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pishuonlain [190]
2 years ago
11

The combine mass of two objects sticking together is 1500 kg and is traveling at 63 m/s. If object 1 has a mass of 400 kg and wa

s traveling at 45 m/s. Then how fast was object 2 moving before the impact?
Physics
1 answer:
Roman55 [17]2 years ago
7 0

Answer:

(a) -16.7 N s; (b) -167 N

Explanation:

Given: m = 0.530 kg; vi = 18.0 m/s; vf = 13.5 m/s; t = 0.100 s

Find: (a) Impulse, (b) Force

(a) Impulse = Momentum Change = m•Delta v = m•(vf - vi)= (0.530 kg)•( -13.5 m/s - 18.0 m/s)

Impulse = -16.7 kg•m/s = -16.7 N•s

where the "-" indicates that the impulse was opposite the original direction of motion.

(Note that a kg•m/s is equivalent to a N•s)

(b) The impulse is the product of force and time. So if impulse is known and time is known, force can be easily determined.

Impulse = F•t

F = Impulse/t = (-16.7 N s) / (0.100 s) = -167 N

where the "-" indicates that the impulse was opposite the original direction of motion.

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Which has the most kinetic energy from its motion?
victus00 [196]

Answer:

option c

Explanation:

K.E=1/2mv(squared)

K.E=1/2x900x80x80

K.E=450x6400=2880000J

convert to KJ=2880000/1000=2880KJ

6 0
4 years ago
At one instant of time a rocket is traveling in outer space at 2500m/s and is exhausting fuel at a rate of 100 kg/s. If the spee
lawyer [7]

Thrust is a reaction force described quantitatively by Newton's third law. When a system expels or accelerates mass in one direction, the accelerated mass will cause a force of equal magnitude but opposite direction on that system. Mathematically can be written as,

T = v\frac{dm}{dt}

Here,

v = speed of the exhaust gases measured relative to the rocket.

\frac{dm}{dt}= Rate of change of mass with respect to time

Our values are given as,

v = 1500m/s

\frac{dm}{dt} = 100kg/s

Replacing we have that

T = 1500*100

\therefore T = 1.5*10^5N

7 0
3 years ago
The speed or celerity of deep-water waves... A. increases with increasing water depth B. increases as water becomes blue in colo
antoniya [11.8K]

Answer:

A.

Explanation:

The speed or celerity of deep-water waves increases with increasing water depth.

With increase in depth, the pressure inside the increases as p=ρgh.

And with this increase in pressure force associated with waves increase because force is directly proportional to the pressure. Now greater the force greater will be accleration and greater velocity is obtained.

4 0
3 years ago
Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24
Oksi-84 [34.3K]
Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}
Now At xp we have:
\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0
\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0
Which is a second order equation, using the quadratic formula to solve for xp would give us:
xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}
or
xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.

6 0
3 years ago
4. Identify Problems A student is describing a longitudinal wave in his notebook. He
andrew11 [14]

Answer:

 "The distance between crests is 3 cm." 

Explanation:

If he writes down "The distance between crests is 3 cm." 

That means he is describing the wavelength of a wave and not longitudinal wave. He ought to write something about " direction "

Longitudinal waves are waves in which the displacement of the medium is in the same direction as, or parallel to, the direction of propagation of the wave. While

Wavelength is the distance between the two successfully Crest or trough

3 0
3 years ago
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