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3 years ago

15

A 6.0-kg box slides down an inclined plane that slopes 39° to the horizontal. The box accelerates at a rate of 3.1 m/s^2. What i

s the coefficient of kinetic friction between the box and the surface of the inclined plane?
0.55
0.40
0.16
0.48

1 answer:

Lynna [10]3 years ago

4 0

I'm pretty sure its b. 0.40 no doubt about it XD

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Two people stand facing each other at roller skating rink then push off each other

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a) 0 kg m/s

b) 0 kg m/s

c) +3 m/s

d) 60 N

Explanation:

a)

The momentum of an object is a vector quantity given by:

$p=mv$

where

m is the mass of the object

v is the velocity of the object

In this problem, we have a system of two people, so the total momentum will be the sum of the individual momenta of the two people:

$p=p_1 + p_2$

Which can be rewritten as

$p=m_1 u_1 + m_2 u_2$

where $m_1,m_2$ are the masses of the two people and $u_1,u_2$ their initial velocities.

However, the two people are initially at rest, so

$u_1 = 0\\u_2 = 0$

Therefore the total momentum is

$p=0+0=0$

b)

The principle of conservation of momentum states that when there are no external forces acting on a system, the total momentum of the system is conserved, so we can write:

$p_i = p_f$

where

$p_i$ is the total momentum of the system before

$p_f$ is the total momentum of the system after

In this problem,

$p_i = 0$

As we calculated in part a: this is because the total momentum of the two people before they push off each other is zero.

Therefore, according to the law of conservation of momentum,

$p_f = p_i = 0$

So the total momentum is zero also after they push off each other.

c)

The total momentum of the girl and the boy after they push off each other can be written as:

$p_f = m_1 v_1 + m_2 v_2$ (1)

where:

$m_1 = 30 kg$ is the mass of the girl

$v_1 = -5 m/s$ is her velocity (she moves backward, so the negative sign)

$m_2 = 50 kg$ is the mass of the boy

$v_2$ is the velocity of the boy

As calculated in part b), we also know that the total momentum of the girl and the boy is

$p_f = 0$ (2)

By combining eq(1) and eq(2) we get

$0=m_1 v_1 + m_2 v_2$

And solving for v2 we find the velocity of the boy:

$v_2=-\frac{m_1 v_1}{m_2}=-\frac{(30)(-5)}{50}=+3 m/s$

and the positive sign means he is moving forward.

d)

We can solve this part by applying the impulse theorem, which states that the change in momentum of an object is equal to the product between the force applied on it and the duration of the collision:

$\Delta p = F\Delta t$

where

$\Delta p$ is the change in momentum

F is the force

$\Delta t$ is the time during which the force is applied

In this problem:

$\Delta t = 2.5 s$

For the boy, the change in momentum is:

$\Delta p = m_2 (v_2 - u_2)$

And since

$m_2 = 50 kg\\u_2 = 0 m/s\\v_2 = 3 m/s$

We have

$\Delta p = (50)(3-0)=150 kg m/s$

So, the force exerted between the boy and the girl is:

$F=\frac{\Delta p}{\Delta t}=\frac{150}{2.5}=60 N$

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3 years ago

Vygotsky’s theory of cognitive development states that an individual constructs knowledge primarily through interaction with imp

defon

Answer:

false

Explanation:

trust me i j took the test

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3 years ago

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Can someone pls help me and give me the answers i will mark brainliest :3

Katyanochek1 [597]

Answer:

with what subject

Explanation:

i really only do ELA and sometimes science

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2 years ago

Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for sl

ikadub [295]

Answer:

Please see below as the answer is self-explanatory.

Explanation:

We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).
Both movements are independent each other, due to they are perpendicular.
In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.
Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:

$v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)$

Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:

$t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s (2)$

In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.
Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:

$\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (3)$

In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:

$v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)$

Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:

$\Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)$

Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.

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How did amphibians adapt to their changing environment ?

kobusy [5.1K]

Answer:

I think it is B

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