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Answer:

Endothermic, because the reactants are lower in energy (C)

Explanation:

From the graph, you can see the energy of the products is higher than the energy of the reactants. If you recall that  when the enthalpy change  Eproducts is gretater than  Ereactants, the reaction is said to be endothermic.

Answer:

I hear points of low volume sound and points of high volume of sound.

Explanation:

This is because, since the two sources of sound have the same frequency and are separated by a distance, d = 10 mm, there would be successive points of constructive and destructive interference.

Since their frequencies are similar, we should have beats of high and low frequency.

So, at points of low frequency, the amplitude of the wave is smallest and there is destructive interference. The frequency at this point is the difference between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, f - f' = 400 Hz - 400 Hz = 0 Hz. So, the volume of the sound is low(zero) at these points.

Also, at points of high frequency, the amplitude of the wave is highest and there is constructive interference. The frequency at this point is the sum between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, (f + f') = 400 Hz + 400 Hz = 800 Hz. So, the volume of the sound is high at these points.

So, as you wander around the room, I should hear points of high and low sound across the room.

Answer:

d = 29.89 m

Explanation:

To solve this, we need to separate this problem in two parts.

One part would be the the time taken by the rock to actually hit the bottom, and the other part would be the time taken by the sound to reach the inspector.

Joining these two times we have:

t = t₁ + t₂    (1)

This time is 2.56 s.

Now, as we are asked to determine the distance from the top floor to the bottom, and we have two times taken in different ways, one by sound and the other the actual, we can say the same thing on distance, we need a distance relationed to the time taken by rock to hit the bottom, and the other distance relationet to the time taken by sound to reach the inspector.

Doing this we have that the distance traveled by the rock is:

y₁ = gt²/2

y₁ = 9.8t²/2 = 4.9t₁²    (2)

Now, the distance traveled by sound would be:

y₂ = v * t₂ = 336t₂   (3)

Remember that the speed of the sound is 336 m/s

From this last expression (3), we can actually write t₂ in function of t₁, using (1):

2.56 = t₁ + t₂

t₂ = 2.56 - t₁    (4)

Replacing (4) in (3):

y₂ = 336(2.56 - t₁)    (5)

Now that we have y₁ and y₂, we can equal (2) and (5), both expressions to get the value of t₁, and then, calculate the distance:

4.9t₁² = 336(2.56 - t₁)

4.9t₁² = 860.16 - 336t₁

4.9t₁² + 336t₁ - 860.16 = 0

Using the quadractic formula, we can calculate t₁:

t₁ = -336 ±√(336)² + 4*4.9*860.16 / (2*4.9)

t₁ = -336 ±√129,7555.136 / 9.8

t₁ = -336 ± 360.21 / 9.8     Using only the positive value we have:

t₁ = 2.47 s

This means that the rock hits the bottom in 2.47 s, and the remaining 0.09 s belongs to the time taken by sound. (2.47 + 0.09 = 2.56 s)

With this, we can calculate the distance of the rock using expression (2):

y₁ = 4.9 * (2.47)²

Hope this helps

Solution :

The volume rate of flow is given by : R = 5.0 L/s

                                                                 [tex]$ = 5.0 \times 10^{-3} \ m^3/s$[/tex]

The radius of the pipe, [tex]$r_1= 5 \times 10^{-2} \ m$[/tex]

∴ [tex]$ 5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]

then, [tex]$v_1 = \frac{5.0 \times 10^{-3}}{(3.14)(5 \times 10^{-2})^2}$[/tex]

             = 0.637 meter per second

Then the speed of the water at wider section,

[tex]$R=A_1v_1$[/tex]

Similarly, the speed of water at narrow pipe.

