9 \ kg\cdot m^3 \cdot s^{-4} \cdot A^{-2}

       Substituting into the equation

            Q =\frac{832 * 0.315}{9*10^9}

               Q= 2.912*10^{-8}C

According to  Gauss law  the electric field from  outside a sphere is taken to be an electric field from a point charge this mean that the potential outside a sphere is also taken as electric potential inside a sphere

The magnitude of a electric field from a sphere (point charge ) is mathematically represented as

                  E = \frac{kQ}{r_1^2}

Substituting values

                 E = \frac{9*10^{9} * 2.912*10^{-8}}{0.315^2}

                     E= 2641.3 V/m

 According the the law of energy conservation

  The electric potential energy at the point outside the surface where the second potential was measured(21 cm from the sphere surface) = The electric potential energy at the surface + The kinetic energy of the charge (electron) at that the surface

Generally Electric potential energy is mathematically represented as

         EPE = V * e

Where is e is an electron

And Kinetic energy is mathematically represented as

         KE = \frac{1}{2} m v^2

From the statement above

          V_2 e = V_1 e + \frac{mv^2}{2}

But from the question we can deduce that the potential at the surface is zero

So the equation becomes

            V_2 e = \frac{mv^2}{2}

The charge an electron has a value  e = 1.602*10^{-19}C

And the mass of an electron is m = 9.109 *10^{-31}kg

     Making v the subject

       v = \sqrt{\frac{2V_2 e}{m} }

Substituting value

      v = \sqrt{\frac{2 * 499 * 1.602 *10^{-19}}{9.109*10^{-31}} }

         v= 1.76 *10^{14} m/s

           

7 0
3 years ago
I’m still lost on this please help me on this I know it’s not D I got it wrong
RSB [31]

Before you even look at the questions, look over the graph, so you know what kind of information is there.

The x-axis is "time".  OK.  You know that as the graph moves from left to right, it shows what's happening as time goes on.

The y-axis is "speed" of something.  OK.  When the graph is high, the thing is moving fast.  When the graph is low, the thing is moving slow.  When the graph slopes up, the thing is gaining speed.  When the graph slopes down, the thing is slowing down.  When the graph is flat, the speed isn't changing, so the thing is moving at a constant speed.

NOW you can look at the questions.

OMG !  It's only ONE question:  What's happening from 'c' to 'd' ?  Well I don't know.  Perhaps we can figure it out if we LOOK AT THE GRAPH !

-- Between c and d, the graph is flat.  The speed is not changing.  It's the same speed at d as it was back at c .

What speed is it ?

-- Look back at the y-axis.  The speed at the height of c and d is 'zero' .

-- The 2nd and 4th choices are both correct.  From c to d, <em>the speed is constant</em>.  The constant speed is zero.  <em>The car is not moving</em>.

5 0
3 years ago
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