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3 years ago

Please solve this for 15 points please dont put in a link.

Physics

2 answers:

Luda [366]3 years ago

7 0

Answer:

Explanation:

The reactant, 2H2O, has been divided into its elements in the product, which are 2H2 and O2.

s2008m [1.1K]3 years ago

6 0

The answer to your question is E

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A projectile is fired with an initial speed of 40 m/s at an angle of elevation of 30∘. Find the following: (Assume air resistanc

BabaBlast [244]

Answer:

a. 2.0secs

b. 20.4m

c. 4.0secs

d. 141.2m

e. 40m/s, ∅= -30°

Explanation:

The following Data are giving

Initial speed U=40m/s

angle of elevation,∅=30°

a. the expression for the time to attain the maximum height is expressed as

$t=\frac{usin\alpha }{g}$

where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at

$t=40sin30/9.81\\t=2.0secs$

b. the expression for the maximum height is expressed as

$H=\frac{u^{2}sin^{2}\alpha }{2g} \\H=\frac{40^{2}0.25 }{2*9.81} \\H=20.4m$

c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,

Hence T=2t

T=2*2.0

T=4.0secs

d. The range of the projectile is expressed as

$R=\frac{U^{2}sin2\alpha}{g}\\R=\frac{40^{2}sin60}{9.81}\\R=141.2m$

e. The landing speed is the same as the initial projected speed but in opposite direction

Hence the landing speed is 40m/s at angle of -30°

3 0

3 years ago

What type of light can go through aluminum foil

EleoNora [17]

Answer:

DmxmxmdmdExplanation:sejwjsjskdkdkdekskekememd

8 0

3 years ago

Solve using correct significant figures and indicating maximum absolute uncertainty.

Vera_Pavlovna [14]

We use the criterion of significant figures to find the result with reliable figures

X = 9.2 10-5

now with the propagation of errors we obtain the result with its uncertainty

X ± ΔX = (9.2 ± 0.5) 10⁻⁵

given Parameter

* expression values with their absolute errors

to find

* the result with the correct significant figures

* the absolute error of the expression

Significant figures are defined with the number of decimals that give information, the number of figures in a quantity gives information about the uncertainty of this quantity.

There are two criteria for applying significant figures:

* Add and subtract the result of going with the number of decimal places of the figure that has the least

* Product and division as a result of going with the least number of significant figures than the value that has the least.

Remember that the zero to the left do not form a pair of the significant figures

Let's apply this belief to the case presented, let's write the precaution

$x = \frac{a-b}{c}$

where in this case they are worth

a = 0.0336 ± 0.0002

b = 0.010 ± 0.001

c = 255.4 ± 0.4

We see that the significant figures of each parameterize (a, b, c) and their absolute errors are correct.

Let's apply the criteria to the operation

a-b = 0.0336 - 0.010

a- b = 0.0236

we apply the criterion of significant figures for the subtraction, the result must be with 3 decimal places

a - b = 0.024

let's do the other operation

X = $\frac{a-b}{c}$

X = 0.024 / 255.4

X = 9.24 10⁻⁵

We apply the criterion of significant figures for the division, in this case the result is left with two significant figures

X = 9.2 10⁻⁵

The uncertainty or error of the measurements is of most importance as it determines how many significative figures are reliable at a given magnitude.

If the magnitudes are measured with some type of instrument, the absolute error is given by the appreciation of the instrument, if the magnitude is calculated using some equation, the errors must be propagated using the variations of each parameter in the worst case.

the uncertainty of the calculated quantity (X) is

$\Delta X = | \frac{dX}{da}| \Delta a + | \frac{dX}{db} | \Delta b + | \frac{dX}{dc}| \Delta c$

let's perform the derivatives

$\frac{dX}{da} = \frac{1}{c}$

$\frac{dX}{db} = - \frac{1}{c}$

$\frac{dX}{dc} = - \frac{a-b}{c^2}$

we substitute

remember that the bulk value guarantees that we tune the worst case. So all the mistakes add up

ΔX = $\frac{1}{c}$ Δa + $\frac{1}{c}$ Δb + $\frac{a-b}{c^2}$ Δc

ΔX = $\frac{1}{c}$ (Δa + Δb) + $\frac{a-b}{c^2}$ Δc

we substitute

ΔX = $\frac{1}{255.4}$ (0.0002 + 0.001) + $\frac{0.0336-0.010}{255.4^2}$ 0.4

ΔX = 4.698 10⁻⁶ + 1.45 10⁻⁷

DX = 4.8 10-6

Absolute errors must be given with a single significant figure

ΔX = 5 10⁻⁶

The result of the requested quantity using the criterion of significant figures and propagation of errors is

X ± ΔX = (9.2 ± 0.5) 10⁻⁵

learn more about significative figure here:

brainly.com/question/18955573

8 0

3 years ago

At each corner of a square of side there are point charges of magnitude Q, 2Q, 3Q, and 4Q

Bad White [126]

Answer:

$\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}$

$\displaystyle \theta =68^o$

Explanation:

Electrostatic Force

It's the force that appears between two electrical charges q1 q2 when they are placed at a certain distance d. The force can be computed by using the Coulomb's law:

$\displaystyle F=\frac{KQ_1Q_2}{d^2}$

We have an arrangement of 4 charges as shown in the image below. We need to calculate the total force exerted on the charge 2Q by the other 3 charges. The free body diagram is also shown in the second image provided. The total force on 2Q is the vectorial sum of F1, F2, and F3. All the forces are repulsive, since all the charges have the same sign. Let's compute each force as follows:

$\displaystyle |F_1|=\frac{KQ(2Q)}{l^2}=\frac{2KQ^2}{l^2}$

$\displaystyle |F_2|=\frac{K(2Q)(4Q)}{l^2}=\frac{8KQ^2}{l^2}$

The distance between 3Q and 2Q is the diagonal of the rectagle of length l:

$\displaystyle |d_3|=\sqrt{l^2+l^2}=\sqrt{2}\ l$

The force F3 is

$\displaystyle |F_3|=\frac{K(3Q)(2Q)}{(\sqrt{2l)}^2}=\frac{3KQ^2}{l^2}$

Each force must be expressed as vectors. F1 is pointed to the right direction, thus its vertical components is zero

$\displaystyle \vec{F_1}=\left \langle |F_1|,0 \right \rangle=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle$

F2 is pointed upwards and its horizontal component is zero

$\displaystyle \vec{F_2}=\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle$

F3 has two components because it forms an angle of 45° respect to the horizontal, thus

$\displaystyle \vec{F_3}=\left \langle \frac{3KQ^2}{l^2}\ cos45^o,\frac{3KQ2}{l^2} sin45^o\right \rangle$

$\displaystyle \vec{F_3}=\left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

Now we compute the total force

$\displaystyle \vec{F_t}=\vec{F_1}+\vec{F_2}+\vec{F_3}$

$\displaystyle \vec{F_t}=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle +\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle + \left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

$\displaystyle \vec{F_t}=\left \langle \left(2+\frac{3\sqrt{2}}{2}\right)\frac{KQ^2}{l^2},\left(8+\frac{3\sqrt{2}}{2}\right) \frac{KQ^2}{l^2}\right \rangle$