Ask question

Ask question

All categories

2 years ago

9

in an automobile crash, a vehicle that was stopped at a red light is rear-ended by another vehicle. The vehicles have the same m

ass. If the tire marks show that the two vehicles moved after the collision at 4 m/s, what was the speed of the vehicle before the collision

1 answer:

baherus [9]2 years ago

6 0

The initial speed of the vehicle before the collision is 8 m/s.

Let the mass of the vehicle = m
Let the initial speed of the vehicle stopped = u
The initial speed of the vehicle parked at the red light = 0

<h3>Principle of conservation of linear momentum</h3>

The initial speed of the vehicle before the collision is calculated by applying principle of conservation of linear momentum as follows;

$m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\mu + m(0) = 4(m+ m)\\\\mu = 4(2m)\\\\mu = 8m\\\\u = 8 \ m/s$

Thus, the initial speed of the vehicle before the collision is 8 m/s.

Learn more about inelastic collision here: brainly.com/question/7694106

You might be interested in

The core of a certain reflected reactor consist of a cylinder 10 ft high 10 ft in diameter The measured maximum-to-average flux

Westkost [7]

Answer:

The maximum power density in the reactor is 37.562 KW/L.

Explanation:

Given that,

Height = 10 ft = 3.048 m

Diameter = 10 ft = 3.048 m

Flux = 1.5

Power = 835 MW

We need to calculate the volume of cylinder

Using formula of volume

$V =\pi r^2 h$

Put the value into the formula

$V=\pi\times(1.524)^2\times 3.048$

$V= 22.23\m^3$

$V = 22.23\times10^{3}\ Liter$

We need to calculate the maximum power density in the reactor

Using formula of power density

$P=\dfrac{E}{V}$

Where, P = power density

E = energy

V = volume

Put the value into the formula

$P=\dfrac{835\times10^{6}}{22.23\times10^{3}}$

$P=37561.85 = 37.562\times KW/L$

Hence, The maximum power density in the reactor is 37.562 KW/L.

6 0

3 years ago

Suppose a light source is emitting red light at a wavelength of 700 nm and another light source is emitting ultraviolet light at

klasskru [66]

Answer:

b) twice the energy of each photon of the red light.

Explanation:

$\lambda$ = Wavelength

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

Energy of a photon is given by

$E=h\nu\\\Rightarrow E=h\dfrac{c}{\lambda}$

Let $\lambda_1$ = 700 nm

$\lambda_2=350\\\Rightarrow \lambda_2=\dfrac{\lambda_1}{2}$

For red light

$E_1=\dfrac{hc}{\lambda_1}$

For UV light

$E_2=\dfrac{hc}{\dfrac{\lambda_1}{2}}$

Dividing the equations

$\dfrac{E_1}{E_2}=\dfrac{\dfrac{hc}{\lambda_1}}{\dfrac{hc}{\dfrac{\lambda_1}{2}}}\\\Rightarrow \dfrac{E_1}{E_2}=\dfrac{1}{2}\\\Rightarrow E_2=2E_1$

Hence, the answer is b) twice the energy of each photon of the red light.

7 0

2 years ago

Read 2 more answers

lightning is actually an enormous display of the concept of A.chemical heating B.solar energy C.magnetism D.grounding

allsm [11]

Answer:

D) Grounding

Explanation:

The potential difference between cloud and ground leads to ionization of the atmosphere and resulting conduction through the air often to ground (although it can be between clouds at different potentials. I would say grounding, like the spark when you touch a hot battery terminal to ground on a car.

5 0

2 years ago

The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the

FrozenT [24]

Answer:

The electric current in the wire is 0.8 A

Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

$B=\frac{2\pi*a }{u*I}$

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

$I=\frac{2\pi*a*B }{u}$ Formula(2)

$B=0.2*10^{-5} T = 0.2*10^{-5} \frac{weber}{m^{2} }$

a =8cm=0.08m

$u=4*\pi *10^{-7} \frac{Weber}{A*m}$

We replace the known information in the formula (2)

$I=\frac{2\pi*0.08*0.2*10^{-5} }{4\pi *10x^{-7} }$

I=0.8 A

Answer: The electric current in the wire is 0.8 A

4 0

2 years ago

Con lắc lò xo có độ cứng k = 100N/m được gắn vật có khối lượng m=0.1kg, kéo vật ra khỏi vị trí cân bằng 1 đoạn 5cm rồi buông tay

Delvig [45]

Answer:

The maximum velocity is 1.58 m/s.

Explanation:

A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.

Spring constant, K = 100 N/m

mass, m = 0.1 kg

Amplitude, A = 5 cm = 0.05 m

Let the angular frequency is w.

$w = \sqrt{K}{m}\\\\w = \sqrt{100}{0.1}\\\\w = 31.6 rad/s$

The maximum velocity is

$v_{max} = w A\\\\v_{max} = 31.6\times 0.05 = 1.58 m/s$

8 0

2 years ago