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3 years ago

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It is weigh-in time for the local under-85-kg rugby team. The bathroom scale used to assess eligibility can be described by Hook

e's law and is depressed 0.65 cm by its maximum load of 110 kg.

1 answer:

Molodets [167]3 years ago

7 0

<span>F = kx, where F is the force we apply, k is the spring constant, and x is the extension of the material
so the </span><span>spring's effective spring constant k is:
k = F / x
k = ( 110 kg ) ( 9.81 m/s^2) / 0.65 cm
k = 1660 N/cm is the </span>spring's effective spring constant

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The brakes exert a 1,951 N force on a car weighing 18,985 N and moving at 12 m/s. The car finally stops. How long (in seconds) d

nata0808 [166]

Answer:20,000

Explanation:

7 0

3 years ago

Help me with this physics math?

solong [7]

Ok. PEMDAS tells us to take care of the square first. When we do that, the denominator becomes

(6.4)^2 x 10^12

= 40.96 x 10^12 .

Now it's just a matter of mashing out the fraction.

The 'mantissa' (the number part) is

6/40.96 = 0.1465

and the order of magnitude is

10^24 / 10^12 = 10^12 .

Put it all together and you've got

1.465 x 10^11 .

4 0

3 years ago

An object is attached to a hanging unstretched, ideal and massless spring and slowly lowered to its equilibrium position, a dist

Ad libitum [116K]

Answer:

10.6cm

Explanation:

We are given 5.3cm below the starting point (spring extension).

Therefore, to find static vertical equilibrium, we use the equation:

kx = mg

Where:

k = spring constant =

=mg/5.3 kg/s²

We are told the object was dropped from rest.

Therefore:

loss in potential energy = gain in spring p.e

Let's use the expression:

mgx = ½kx²

We are asked to find the stretch at maximum elongation x.

To find x, we make x subject of the formula.

Therefore, we have:

x = 2mg/k (after rearranging the equation above)

x = (2mg) / (mg/5.3)

x = 10.6cm

3 0

2 years ago

Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W

jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

$T^{2}=\frac{4\pi^{2}}{GM}r^{3}$ (1)

Where;

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24}kg$ is the mass of the Earth

$r$ is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period $T_{X}$ is:

$T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}$ (2)

Where $r_{X}=1.2(10)^{6}m$

$T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}$ (3)

$T_{X}=413.712 s$ (4)

For satellite Y, the orbital period $T_{Y}$ is:

$T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}$ (5)

Where $r_{Y}=1.9(10)^{5}m$

$T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}$ (6)

$T_{Y}=26.064 s$ (7)

This means $T_{X}>T_{Y}$

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

$V_{X}=\sqrt{\frac{GM}{r_{X}}}$ (8)

$V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}$

$V_{X}=18224.783 m/s$ (9)

<u>For Satellite Y:</u>

$V_{Y}=\sqrt{\frac{GM}{r_{Y}}}$ (10)

$V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}$

$V_{Y}= 45801.13 m/s$ (11)

This means $V_{Y}>V_{X}$

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

4 0

3 years ago

Read 2 more answers

A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, wh

ladessa [460]

Answer:

a) p₂ = 1.88 kg*m/s

θ = 273.4 º

b) Kf = 37% of Ko

Explanation:

a)

Assuming no external forces acting during the collision, total momentum must be conserved.
Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.
The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.
So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:

$p_{o1x} = 2.00 kg*m/s (1)$

$p_{o1y} = 0 (2)$

We can do the same for the particle moving along the positive y-axis:

$p_{o2x} = 0 (3)$

$p_{o2y} = 4.00 kg*m/s (4)$

Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.
Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:

$p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)$

$p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s (6)$

Now, the total initial momentum, along these directions, must be equal to the total final momentum.
We can write the equation for the x- axis as follows:

$p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x} (7)$

We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:

$p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)$

Now, we can repeat exactly the same process for the y- axis, as follows:

$p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y} (9)$

We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:

$p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)$

Since we have the x- and y- components of the final momentum of the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:

$p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} } = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)$

We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:

$tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)$

The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.

b)

Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:

$\frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)$

So, the final kinetic energy has lost a 37% of the initial one.

6 0

2 years ago