The radius of the [tex]$r_2 = 2.5 \times 10^{-2}$[/tex] m

[tex]$5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]

then, [tex]$v_2 = \frac{5.0 \times 10^{-3}}{(3.14)(2.5 \times 10^{-2})^2}$[/tex]

             = 2.55 meter per sec

Now from Bernoulli's theorem,

[tex]$P_1 + \frac{1}{2} \rho v_1^2 =P_2 + \frac{1}{2} \rho v_2^2 $[/tex]

[tex]$P_1 = P_2 + \frac{1}{2} \rho (v_2^2 - v_1^2)$[/tex]

    [tex]$= 50 \kPa + (0.5)(10^3)[(2.55)^2-(0.637)^2]$[/tex]

    = 50 kPa + 3.05 kPa

    = 53.05 kPa

or 53000 Pa

This question involves the concepts of Bernoulli's Theorem and Volumetric Flowrate.

The pressure reading in the wide section is "53.05 KPa".

First, we will use the volumetric flow rate to find the velocities of the water at wide and narrow sections.

[tex]V = A_1v_1[/tex]

where,

V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s

r₁ = radius of narrow section = 5 cm/2 = 2.5 cm = 0.025 m

A₁ = Area of narrow section = πr₁² = π(0.025 m)²

v₁ = velocity at narrow section = ?

Therefore,

[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.025\ m)^2](v_1)\\\\v_1=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.025\ m)^2}\\\\v_1=2.55\ m/s\\[/tex]

Similarly,

[tex]V = A_2v_2[/tex]

where,

V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s

r₂ = radius of wide section = 10 cm/2 = 5 cm = 0.05 m

A₂ = Area of wide section = πr₁² = π(0.05 m)²

v₂ = velocity at wide section = ?

Therefore,

[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.05\ m)^2](v_2)\\\\v_2=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.05\ m)^2}\\\\v_2=0.64\ m/s\\[/tex]

Now, we will use Bernoulli's Theorem to find out the pressure wide section.

[tex]P_1 + \frac{1}{2}\rho v_1^2=P_2 + \frac{1}{2}\rho v_2^2[/tex]

where,

[tex]\rho[/tex] = density of water = 1000 kg/m³

P₁ = pressure in narrow section = 50 KPa = 50000 Pa

P₂ = pressure in wide section = ?

Therefore,

[tex]50000\ Pa + \frac{1}{2}(1000\ kg/m^3)(2.55\ m/s)^2=P_2 + \frac{1}{2}(1000\ kg/m^3)(0.64\ m/s)^2[/tex]

P₂ = 50000 Pa + 3251.25 Pa - 204.8 Pa

P₂ = 53046.45 Pa = 53.05 KPa

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Answer:

1.70 m/s

Explanation:

The computation of the final speed of the couple is shown below:

initial momentum of A is

= mv

= 68.2 × 2.48

= 169.136 kg

And, the initial momentum of B is

= mv

= 34.4 × 1.18

= 40.592 kg

Now magnitude is

= sqrt( A^2 + B^2)

= sqrt( 28,606.99 + 1,647.71)

= 173.94

Now the final speed is

= 173.94 ÷ (68.2 + 34.4)

= 1.70 m/s

Answer:

amplitude is the answer

Answer:

10 J.

Explanation:

Given that,

Net force acting on the rake, F = 2 N

Distance moved by the rake, d = 5 m

We need to find the kinetic energy gained by the rake. We know that,

Kinetic energy = work done

So,

K = F×d

K = 2 N × 5 m

K = 10 J

So, 10 J of kinetic energy is gained by the rake.

Violet pulls a rake horizontally on a frictionless driveway with a net force of 2.0 N for 5.0 m.

How much kinetic energy does the rake gain?

Answer: 10 J

Answer:

The initial velocity in the x-direction with which the branch was thrown is approximately 10.224 m/s

Explanation:

The given parameters of the motion of the branch are;

The height from which the branch is thrown = 3.00 m

The horizontal distance the branch lands from where it was thrown, x = 8.00 m

The direction in which the branch is thrown = Horizontally

Therefore, the initial vertical velocity of the branch, [tex]u_y[/tex] = 0 m/s

The time it takes an object in free fall (zero initial downward vertical velocity) to reach the ground is given as follows;

s = [tex]u_y[/tex]·t + 1/2·g·t²

Where;

