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2 years ago

5

Which statements describe the book and the forces acting on it? Check all that apply. The forces are balanced. The forces are un

balanced. The net force is zero. The book is at rest. The net force is to the right. The book is moving to the right.

2 answers:

dedylja [7]2 years ago

5 0

Answer:

2.The forces are unbalanced.

5.The net force is to the right.

6.The book is moving to the right.

Explanation:

correct on edge :)

myrzilka [38]2 years ago

3 0

Answer:

2. 5. and 6.

Explanation:

I got it right

YOU'RE WELCOME

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A vertical spring of constant k = 400 N/m hangs at rest. When a 2 kg mass is attached to it, and it is released, the spring exte

Viefleur [7K]

Answer:

4.9 cm

Explanation:

From Hook's Law,

F = ke......................... Equation 1

Where F= force, e = extension, k = spring constant.

Note: the Force acting on the the spring is the weight of the mass.

W = mg.

F = mg.................... Equation 2

Where m = mass, g = acceleration due to gravity

Substitute equation 2 into equation 1

mg = ke

make e the subject of the equation

e = mg/k............... Equation 3.

Given: m = 2 kg, g = 9.8 m/s², k = 400 N/m

e = (2×9.8)/400

e = 19.6/400

e = 0.049 m

e = 4.9 cm

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2 years ago

A 2-kg cart, traveling on a horizontal air track with a speed of 3 m/s, collides with a stationary 4-kg cart. The carts stick to

daser333 [38]

Answer:

Magnitude of impulse, |J| = 4 kg-m/s

Explanation:

It is given that,

Mass of cart 1, $m_1=2\ kg$

Mass of cart 2, $m_2=4\ kg$

Initial speed of cart 1, $u_1=3\ m/s$

Initial speed of cart 2, $u_2=0$ (stationary)

The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$V=\dfrac{2\times 3}{(2+4)}$

V = 1 m/s

The magnitude of the impulse exerted by one cart on the other is given by:

$J=F\times t=m(V-u)$

$J=m(V-u)$

$J=2\times (1-3)$

J = -4 kg-m/s

or

|J| = 4 kg-m/s

So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.

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2 years ago

The electric potential in a particular region of space varies only as a function of y-position and is given by the function V(y)

nikdorinn [45]

Answer:

$E = 55.9583\ Volts/meter$

Explanation:

First let's find the electric potential using y = 22.5:

$V(y) = 1.69y^2 +15.6y+52.5$

$V(22.5) = 1.69(22.5)^2 + 15.6*22.5 + 52.5$

$V(22.5) = 1259.0625\ Volts$

Then, to find the magnitude of the electric field, we just need to divide the electric potential by the distance y:

$E = V/d$

$E = 1259.0625/22.5$

$E = 55.9583\ Volts/meter$

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2 years ago

Ancient Greek philosophers spent lots of time thinking about science and imaging explanations for the natural world. What part o

Illusion [34]

Answer:

Testing

Explanation:

Ancient Greek philosophers lived with the ideology to simply contemplate life. This means that their whole life revolved around thinking and questioning everything. This would include creative thinking, because they would sometimes come up with theories which require creativeness. They would often debate with their friends as to why their theory should be accepted or what their opinions were on the matter. More often than not, they argued a lot, and many philosophers went against some powerful people in the community and some were even sentenced to death.

The main process they didn't/couldn't do was the testing. They could never test certain theories because they did not have the means to.

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2 years ago

Bricks and insulation are used to construct the walls of a house. The

ira [324]

Answer:

$\dot Q=350.438\ W$

Explanation:

Given:

<u>the thermal resistance in the form of </u>

$R_1=\frac{x_1}{k_1} =0.095\ m^2.^{\circ}C.W^{-1}$

$R_2=\frac{x_2}{k_2} =0.704\ m^2.^{\circ}C.W^{-1}$

where:

$x_1\ \&\ x_2$ are the thickness of the respective bricks

$k_1\ \&\ k_2$ are the respective coefficient of conductivity

temperature inside the house, $T_h=24\ ^{\circ}C$

temperature outside the house, $\ T_c=10^{\circ}C$

area of the wall, $A=20\ m^2$

Since the bricks and insulation are used to construct a wall then they must be used in series for better shielding.

<u>Using Fourier's law:</u>

$\dot Q=k.A.\frac{dT}{x}$

$\dot Q={dT}\div {\frac{x}{k.A} }$

in series the resistances get add up

$\dot Q=dT\div (\frac{x_1}{k_1.A}+\frac{x_2}{k_2.A} )$

$\dot Q=(24-10)\div (\frac{0.095 }{20}+ \frac{0.704 }{20} )$

$\dot Q=350.438\ W$

7 0

2 years ago