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amm1812
2 years ago
10

Consider a system to be one train car moving toward another train car at rest. When the train cars collide, the two cars stick t

ogether.
What is the total momentum of the system after the collision?
800 kg • m/s
1,600 kg • m/s
2,400 kg • m/s
4,000 kg • m/s

Physics
2 answers:
sweet [91]2 years ago
5 0

Answer:

C. 2,400 kg • m/s

Explanation:

Doing it now! Good luck

oee [108]2 years ago
3 0

Answer:

Momentum after = 2400 kgm/s

Explanation:

Momentum after = momentum before

Momentum after = m1v1 +m2v2

Momentum after = (600)(4) + (400)(0)

Momentum after = 2400 kgm/s

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If an object is projected upward from ground level with an initial velocity of 144144 ft per​ sec, then its height in feet after
Liula [17]

Answer:

4.5 s, 324 ft

Explanation:

The object is projected upward with an initial velocity of

v_0 = 144 ft/s

The equation that describes its height at time t is

s(t) = -16t^2 + 144 t (1)

where t, the time, is measured in seconds.

In order to find the time it takes for the object to reach the maximum height, we must find an expression for its velocity at time t, which can be found by calculating the derivative of the position, s(t):

v(t) = s'(t) = -32t +144 (2)

At the maximum heigth, the vertical velocity is zero:

v(t) = 0

Substituting into the equation above, we find the corresponding time at which the object reaches the maximum height:

0=-32t+144\\t=\frac{144}{32}=4.5 s

And by substituting this value into eq.(1), we also find the maximum height:

s(t) = -16(4.5)^2+144(4.5)=324 ft

3 0
3 years ago
A toy car accelerates from 3m/s to 5m/s in 5sec . what is the acceleration​
san4es73 [151]

Answer:

0.4

Explanation:

Because 3m/s is the initial velocity(u) and 5m/s is the final velocity(v) and time is 5 sec.

So, acceleration = v-u ÷ t

I'm confused

7 0
2 years ago
Projectiles that strike objects are good examples of inelastic collisions. A 0.1 kg nail driven by a gas powered nail driver col
Ratling [72]
In an inelastic collision, only momentum is conserved, while energy is not conserved.

1) Velocity of the nail and the block after the collision
This can be found by using the total momentum after the collisions:
p_f=(m+M)v_f=4.8 kg m/s
where
m=0.1 kg is the mass of the nail
M=10 kg is the mass of the block of wood
Rearranging the formula, we find v_f, the velocity of the nail and the block after the collision:
v_f= \frac{p_f}{m+M}= \frac{4.8 kg m/s}{0.1 kg+10 kg}=  0.48 m/s

2) The velocity of the nail before the collision can be found by using the conservation of momentum. In fact, the total momentum before the collision is given only by the nail (since the block is at rest), and it must be equal to the total momentum after the collision:
p_i = mv_i = p_f
Rearranging the formula, we can find v_i, the velocity of the nail before the collision:
v_i =  \frac{p_f}{m}= \frac{4.8 kg m/s}{0.1 kg}=48 m/s
6 0
3 years ago
Read 2 more answers
The 2.50 kg cube in the figure has edge lengths d = 6.50 cm and is mounted on an axle
kozerog [31]

Answer:

0.191 s

Explanation:

The distance from the center of the cube to the upper corner is r = d/√2.

When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ.  The new vertical distance from the center to the corner is r cos θ.

Sum of the torques:

∑τ = Iα

Fr cos θ = Iα

(k r sin θ) r cos θ = Iα

kr² sin θ cos θ = Iα

k (d²/2) sin θ cos θ = Iα

For a cube rotating about its center, I = ⅙ md².

k (d²/2) sin θ cos θ = ⅙ md² α

3k sin θ cos θ = mα

3/2 k sin(2θ) = mα

For small values of θ, sin θ ≈ θ.

3/2 k (2θ) = mα

α = (3k/m) θ

d²θ/dt² = (3k/m) θ

For this differential equation, the coefficient is the square of the angular frequency, ω².

ω² = 3k/m

ω = √(3k/m)

The period is:

T = 2π / ω

T = 2π √(m/(3k))

Given m = 2.50 kg and k = 900 N/m:

T = 2π √(2.50 kg / (3 × 900 N/m))

T = 0.191 s

The period is 0.191 seconds.

7 0
2 years ago
How much would it cost to cover the entire land area of the U.S. in dollar bills?
agasfer [191]

Answer:

  $900 trillion

Explanation:

If Alaska is 20% of the contiguous US, then the approximate area of interest is ...

  1200 miles × 3000 miles = 3.6×10^6 square miles.

The size of a dollar bill is about ...

  (6.5 cm)·(15.5 cm) = 100.75 cm^2

One mile is 160,934.4 cm, so 1 square mile is about ...

  1 mi^2 = (160,934.4 cm)^2 ≈ 2.59·10^10 cm^2

The number of dollars of interest is then ...

  (3.6 · 10^6 mi^2)(2.59 · 10^10 cm^2)/(100.75 cm^2) ≈ 9.3·10^14

  ≈ 930 × 10^12 . . . dollars

It would cost about 900 trillion dollars to cover the land area of the US in $1 bills.

7 0
3 years ago
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