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2 years ago

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While looking at calcium (Ca) on the periodic table, a student needs to find an element with a greater atomic mass in the same p

eriod. Where should the student look?

2 answers:

inessss [21]2 years ago

7 0

Directly to the right :)

stiks02 [169]2 years ago

5 0

To the right of calcium in the same period.

Hope this helps!

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A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? A flat metal washe

jenyasd209 [6]

answer;

The hole in the center of the washer will expand

explanation;

<em>A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? The hole in the center will remain the same size. Changes in the hole cannot be determined without know the composition of the metal. The hole in the center of the washer will expand. The hole in the center of the washer will contract.</em>

this is an example of area expansivity.

coefficient of area expansivity is change in area per area per degree rise in temperature

a=dA/(A*dt)

as the temperature rises , there will be volumetric and area expansivity on the body. volume also increases because of the intermolecular forces of attraction between the molecule is now getting apart.

7 0

2 years ago

A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .

nevsk [136]

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

v is image distance

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm$

Magnification,

$m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3$

So, the magnification of the lens is 3.

3 0

2 years ago

A 65 kg students is walking on a slackline, a length of webbing stretched between two trees. the line stretches and sags so that

polet [3.4K]

Answer : Tension in the line = 936.7 N

Explanation :

It is given that,

Mass of student, m = 65 kg

The angle between slackline and horizontal, $\theta=20^0$

The two forces that acts are :

(i) Tension

(ii) Weight

So, from the figure it is clear that :

$mg=2T\ sin\ \theta$

$T=\dfrac{mg}{2\ sin\theta}$

$T=\dfrac{65\ kg\times 9.8\ m/s^2}{2(sin\ 20)}$

$T=936.7\ N$

Hence, this is the required solution.

5 0

2 years ago

Read 2 more answers

To practice Problem-Solving Strategy 10.1 for energy conservation problems. A sled is being held at rest on a slope that makes a

Gwar [14]

Answer:

μk = (d1)sin(θ) / [(cosθ)(d1) + (d2)]

Explanation:

To solve this, let's use the work/energy theorem which states that: The change in an object's Kinetic energy is equal to the total work (positive and/or negative) done on the system by all forces.

Now, in this question, the change in the object's KE is zero because it starts at rest and ends at rest. (ΔKE = KE_final − KE_initial = 0). Thus, it means the sum of the work, over the whole trip, must also be zero.

Now, if we consider the work done during the downhill slide,there will be three forces acting on the sled:

1. Weight (gravity). This force vector has magnitude "mg" and points points straight down. It makes an angle of "90°–θ" with the direction of motion. Thus;

Wgrav = (mg)(d1)cos(90°–θ)

From trigonometry, we know that cos(90°–θ) = sinθ, thus:

Wgrav = (mg)(d1)sin(θ)

2. Normal force, Fn=(mg)cosθ. This force vector is perpendicular to the direction of motion, so it does zero work.

3. Friction, Ff = (Fn)μk = (mg) (cosθ)μk and it points directly opposite of the direction of motion,

Thus;

Wfric = –(Fn)(d1) = –(mg)(cosθ)(μk)(d1)

(negative sign because the direction of force opposes the direction of motion.)

So, the total work done on the sled during the downhill phase is:

Wdownhill = [(mg)(d1)sin(θ)] – [(mg)(cosθ)(μk)(d1)]

Now, let's consider the work done during the "horizontal sliding" phase. The forces here are:

1. Gravity: it acts perpendicular to the direction of motion, so it does zero work in this phase.

2. Normal force, Fn = mg. It's also perpendicular to the motion, so it also does zero work.

3. Friction, Ff = (Fn)(μk) = (mg)(μk). Thus; Wfric = –(mg)(μk)(d2) (negative because the direction of the friction force opposes the direction of motion).

The total work done during this horizontal phase is:

Whoriz = –(mg)(μk)(d2)

Hence, the total work done on the sled overall is:

W = Wdownhill + Whoriz

= (mg)(d1)sin(θ) – (mg)(cosθ)(μk)(d1) – (mg)(μk)(d2)

I have deduced that the total work is zero (because change in kinetic energy is zero), thus;

(mg)(d1)sin(θ) – (mg)(cosθ)(μk)(d1) – (mg)(μk)(d2) = 0

Now, let's make μk the subject of the equation:

First of all, divide each term by mg;

(d1)sin(θ) – (cosθ)(μk)(d1) – (μk)(d2) = 0

Rearranging, we have;

(d1)sin(θ) = (cosθ)(μk)(d1) + (μk)(d2)

So,

(d1)sin(θ) = [(cosθ)(d1) + (d2)](μk)

And

μk = (d1)sin(θ) / [(cosθ)(d1) + (d2)]

5 0

2 years ago

A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle

liq [111]

Answer:

0.0031792338 rad/s

Explanation:

$\theta$ = Angle of elevation

y = Height of balloon

Using trigonometry

$tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta$

Differentiating with respect to t we get

$\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta$

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

$h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft$

Now y = 2500 ft

$cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974$

$\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s$

The angle is changing at 0.0031792338 rad/s

6 0

2 years ago