[tex]u_y[/tex] = 0 m/s

s = The initial height of the object = 3.00 m

g = The acceleration due to gravity = 9.8 m/s²

∴ s = 0·t + 1/2·g·t² = 0 × t + 1/2·g·t² = 1/2·g·t²

t = √(2·s/g) = √(2 × 3/9.8) = (√30)/7 ≈ 0.78246

The horizontal distance covered before the branch touches the ground, x = 8.00 m

Therefore, the initial velocity in the horizontal, x-direction with which the branch was thrown, 'uₓ', is given as follows;

uₓ = x/t = 8.00 m/((√30)/7 s)

Using a graphing calculator, we get;

uₓ = 8.00 m/((√30)/7 s) = (28/15)·√30 m/s ≈ 10.224 m/s

The initial velocity in the horizontal, x-direction with which the branch was thrown, uₓ ≈ 10.224 m/s.

Answer:

a)    F₁ = 267.3 N,   N₁ = 1300 N,  b)    μ = 0.324

Explanation:

For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces

let's use subscript 1 for the ladder and 2 for the firefighter

            ∑ τ = 0

          -W₁ x₁ - W₂ x₂ + N₁ y = 0

           N₁ = [tex]\frac{W_1 x_1 + W_2 x_2}{y}[/tex]          (1)

the center of mass of the ladder is at its geometric center,

d = L / 2 = 15/2 = 7.5 m

         cos 60 = x₁ / d₁

         x₁ = d₁ cos 60

         x₁ = 7.5 cos 60

         x₁ = 3.75 m

for the firefighter d₂ = 4 m

         cos 60 = x₂ / d₂

         x₂ = d₂ cos 60

          x₂ = 4 cos 60 = 2 m

for the fulcrum d₃ = 15 m

         sin 60 = y / d₃

         y = d₃ sin 60

         y = 15 sin 60

         y = 13 m

we look for the Normal by substituting in equation 1

         N₂ = [tex]\frac{500 \ 3.75 \ + 800 \ 2}{13}[/tex]

         N₂ = 267.3 N

now let's use the translational equilibrium relations

 X axis

           F₁ - N₂ = 0

           F₁ = N₂

           F₁ = 267.3 N

Axis y

          N₁ - W₁ -W₂ = 0

          N₁ = W₁ + W₂

          N₁ = 500 + 800

          N₁ = 1300 N

b) for this case change the firefighter's distance d₂ = 9 m

          x₂ = 9 cos 60

          x₂ = 4.5 m

we substitute in 1

          N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}  

          N₂ = 421.15 N

of the translational equilibrium equation on the x-axis

          fr = F₁ = N₂

          fr = 421.15 N

friction force has the expression

          fr = μ N

in this case the reaction of the Earth to the support of the ladder is N1 = 1300N

          μ = fr / N₁

          μ = 421.15 / 1300

          μ = 0.324

Time taken : 4.77 s

Given

a = 13 m/s²

vf = final velocity = 62 m/s

Required

time taken

Solution

An equation of uniformly accelerated motion  

[tex]\large {\boxed {\bold {x=xo+vo.t+\frac {1} {2} at ^ 2}}}[/tex]

vf = vo + at  

vf² = vo² + 2a (x-xo)  

x = distance on t  

vo / vi = initial speed  

vt / vf = speed on t / final speed  

a = acceleration  

Input the value :

vo =0 ⇒from rest

vf=at

62 m/s = 13 m/s².t

t = 62 : 13

t = 4.77 s

Answer:45 m/s north

Explanation:

Answer:The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would..

Explanation: It would decrease.

The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would decrease because the acceleration due by gravity on the moon is less than the acceleration due to gravity on the earth, therefore the correct answer is option C.

It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another.

In comparison to the Earth, the Moon is less massive. Your weight would drop if you traveled to the moon because the acceleration caused by gravity there is lower than that caused by gravity here on Earth.

As a result of the less gravity on the moon, the weight would decrease.

Thus, the correct answer is option C.

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Answer:

0.392

Explanation:

Mm = 0.11Me

Rm = 0.53Re

g = GM / r^2

G = 6.67 * 10^-11

gmars = (G * 0.11Me) / (0.53Re)^2

Recall:

gearth = GMe /Re^2

Hence, gmars in terms of gearth equals

gmars = gearth * (0.11 / 0.53^2)

gmars = gearth * 0.3915984

gmars = 0.392gearth

Answer:

Only the car transforms electrical energy into more than one form of energy.

