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Setler79 [48]
3 years ago
10

Two pop cans are at rest on a stand. A firecracker is placed between the cans and lit. The firecracker explodes and exerts equal

and opposite forces on the two cans. Assuming the system of two cans to be isolated, the post-explosion momentum of the system ____.
a. is dependent upon the mass and velocities of the two cans
b. is dependent upon the velocities of the two cans (but not their mass)
c. is typically a very large value
d. can be a positive, negative or zero value
e. is definitely zero
Physics
1 answer:
erastovalidia [21]3 years ago
7 0

Answer:

e. is definitely zero

Explanation:

Given that

At initial condition the speed of the pop cans is zero.

We know that linear momentum

P = Mass x velocity

P =  m v

At initial condition v = 0

P= 0

If there is no any external force then the linear momentum of the system will be conserve.And given that ,consider the system isolated.

Therefore the answer is e.

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Which pair of elements has the most similar properties?
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The answer is B, because oxygen and sulfur are in the same group (group 6A)
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if the current in a wire is 2.0 amperes and the potential difference across the wire is 10 volts what is the resistance of the w
Pavlova-9 [17]

Answer:

R = 2Ω

Explanation:

Potential difference (V) = current (I) * Resistance (R)

V = IR

I = 2.0A

V = 10v

R = ?

V = IR

R = V / I

R = 10 / 2

R = 2Ω

The resistance across the wire is 2Ω

3 0
3 years ago
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Alguien me ayuda con calculos estequiometricos?
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Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay
kaheart [24]

Answer:

0.278 m/s

Explanation:

We can answer the problem by using the law of conservation of momentum. In fact, the total momentum before the collision must be equal to the total momentum after the collision.

So we can write:

mu=(m+M)v

where

m = 0.200 kg is the mass of the koala bear

u = 0.750 m/s is the initial velocity of the koala bear

M = 0.350 kg is the mass of the other clay model

v is their final combined velocity

Solving the equation for v, we get

v=\frac{mu}{m+M}=\frac{(0.200)(0.750)}{0.200+0.350}=0.278 m/s

8 0
3 years ago
Holding onto a tow rope moving parallel to a frictionless ski slope, a 68.7 kg skier is pulled up the slope, which is at an angl
Furkat [3]

Answer:

a) F = 78.606\,N, b) F = 88.911\,N

Explanation:

a) Let consider two equations of equilibrium, the first parallel to ski slope and the second perpendicular to that. The equations are, respectively:

\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = 0\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0

The force on the skier is:

F = m \cdot g \cdot \sin \theta

F = (68.7\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 6.7^{\textdegree}

F = 78.606\,N

b) The equations of equilibrium are the following:

\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = m\cdot a\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0

The force on the skier is:

F = m\cdot (a + g \cdot \sin \theta)

F = (68.7\,kg)\cdot (0.150\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}}\cdot \sin 6.7^{\textdegree})

F = 88.911\,N

3 0
3 years ago
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