5. What is the elevation along the shoreline (sea level)?<br> O ft<br> 1 ft<br> 10 ft<br> 100 ft

Two 5000-kg passenger cars roll without friction (one at 1 m/s, the other at 2 m/s) toward one another on a level track. They co

balu736 [363]

The combined momentum of the passengers is 5000 kgm/s.

<h3>Combined momentum of the passenger</h3>

The combined momentum of the passengers is calculated as follows;

P = mv1 + mv2

where;

m is mass of the passengers
v1 is velocity of the first passenger
v2 is velocity of the second passenger

P = m(v1 + v2)

P = 5000(-1 + 2)

P = 5000 kgm/s

Thus, the combined momentum of the passengers is 5000 kgm/s.

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5 0

1 year ago

If a car change the velocity from 36M/S to 28M/S with an acceleration of -2.0 M/S then how much time does it take for the vehicl

leonid [27]

The time it will take if a car change the velocity from 36m/s to 28m/s with an acceleration of -2.0m/s² is 4s.

<h3>How to calculate deceleration?</h3>

Deceleration is the amount by which a speed or velocity decreases (and so a scalar quantity or a vector quantity).

The deceleration of a body can be calculated using the following formula:

-a = (v - u)/t

Where;

-a = deceleration
v = final velocity
u = initial velocity
t = time in seconds

-2 = (28 - 36)/t

-2t = -8

t = 4s

Therefore, the time it will take if a car change the velocity from 36m/s to 28m/s with an acceleration of -2.0m/s² is 4s.

Learn more about deceleration at: brainly.com/question/4403243

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6 0

2 years ago

How much stronger is the gravitational pull of the sun on earth

Alex777 [14]

28 times stronger than the force of gravity on the surface of the Earth.

6 0

3 years ago

A car initially at rest accelerates at 10m/s^2. The car’s speed after it has traveled 25 meters is most nearly... A.) 0.0m/s B.)

STALIN [3.7K]

The car traverses a distance $x$ after time $t$ according to

$x=\dfrac12at^2$

where $a$ is its acceleration, 10 m/s^2. The time it takes for the car to travel 25 m is

$25\,\mathrm m=\left(5\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t=\sqrt 5\,\mathrm s$

5 is pretty close to 4, so we can approximate the square root of 5 by 2. Then the car's velocity $v$ after 2 s of travel is given by

$v=\left(10\dfrac{\rm m}{\mathrm s^2}\right)(2\,\mathrm s)\approx20\dfrac{\rm m}{\rm s}$

which makes C the most likely answer.

3 0

3 years ago

An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field

lawyer [7]

Answer:

a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

Em₀ = U = e DV

final point. Where they hit the target

Em_f = K = ½ m v2

energy is conserved

Em₀ = Em_f

e ΔV = ½ m v²

ΔV = $\frac{1}{2}$ mv²/e (1)

If the speed of light is c and this is 100% then 1% is

v = 1% c = c / 100

v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

ΔV = $\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }$

ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

$K_e = K_p$

K_p = ½ m v_e²

K_p = $\frac{1}{2}$ 9.1 10⁻³¹ (3 10⁶)²

K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

ΔV = Kp / M

ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

e ΔV = ½ M v²

v² = 2 e ΔV / M

v² = $\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }$

v² = 49,035 10⁸

v = 7 10⁴ m / s

Relation

v/c = $\frac{7 \ 10^4 }{ 3 \ 10^8}$

v/c= 2.33 10⁻⁴

8 0

3 years ago

5. What is the elevation along the shoreline (sea level)? O ft 1 ft 10 ft 100 ft