Explanation:

The motion of the car is mechanical energy but it can also transform into another energy witch is electrical energy

Answer:

A

Explanation:

Answer:

27872.2 miles per hour

Explanation:

Given that :

Radius of Jupiter is approximately = 44,360 miles

Revolution is 10 hours ;

Jupiter makes one revolution in 10 hours :

Using the relation to obtain the velocity :

V = re

r = radius

w = 2π/T

Hence,

V = r * 2π/ T

V =44360 * 2 * π/10

V = 88720 * π/10

V = 278722.10 / 10

V = 27872.210

V = 27872.2 miles per hour

Answer:

am not sure about the answer

Explanation:

you need to find out the amount of force it's going in for example 10n or 100n then you need to times it the distance then devide by the time

Answer:

14.1seconds approx

Explanation:

Given data

Distance= 59.1m

Your velocity= 2.35m/s

Walkway velocity= 1.85m/s

Total velocity= 2.35+1.85= 4.2m/s

We know that

Velocity= distance/time

time= distance/velocity

substitute

time= 59.1/4.2

time= 14.07

time=14.1seconds approx

Hence the time is 14.1seconds approx

Answer:

a) v₀ₓ = 62.76 m / s, b)   θ₁ = 17.6º,   θ₂ = 67.0º

Explanation:

We can solve this exercise using the projectile launch ratios

a) Let's find the time it takes for the bullet to reach the water level

       y = y₀ + v_{oy} t - ½ g t²

when it reaches the water its height is zero y = 0, as the bullet is fired horizontally its initial vertical velocity is zero

       0 = y₀ + 0 - ½ g t²

       t =[tex]\sqrt{2y_o/g}[/tex]

       t = [tex]\sqrt{2 \ 7 /9.8}[/tex]          

       t = 1,195 s

now we can calculate the speed with the horizontal movement

        x = v₀ₓ t

        v₀ₓ = x / t

        v₀ₓ = 75.0 / 1.195

        v₀ₓ = 62.76 m / s

b) if the speed of the bullets is half of that found

         v₀ = 62.76 / 2 = 31.38 m / s

let's write the expressions for the distance

          x = v₀ cos θ t

          y = y₀ + v_{oy} sin θ t - ½ g t²

          t = [tex]\frac{x}{v_o \ cos \theta}[/tex]

we substitute

          [tex]0 = y_o + v_o sin \theta \ \frac{x}{v_o \cos \thetay} - 1/2 g \ (\frac{x}{v_o \ cos \theta})^2[/tex]

          [tex]0 = y_o + x tan \theta - \frac{1}{2} g \ \frac{x^2}{ v_o^2 \ cos^2 \theta}[/tex]    

let's use the identified trigonometry

          sec² θ = 1 + tan² θ

         sec θ = 1 / cos θ

we substitute

          [tex]0 = y_o + x tan \theta - \frac{g x^2}{2 v_o^2} ( 1 + tan^2 \theta)[/tex]

          [tex]\frac{g x^2}{2v_o^2} tan^2 \theta - x tan \theta + \frac{gx^2}{2v_o^2} - y_o = 0[/tex]

we change variable

         tan θ = H

         [tex]\frac{gx^2}{2 v_o^2 } H^2 - x H + \frac{gx^2}{2v_o^2}-y_o =0[/tex]

we subtitle the values

         [tex]\frac{9.8 \ 75^2}{2 \ 31.38^2} H^2 - 75 H + \frac{9.8 \ 75^2}{2 \ 31.38^2}-7 =0[/tex]

         27.99 H² - 75 H + 20.99 = 0

         H² - 2.679 H + 0.75 = 0

we solve the quadratic equation

         H = [2.679 ± [tex]\sqrt{2.679^2 - 4 0.75}[/tex]] / 2

         H = [2,679 ± 2,044] / 2

         H₁ = 0.3175

         H₂ = 2.3615

now we can find the angles

          H₁ = tan θ₁

          θ₁ = tan⁻¹ H₁

          θ₁ = tan⁻¹ 0.3175

          θ₁ = 17.6º

          θ₂ = 67.0º

for these two angles the bullet hits the boat

Answer:

t = 2.01 s

Vf = 19.7 m/s

Explanation:

