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3 years ago

15

How much work must be done to stop a 1075-kg car traveling at 115 km/h ? Express your answer to two significant figures and incl

ude the appropriate units

1 answer:

jok3333 [9.3K]3 years ago

6 0

Answer:

Work done on the car will be 548440.9463 J

Explanation:

We have given mass of the car m = 1075 kg

As the car finally stops so final velocity of the car $v_f=0m/sec$

Initial velocity of car u = 115 km/hr $=115\times \frac{5}{18}=31.943m/sec$

Work done is equal to change in kinetic energy

So work done $=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$

= $=\frac{1}{2}m(v_f^2-v_i^2)=\frac{1}{2}\times 1075\times (0^2-31.943^2)=-548440.946J$

As the work done is negative so work is done on the car to stop it .

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Sphere A of mass 0.600 kg is initially moving to the right at 4.00 m/s. sphere B, of mass 1.80 kg is initially to the right of s

anzhelika [568]

A) The velocity of sphere A after the collision is 1.00 m/s to the right

B) The collision is elastic

C) The velocity of sphere C is 2.68 m/s at a direction of $-5.2^{\circ}$

D) The impulse exerted on C is 4.29 kg m/s at a direction of $-5.2^{\circ}$

E) The collision is inelastic

F) The velocity of the center of mass of the system is 4.00 m/s to the right

Explanation:

A)

We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:

$p_i = p_f\\m_A u_A + m_B u_B = m_A v_A + m_B v_B$

$m_A = 0.600 kg$ is the mass of sphere A

$u_A = 4.00 m/s$ is the initial velocity of the sphere A (taking the right as positive direction)

$v_A$ is the final velocity of sphere A

$m_B = 1.80 kg$ is the mass of sphere B

$u_B = 2.00 m/s$ is the initial velocity of the sphere B

$v_B = 3.00 m/s$ is the final velocity of the sphere B

Solving for vA:

$v_A = \frac{m_A u_A + m_B u_B - m_B v_B}{m_A}=\frac{(0.600)(4.00)+(1.80)(2.00)-(1.80)(3.00)}{0.600}=1.00 m/s$

The sign is positive, so the direction is to the right.

B)

To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

Before the collision:

$K_i = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 =\frac{1}{2}(0.600)(4.00)^2 + \frac{1}{2}(1.80)(2.00)^2=8.4 J$

After the collision:

$K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(0.600)(1.00)^2 + \frac{1}{2}(1.80)(3.00)^2=8.4 J$

The total kinetic energy is conserved: therefore, the collision is elastic.

C)

Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.

Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:

$p_x = m_B v_B = (1.80)(3.00)=5.40 kg m/s$

While the total momentum along the y-axis is zero:

$p_y = 0$

We can now write the equations of conservation of momentum along the two directions as follows:

$p_x = p'_{Bx} + p'_{Cx}\\0 = p'_{By} + p'_{Cy}$ (1)

We also know the components of the momentum of B after the collision:

$p'_{Bx}=(1.20)(cos 19)=1.13 kg m/s\\p'_{By}=(1.20)(sin 19)=0.39 kg m/s$

So substituting into (1), we find the components of the momentum of C after the collision:

$p'_{Cx}=p_B - p'_{Bx}=5.40 - 1.13=4.27 kg m/s\\p'_{Cy}=p_C - p'_{Cy}=0-0.39 = -0.39 kg m/s$

So the magnitude of the momentum of C is

$p'_C = \sqrt{p_{Cx}^2+p_{Cy}^2}=\sqrt{4.27^2+(-0.39)^2}=4.29 kg m/s$

Dividing by the mass of C (1.60 kg), we find the magnitude of the velocity:

$v_c = \frac{p_C}{m_C}=\frac{4.29}{1.60}=2.68 m/s$

And the direction is

$\theta=tan^{-1}(\frac{p_y}{p_x})=tan^{-1}(\frac{-0.39}{4.27})=-5.2^{\circ}$

D)

The impulse imparted by B to C is equal to the change in momentum of C.

The initial momentum of C is zero, since it was at rest:

$p_C = 0$

While the final momentum is:

$p'_C = 4.29 kg m/s$

So the magnitude of the impulse exerted on C is

$I=p'_C - p_C = 4.29 - 0 = 4.29 kg m/s$

And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore $-5.2^{\circ}$ .

E)

To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:

$K_i = \frac{1}{2}m_B u_B^2 = \frac{1}{2}(1.80)(3.00)^2=8.1 J$

The total kinetic energy after the collision is the sum of the kinetic energies of B and C:

$K_f = \frac{1}{2}m_B v_B^2 + \frac{1}{2}m_C v_C^2 = \frac{1}{2}(1.80)(1.20)^2 + \frac{1}{2}(1.60)(2.68)^2=7.0 J$

Since the total kinetic energy is not conserved, the collision is inelastic.

F)

Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:

$F=Ma$

Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.

Therefore, the centre of mass after the 2nd collision must be equal to the velocity of the centre of mass before the 1st collision: which is the velocity of the sphere A before the 1st collision (because the other 2 spheres were at rest), so it is simply 4.00 m/s to the right.

Learn more about momentum and collisions:

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8 0

3 years ago

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

A long, nonconducting cylinder (radius = 6.0 mm) has a nonuniform volume charge density given by αr2, where α = 6.2 mC/m5 and r

erastovalidia [21]

Answer: 2.80 N/C

Explanation: In order to calculate the electric firld inside the solid cylinder

non conductor we have to use the Gaussian law,

∫E.ds=Q inside/ε0

E*2πrL=ρ Volume of the Gaussian surface/ε0

E*2πrL= a*r^2 π* r^2* L/ε0

E=a*r^3/(2*ε0)

E=6.2 * (0.002)^3/ (2*8.85*10^-12)= 2.80 N/C

5 0

4 years ago

If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far

zavuch27 [327]

Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

So for ∑Fx=0, ∑Fy=0 and ∑M=0

We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

And Between somewhere we have the weight of painter acting downward equal to= mg

So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d= $\frac{N2}{mg}*l$ * tan∅

Also, we have f₁=N₂

As f₁= чN₁

So f₁= 0.357 * 69.1 * 9.8

f₁= 241.75

Putting in d equation, we have

d= $\frac{241.75}{69.1*9.8} *l$ * tan 58

d= 0.57* l

So painter can be along the 57% of length before the ladder begins to slip

3 0

3 years ago

Read 2 more answers

A 2.02 nF capacitor with an initial charge of 4.55 µC is discharged through a 1.22 kΩ resistor. (a) Calculate the current in the

Mandarinka [93]

Answer:

a) 0.048A

b) 0.18µC

c) 1.85A

Explanation:

The discharged current of the capacitor as a function of time is given by:

$i=\frac{q_o}{RC}*e^{-\frac{t}{\tau}}\\where:\\\tau=RC\\$

$\tau=1.22*10^3*2.02*10^{-9}\\\tau=2.46*10^{-6}s$

a)

$i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{9\µs}{2.46\µs}}\\\\i=0.048A$

b)

$q=q_o*e^{-\frac{t}{\tau}}$

$q=4.55\µC*e^{-\frac{8\µs}{2.46\µs}}\\q=0.18\µC$

c) the maximum current occurs when t=0

$i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{0\µs}{2.46\µs}}\\\\i=1.85A$

6 0

3 years ago