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Alborosie
3 years ago
12

A 3.0-kg and a 1.0-kg box rest side-by-side on a smooth, level floor. A horizontal force of 32 N is applied to the 1.0-kg box pu

shing it against the 3.0-kg box, and, as a result, both boxes slide along the floor. How hard do the two boxes push against each other
Physics
1 answer:
stira [4]3 years ago
5 0

Considering both boxes as one body, it would have a total mass of 4.0 kg. By Newton's second law, the 32 N force applies an acceleration <em>a </em>such that

∑ <em>F</em> = 32 N = (4.0 kg) <em>a</em>   →   <em>a</em> = 8.0 m/s²

and both boxes share this acceleration. (There is no friction, so the given force is the only one involved in the direction of the boxes' motion.)

Now consider just the smaller box. It is feeling the effect of the 32 N push in one direction and, as it comes into contact with the larger box, a normal force that points in the opposite direction. Let <em>n</em> be the magnitude of this normal force; this is what you want to find. By Newton's second law,

∑ <em>F</em> = 32 N - <em>n</em> = (1.0 kg) (8.0 m/s²)

<em>n</em> = 32 N - 8.0 N

<em>n</em> = 24 N

Just to make sure that this is consistent: by Newton's third law, the larger box feels the same force but pointing in the opposite direction. On the smaller box, <em>n</em> opposes the pushing force, so points backward. So from the larger box's perspective, <em>n</em> acts on it in the forward direction. This is the only force acting on the larger box, so Newton's second law gives

∑ <em>F</em> = 24 N = (3.0 kg) (8.0 m/s²)

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a proton of mass 1 u travelling with a speed of 3.6 x 10 ^4 m/s has an elastic head on collision with a helium nucleus initially
CaHeK987 [17]

Answer:

Velocity of the helium nuleus  = 1.44x10⁴m/s

Velocity of the proton = 2.16x10⁴m/s

Explanation:

From the conservation of linear momentum of the proton collision with the He nucleus:

P_{1i} + P_{2i} = P_{1f} + P_{2f] (1)

<em>where P_{1i}: is the proton linear momentum initial, P_{2i}: is the helium nucleus linear momentum initial, P_{1f}: is the proton linear momentum final, P_{2f}: is the helium nucleus linear momentum final </em>

<u>From (1):</u>

m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f} (2)

<em>where m₁ and m₂: are the proton and helium mass, respectively, v_{1i} and v_{2i}: are the proton and helium nucleus velocities, respectively, before the collision, and v_{1f} and v_{2f}: are the proton and helium nucleus velocities, respectively, after the collision </em>

By conservation of energy, we have:

K_{1i} + K_{2i} = K_{1f} + K_{2f} (3)

<em>where K_{1i} and  K_{2i}: are the kinetic energy for the proton and helium, respectively, before the colission, and K_{1f} and  K_{2f}: are the kinetic energy for the proton and helium, respectively, after the colission </em>

<u>From (3):</u>

\frac{1}{2}m_{1}v_{1i}^{2} + 0 = \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2} (4)  

<u>Now we have two equations: (2) ad (4), and two incognits: v_{1f} and v_{2f}. </u>

Solving equation (2) for v_{1f}, we have:

v_{1f} = v_{1i} -\frac{m_{2}}{m_{1}} v_{2f} (5)

<u>From getting (5) into (4) we can obtain the v_{2f}:</u>

v_{2f}^{2} \cdot (\frac{m_{2}^{2}}{m_{1}} + m_{2}) - 2v_{2f}v_{1i}m_{2} = 0

v_{2f}^{2} \cdot (\frac{(4u)^{2}}{1u} + 4u) - v_{2f}\cdot 2 \cdot 3.6 \cdot 10^{4} \cdot 4u = 0

From solving the quadratic equation, we can calculate the velocity of the helium nucleus after the collision:

v_{2f} = 1.44 \cdot 10^{4} \frac{m}{s} (6)

Now, by introducing (6) into (5) we get the proton velocity after the collision:

v_{1f} = 3.6 \cdot 10^{4} -\frac{4u}{1u} \cdot 1.44 \cdot 10^{4}

v_{1f} = -2.16 \cdot 10^{4} \frac{m}{s}

The negative sign means that the proton is moving in the opposite direction after the collision.

I hope it helps you!

7 0
4 years ago
Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init
Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ=(m+\frac{1}{2})\frac{lamda}{n}  ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= \frac{3*lamda}{2}

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= \sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }

δ= \sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}

δ= \sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }

δ= \sqrt{4900+43^2}-\sqrt{4900+23^2}

δ= \sqrt{4900+1849}-\sqrt{4900+529}

δ= \sqrt{6749}-\sqrt{5429}

δ=  82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= \frac{3*lamda}{2}

δ= 8.47

8.47 = \frac{3*lamda}{2}

λ = \frac{8.47*2}{3}

λ = 5.65m

3 0
4 years ago
A large truck has more inertia than small car because it has more__
taurus [48]
The large truck can rest at a stable equilibrium as compare to the small,from the definition of inertia, <span>Inertia is the tendency of an object to remain at rest or in motion.</span>
6 0
3 years ago
PLEASE HELP
Gennadij [26K]

Answer:

3.675 m

Explanation:

a_{x} =0 v_{xo}=100 a_{y} =-g  v_{yo}=0

X-direction     | Y-direction

R=x_{o}+ v_{xo} t  | y=y_{o}+v_{yo}t+\frac{1}{2}a_{y}t^2

75=100t         |y=0+0+\frac{1}{2} (9.8)(0.75)

\frac{75}{100} =t             | y=3.675 m

0.75s=t              

Hope it helps

3 0
3 years ago
What force is produced when a battery causes an electric current to flow trough a circuit made of copper wire
zzz [600]
You'll hear that force called different things in different places. It
may be called "electromotive force", "EMF", "potential difference",
or "voltage".

It's just a matter of somehow causing the two ends of the wire
to have different electrical potential. When that happens, the
free electrons in the copper suddenly have a burning desire to
travel ... away from the end that's more negative, toward the end
that's more positive, and THAT's an "electric current".
7 0
4 years ago
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