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Vladimir [108]
3 years ago
11

What velocity must a car with a mass of 1110 kg have in order to have the same momentum as a 2280 kg pickup truck traveling at 2

4 m/s to the east? Answer in units of m/s.
Physics
1 answer:
shutvik [7]3 years ago
6 0

Answer:

49.3 m/s

Explanation:

The momentum is defined as the product of the object velocity and its mass.

So the momentum of the truck is

P_t = 2280 * 24 = 54720 kgm/s

For the car to have the same momentum, its speed must be

v_c = P_t/m_c = 54720 / 1110 = 49.3 m/s

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Photons of wavelength 65.0 pm are Compton-scattered from a free electron which picks up a kinetic energy of 0.75 keV from the co
DIA [1.3K]

0.6764*10^-10m

Explanation:

Using E= hc/wavelength

(4.14x10^-15)x(3.0x10^8)/(65x10^-12)=0.1911x10^5 eV=19.1 keV

So subtract the calculated energy from the given energy of scattered photons

9.11-0.75=18.36 keV

To find wavelength

Wavelength= hc/ E

[(4.14x 10^-15)x (3.0x10^8)]/(18.36*10^3) =0.6764^-10 m

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What element shares the property of forming ions with a 1+ charge?
Oduvanchick [21]

Answer:

the Group 1A metals such as sodium and potassium form +1 charges,

Explanation:

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Calculate the impulse imparted when a 3,000-kg car hits a wall at 60 . m/s and comes to a stop.
Pani-rosa [81]

Impulse = change in momentum


The answer is 0.

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3 years ago
Which circuit is a series circuit
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Answer:

The answer is A

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3 years ago
Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t
marshall27 [118]

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

         

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

     

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

        t² - 12.5 t - 150 = 0

we solve the quadratic equation

       t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150}  }{2}

       t = \frac{12.5 \  \pm 27.5}{2}

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

8 0
3 years ago
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