It's know through the International System that the earth's gravity is 9.8 m/s², then we have;

Data:

==================================================================

Time

Use formula:

Replace:

Everything inside the root is solved first. So, we solve the multiplication of the numerator:

It divides:

The square root is performed:

==================================================================

Final Velocity

use formula:

Replace:

Multiply:

==================================================================

How long does it take to reach the ground?

Takes time to reach the ground in 2.01 seconds.

How fast does it hit the ground?

Hits the ground with a speed of 19.7 meters per seconds.

The car's rate of  acceleration : a = 2.04 m/s²

Given

speed = 110 km/hr

time = 15 s

Required

The acceleration

Solution

110 km/hr⇒30.56 m/s

Acceleration is the change in velocity over time

a = Δv : Δt

Input the value :

a = 30.56 m/s : 15 s

a = 2.04 m/s²

Answer:

3

Explanation:

my family i hope thinks of me.  And I don't have friends for them to think of me.

Answer:

Length, l = 0.126 meters.

Explanation:

Given the following data;

Period = 6.98s

Acceleration due to gravity, g = 9.8m/s²

To find the length, l;

[tex] Period, T = 2 \pi \sqrt {lg} [/tex]

Substituting into the equation, we have;

[tex] 6.98 = 2*3.142 \sqrt {l*9.8} [/tex]

[tex] 6.98 = 6.284 \sqrt {9.8l} [/tex]

[tex] \frac {6.98}{6.284} = \sqrt {9.8l} [/tex]

[tex] 1.1108 = \sqrt {9.8l} [/tex]

Taking the square of both sides

[tex] 1.1108^{2} = 9.8l [/tex]

[tex] 1.2339 = 9.8l [/tex]

[tex] l = \frac {1.2339}{9.8} [/tex]

Length, l = 0.126m.

Answer:

Postural deviations can cause poor balance, muscle pain and skeletal stress.

Explanation:

The mass percent of hydrogen in CH₄O is 12.5%.

Mass percent is the mass of the element divided by the mass of the compound or solute.

Step 1: Calculate the mass of the compound.

mCH₄O = 1 mC + 4 mH + 1 mO = 1 (12.01 amu) + 4 (1.00 amu) + 1 (16.00 amu) = 32.01 amu

Step 2: Calculate the mass of hydrogen in the compound.

mH in mCH₄O = 4 mH = 4 (1.00 amu) = 4.00 amu

Step 3: Calculate the mass percent of hydrogen in the compound.

%H = (mH in mCH₄O / mCH₄O) × 100%

%H = 4.00 amu / 32.01 amu × 100% = 12.5%

The mass percent of hydrogen in CH₄O is 12.5%.

CO2 = 1.580 grams H2O = 0.592 grams Lookup the molar mass of each element in the compound Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Calculate the molar mass of CH4O by adding the total masses of each element used. 12.0107 + 4 * 1.00794 + 15.999 = 32.04146 Now calculate how many moles of CH4O you have by dividing by the molar mass. m = 1.15 g / 32.04146 g/mole = 0.035891 mole Now figure out how many moles of carbon and hydrogen you have. Carbon = 0.035891 moles Hydrogen = 0.035891 moles *

Therefore, The mass percent of hydrogen in CH₄O is 12.5%.

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Answer:

The answer is a 4.2m

Explanation:

Given data

Please see attached the rough drawing for your reference.

From the drawing, you ran 18m west and 2.4m south

The displacement is

= 1.8+2.4

=4.2m

Answer:

mas

Explanation:

mass is the amount of space something occupies.

Answer:

nique ta mama  

Explanation:

Answer:

A. Doppler ultrasound collects data from moving objects

Explanation:

Did the test !!

Answer:A. Doppler ultrasound collects data from moving objects.

Explanation: just got it right

on my test