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3 years ago

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A piece having a length of 4.0 cm was cut from a much longer, uniform rod. The piece has a volume of 3.0 cm3 and a mass of 24 g.

Suppose another piece from the same rod is four times as long. What is its mass in grams?

1 answer:

Gre4nikov [31]3 years ago

6 0

Answer: 96

Explanation:

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Doyle learned that only one side of the Moon is ever visible from Earth. He wants to investigate how the moon moves around the E

alisha [4.7K]

Your answer is C. Development of a model

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3 years ago

Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t

salantis [7]

Explanation:

Mass of bumper cars, $m_1=m_2=110\ kg$

Initial speed of car A, $u_1=8\ m/s$

Initial speed of car Z, $u_2=-10\ m/s$

Final speed of car A after the collision, $v_1=-10\ m/s$

We need to find the velocity of car Z after the collision. Let it is equal to $v_2$ . Using the conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$110\times 8+110\times (-10)=110\times (-10)+110v_2$

$v_2=\dfrac{-1320}{110}\ m/s$

$v_2=-12\ m/s$

So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.

5 0

3 years ago

What is the weight of a 5.00 kg object on Earth? Assume g=9.81 m/s^2.

Softa [21]

<em>weight = (mass) x (gravity)</em>

Weight = (5.00 kg) x (9.81 m/s²)

weight = (5.00 x 9.81) (kg-m/s²)

<em>Weight = 49.05 Newton</em>

7 0

3 years ago

A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium p

saul85 [17]

Answer:

0.67 s

Explanation:

This is a simple harmonic motion (SHM).

The displacement, $x$ , of an SHM is given by

$x = A\cos(\omega t)$

A is the amplitude and $\omega$ is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or $\frac{\pi}{4}$ radian.

From trigonometry, $\sin A =\cos B$ if A and B are complementary.

At $t = 0$ , $x = 3.5$

$3.5 = A\cos(\omega \times0)$

$A =3.5$

So

$x = 3.5\cos(\omega t)$

At $t = 0.12$ , $x = 1.5$

$1.5 = 3.5\cos(0.12\omega)$

$\cos(0.12\omega)=\dfrac{1.5}{3.5}=0.4286$

$0.12\omega =\cos^{-1}0.4286$

$0.12\omega = 1.13$

$\omega = 9.4$

The period, $T$ , is related to $\omega$ by

$T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67$

5 0

3 years ago

Calcite (CaCO_3) is a crystal with abnormally large birefringence. The index of refraction for light with electric field paralle

katrin2010 [14]

Answer:

(a) 42.28°

(b) 37.08°

Explanation:

From the principle of refraction of light, when light wave travels from one medium to another medium, we have:

$\frac{n_{b} }{n_{a} }$ = sinθ $_{a}$ /sinθ $_{b}$

In the given problem, we are given the refractive indices of light which are parallel and perpendicular to the axis of the optical lens as 1.4864 and 1.6584 respectively.

For critical angle θ $_{a}$ = θ $_{c}$ , θ $_{b}$ = 90°; $n_{b} = 1$

(a) $n_{a} = 1.4864$

$\frac{1 }{1.4864 }$ = sinθ $_{c}$ /sin90°

0.6728 = sinθ $_{c}θ[tex]_{c} = sin^(-1) 0.6728 = 42.28°(b) [tex]n_{a} = 1.6584$

$\frac{1 }{1.6584}$ = sinθ $_{c}$ /sin90°

0.60299 = sinθ[tex]_{c}

θ[tex]_{c} = sin^(-1) 0.60299 = 37.08°

7 0

3 years ago