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Butoxors [25]
4 years ago
6

A 2,000 kg car is moving at 15m/s when it collides with a 1200kg car sitting still. Briefly compare the impulse imparted on the

1200kg car by the 2000 kg car to the impulse on the 2000 kg car by the 1200 kg car which car under goes the greater change in momentum
Physics
2 answers:
mihalych1998 [28]4 years ago
8 0

Answer:

Explanation:

Given:

mass of car 1 = 2000 kg

speed of car 1 = 15 m/s

mass of car 2 = 1200 kg

car 2 is initially at rest.

Impulse (F Δt) = change in momentum = Δ(mv)

Change in Momentum depends on the mass and velocity.

When car 1 collides with car 2, car 1 would under go greater change in the momentum because it has greater mass and initial velocity while car 2 has smaller mass and is initially at rest.

I am Lyosha [343]4 years ago
4 0

impulse = F × t

The greater the impulse exerted on something, the greater will be the change in momentum.

impulse = change in momentum

Ft = ∆(mv)

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A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is paralle
Alex73 [517]

Answer:

715 N

Explanation:

Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.

Let g = 9.8 m/s2

Gravity and equalized normal force is:

N = P = mg = 107*9.8 = 1048.6 N

Kinetic friction force and equalized tension force on the rope is

T = F_{\mu} = N\mu = 1048.6 * 0.682 = 715.1452 N

6 0
3 years ago
Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as
MAXImum [283]

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m

or in kilometers,

h_2 = 47 km

the first altitude in kilometers is

h_1 = 155 km

so the difference between the two altitudes is

\Delta h = 155 km - 47 km = 108 km

8 0
3 years ago
A 55 kg skier is at the top of a slope. The vertical distance between start and finish is 10m. (a) What is her Potential Energy?
Blababa [14]

Explanation:

Given that,

Mass of a skier, m = 55 kg

The vertical distance between start and finish is 10 m.

(a) The potential energy of a skier is given by :

E=mgh\\\\E=55\times 9.8\times 10\\\\=5390\ J

(b) When she reaches the bottom of the hill, the potential energy is converted to kinetic energy. So,

5390=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2\times 5390}{55}} \\\\v=14\ m/s

Hence, her potential energy is 5390 J and at the bottom of the hill she is moving with a speed of 14 m/s.

6 0
3 years ago
Continuous sinusoidal perturbation Assume that the string is at rest and perfectly horizontal again, and we will restart the clo
Elena-2011 [213]

a) 3.14 \cdot 10^{-4} s

b) See plot attached

c) 10.0 m

d) 0.500 cm

Explanation:

a)

The position of the tip of the lever at time t is described by the equation:

y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t] (1)

The generic equation that describes a wave is

y(t)=A sin (\frac{2\pi}{T} t) (2)

where

A is the amplitude of the wave

T is the period of the wave

t is the time

By comparing (1) and (2), we see that for the wave in this problem we have

\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}

Therefore, the period is

T=\frac{2\pi}{2.00\cdot 10^4}=3.14 \cdot 10^{-4} s

b)

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

The maximum displacement of the wave is given by the value of y when sin(...)=1, and from eq(1), we see that this is equal to

y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

c)

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

\lambda=2L

where

\lambda is the wavelength of the wave

L is the length of the string

In this problem,

L = 5.00 m is the length of the string

Therefore, the wavelength is

\lambda =2(5.00)=10.0 m

d)

The amplitude of a wave is the magnitude of the maximum displacement of the wave, measured relative to the equilibrium position.

In this problem, we can easily infer the amplitude of this wave by looking at eq.(1).

y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]

And by comparing it with the general equation of a wave:

y(t)=A sin (\frac{2\pi}{T} t)

In fact, the maximum displacement occurs when the sine part is equal to 1, so when

sin(\frac{2\pi}{T}t)=1

which means that

y(t)=A

And therefore in this case,

y=0.500 cm

So, this is the displacement.

6 0
3 years ago
6. When sodium and oxygen react to form an ionic compound, the mass of the compound ____ the sum of the masses of the individual
arsen [322]
Because there’s no elemental degradation, the mass of the compound EQUALS the sum of the masses.
3 0
3 years ago
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