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2 years ago

Which statements describe how chemical formulas, such as H2O, represent compounds? Check all that apply.

Physics

2 answers:

Pani-rosa [81]2 years ago

5 0

The statements that apply in this case are:

They show the elements that make up a compound.

They show the types of atoms that make up a molecule.

They show the number of each type of atom in a molecule.

posledela2 years ago

3 0

Answer:its the second one

Explanation:

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. A girl runs and jumps horizontally off a platform 10m above a pool with a speed of 4.0m/s. As soon as she leaves the platform,

faust18 [17]

Answer:

2.39 revolutions

Explanation:

As she jumps off the platform horizontally at a speed of 10m/s, the gravity is the only thing that affects her motion vertically. Let g = 10m/s2, the time it takes for her to fall 10m to water is

$h = gt^2/2$

$10 = 10t^2/2$

$t^2 = 2$

$t = \sqrt{2} = 1.414 s$

Knowing the time it takes to fall to the pool, we calculate the angular distance that she would make at a constant acceleration of 15 rad/s2:

$\theta = \alpha t^2/2$

$\theta = 15 * 2/2 = 15 rad$

As each revolution is 2π, the total number of revolution that she could make is: 15 / 2π = 2.39 rev

3 0

2 years ago

Disturbed by speeding cars outside his workplace, Nobel laureate Arthur Holly Compton designed a speed bump (called the "Holly h

Bezzdna [24]

: 30.50 km/h = 30.50^3 m / 3600s = 8.47 m/s

At the top of the circle the centripetal force (mv²/R) comes from the car's weight (mg)

So, the net downward force from the car (Fn) = (weight - centripetal force) .. and by reaction this is the upward force provided by the road ..

Fn = mg - mv²/R
Fn = m(g - v²/R) .. .. 1800kg (9.80 - 8.47²/20.20) .. .. .. ►Fn = 11 247 N (upwards)
(b)
When the car's speed is such that all the weight is needed for the centripetal force .. then the net downward force (Fn), and the reaction from the road, becomes zero.

ie .. mg = mv²/R .. .. v² = Rg .. .. 20.20m x 9.80 = 198.0(m/s)²

►v = √198 = 14.0 m/s

3 0

2 years ago

Slick Willy is in traffic court (again) contesting a $50.00 ticket for running a red light. "You see, your Honor, as I was appro

Masteriza [31]

Answer:

61578948 m/s

Explanation:

λ $_{actual}$ = λ $_{observed}$ $\frac{c+v_{o}}{c}$

687 = 570 $(\frac{3 * 10^{8} +v_{o} }{3 * 10^{8}} )$

$v_{o}$ = 61578948 m/s

So Slick Willy was travelling at a speed of 61578948 m/s to observe this.

8 0

2 years ago

Read 2 more answers

The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f

zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

584 = Ha + Lg

an also this mean that torques acting on the body is balanced

So, 0.410 Ha = 0.840 Lg

Making Lg the subject of formula in the equation above we

Lg = 0.4881 Ha

Considering the first equation and replacing Lg with this recent equation we have

584 = Ha + 0.4881 Ha

Therefore Ha = 392.44 N

This value obtained is for both hands for each hand we divide by 2

Therefore we have for each hand $= 392.44/2$ =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

Lg = 0.4881 × 392.44

= 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

191.22/2 = 95.8 N.

8 0

2 years ago

The position of a particle moving along the x axis may be determined from the expression x(t) = btu + ctv, where x will be in me

KIM [24]

As per given equation we have

$x = bt^u + ct^v$

now as per the dimensional analysis we can say that dimension of right side of equation must be equal to left side of the equation

now as per left side of equation its dimension is same as length or meter

now we can say it should be meter on right side also

$bt^u = M^0L^1T^0$

$b*T^8 = M^0L^1T^0$

$b = M^0L^1T^{-8}$

similarly for other term we have

$ct^v = M^0L^1T^0$

$c*T^7 = M^0L^1T^0$

$c = M^0L^1T^{-7}$

so above are the dimensions of b and c

8 0

1 year ago