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Ratling [72]
3 years ago
5

A car with a mass of 1380 Kg is traveling at 23 m/s to the north. A truck with a mass of 1625 Kg is traveling at 26 m/s to the s

outh. Imagine that both of these vehicles crash into each other head-on and get stuck together. What will their velocity be immediately after the collision
Physics
1 answer:
trasher [3.6K]3 years ago
8 0

Answer: -3.49 m/s (to the south)

Explanation:

This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum p_{i} must be equal to the final momentum p_{f}, and taking into account this is aninelastic collision:

Before the collision:

p_{i}=mV_{o}+MU_{o} (1)

After the collision:

p_{f}=(m+M)V_{f} (2)

Where:

m=1380 kg is the mass of the car

V_{o}=23 m/s is the velocity of the car, directed to the north

M=1625 kg is the mass of the truck

U_{o}=-26 m/s is the velocity of the truck, directed to the south

V_{f} is the final velocity of both the car and the truck

p_{i}=p_{f} (3)

mV_{o}+MU_{o}=(m+M)V_{f} (4)

Isolating V_{f}:

V_{f}=\frac{mV_{o}+MU_{o}}{m+M} (5)

V_{f}=\frac{(1380 kg)(23 m/s)+(1625 kg)(-26 m/s)}{1380 kg+1625 kg} (6)

Finally:

V_{f}=-3.49 m/s The negative sign indicates the direction of the velocity is to the south

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Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su
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Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

      0 V/m

E⃗ 2

      0 V/m

E⃗ 3

         E_3 =  2.153 *10^{9} \  V/m

E⃗ 4

E_4 =  5.754 *10^ {8} \  V/m

Explanation:

From the question we are told that

The diameter is d =  6.53 \mu m  = 6.53*10^{-6}\  m

The charge is Q =  -.2.55 *10^{-12} \  C

The permittivity of free space is \epsilon_o  =  8.85* 10^{-12}\  C / V.m

The distance considered is d =  3.05 \mu m  =  3.05 *10^{-6} \ m

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

E_3 =  \frac{ k  *  |Q|}{ r^2 }

Here k is the coulomb constant with value

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

r is the radius of the sphere which is mathematically as

r =  \frac{d}{2} =   \frac{6.53*10^{-6}}{2}  = 3.265 *10^{-6} \  m

E_3 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }

E_3 =  2.153 *10^{9} \  V/m

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

E_4 =  \frac{ k  *  |Q|}{ R^2 }

Here R is mathematically represented as

R  =  3.265 *10^{-6} +  3.05 *10^{-6}

=>       R  =  6.315 *10^{-6}

So

E_4 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }

E_4 =  5.754 *10^ {8} \  V/m

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Answer: 5 \frac{g}{cm^{3}}

Explanation:

The density \rho of a material is given by:

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m=3645 g is the mass of the cube

V is the volume of the cube

Now, the volume of a cube is equal to the length L of its edge to the power of 3:

V=L^{3} (2)

If we know L=9 cm, the volume of this cube is:

V=(9 cm)^{3}=729 cm^{3} (3)

Substituting (3) in (1):

\rho=\frac{3645 g}{729 cm^{3}} (4)

\rho=5 \frac{g}{cm^{3}} This is the density of the cube

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