A 0.55 kg projectile is launched from the edge of a cliff with an initial kinetic energy of 1550 J and at its highest point is 140 m above the launch point. (a) Calculate the horizontal component of its velocity. (b) Calculate the vertical component of its velocity just after launch. (c) At one instant during its flight the vertical component of its velocity is found to be 65 m/s. At that time, how far is it above or below the launch point

Answer:

The answer is "39.95 J".

Explanation:

Please find the complete question in the attached file.

[tex]\to W_{AC}=(\mu \ m \ g \ \cos \theta ) d[/tex]

            [tex]=(0.45 \times 1.60 \times 9.8 \times \cos 26^{\circ}) 6.30 \\\\=(7.056 \times \cos 26^{\circ}) 6.30 \\\\=6.34189079\times 6.30\\\\=39.95 \ J\\\\[/tex]

[tex]\therefore \\\\\bold{\Delta E =39.95 \ J}[/tex]

Answer:

Angular momentum

Explanation:

Stars are formed as a result of a collapse of a low-temperature cloud of gas and dust. During the colapse conservation of angular momentum causes any small net rotation of the cloud to increase thus forcing the material into rotating

Answer:

the resistance of the wire has no effect on the brightness of the bulb.

Explanation:

Let's apply ohm's law for your light bulb circuit plus wires plus switch

             V = I R_{bulb} + I R_ {wire}

the current in a series circuit is constant

             V = I (R_{bulb} + R_{wire})

To know the effect of the wires on the brightness of the bulb, we must look for the value of the typical resistance of these elements.

Incandescent bulb

Power 60 W

let's use the power ratio

            P = V I = V2 / R

            R = V2 / P

the voltage value for this power is V = 120 V

            R = 120 2/60

            R_bulb = 240 Ω

Resistance of a 14 gauge copper wire (most used), we look for it on the internet

            R = 8.45 Ω/ km

in a laboratory circuit approximately 2 m is used, so the resistance of our cable is

            R = 8.45 10⁻³ 2

            R_wire = 0.0169 Ω

let's buy the two resistors

            R_{bulb} = 240

            R_{wire} = 0.0169

            [tex]\frac{R_{bulb} }{R_{wire} } = \frac{240 }{ 0.0169}[/tex]

              [tex]\frac{ R_{bulb} }{ R_{wire} } = 1.4 \ 10^4[/tex]

therefore resistance of the bulb is much greater than that of the wire, therefore almost all the power is dissipated in the bulb.

In summary, the resistance of the wire has no effect on the brightness of the bulb.

Answer:

7.55 g

Explanation:

Using the relation :

Δt = temperature change = (6° - 0°) = 6°

Q = quantity of heat

C = specific heat capacity = 4190 j/kg/k

1000 J = 1kJ

333 KJ = 333000 j

The quantity of ice that will melt ;

= 0.419 * 6 * 100 / 333000

= 2514000 / 333000

= 7.549 g

The mass of ice that will melt :

2.514 / 0.333

= 7.549 g

Answer:

extreme heat, because no physical damage can demagnetize a magnet

Explanation:

Answer:

the 3rd one

Explanation:

machinery part :nickel or chromium

ornamentation and decoration pieces :silver and gold

processed food :tin coated iron can

bridges and automobiles :zinc metal

distilled water:bad conductor

Answer:

iron metal :chromium

machinery part :nickel or chromium

ornamentation and decoration pieces :silver and gold

processed food :tin coated iron can

bridges and automobiles :zinc metal

distilled water:bad conductor

Explanation:

Answer:

B as distance increase force decrease, but it is not a linear relationship.

Answer:

v = 28.98 ft / s

Explanation:

For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision

In the exercise they indicate the weight of each car

          Wₐ = 1500 lb

          W_b = 1125 lb

Car B's velocity from v_b = 42.0 mph westward, car A travels east

let's find the mass of the vehicles

             W = mg

             m = W / g

             mₐ = Wₐ / g

             m_b = W_b / g

             mₐ = 1500/32 = 46.875 slug

             m_b = 125/32 = 35,156 slug

Let's reduce to the english system

             v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s

We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved

we assume the direction to the east (right) positive

initial instant. Before the crash

           p₀ = mₐ v₀ₐ - m_b v_{ob}

final instant. Right after the crash

           p_f = (mₐ + m_b) v

the moment is preserved

           p₀ = p_f

           mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v

           v = [tex]\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}[/tex]

we substitute the values

           v = [tex]\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6[/tex]

           v = 0.559 v₀ₐ - 26.40                  (1)

Now as the two vehicles united we can use the relationship between work and kinetic energy

the total mass is

              M = mₐ + m_b

              M = 46,875 + 35,156 = 82,031 slug

starting point. Jsto after the crash

              K₀ = ½ M v²

final point. When they stop

             K_f = 0

The work is

             W = - fr x

the negative sign is because the friction forces are always opposite to the displacement

Let's write Newton's second law

Axis y

           N-W = 0

           N = W

the friction force has the expression

            fr = μ N

we substitute

            -μ W x = Kf - Ko

            -μ W x = 0 - ½ (W / g) v²

            v² = 2 μ g x  

            v = [tex]\sqrt{ 2 \ 0.750 \ 32 \ 17.5}[/tex]Ra (2 0.750 32 17.5  

            v = 28.98 ft / s

Answer:

"0.25 kg-m²" is the appropriate answer.

Explanation:

The diagram of the question is missing. Find the attachment of the diagram below.

According to the diagram, the values are:

m₁ = 0.2

m₂ = 0.3

m₃ = 0.3

m₄ = 0.2

d₁ = d₂ = d₃ = d₄ = 0.5 m

As we know,

The moment of inertia is:

⇒  [tex]I=\Sigma M_id_i^2[/tex]

then,

⇒  [tex]I=m_1d_1^2+m_2d_2^2+m_3d_3^2+m_4d_4^2[/tex]

⇒     [tex]=d^2(m_1+m_2+m_3+m_4)[/tex]

On substituting the values, we get

⇒     [tex]=0.5^2\times (0.2+0.3+0.3+0.2)[/tex]

⇒     [tex]=0.25\times 1[/tex]

⇒     [tex]=0.25 \ Kg-m^2[/tex]

Answer:

#1 Yes

Explanation: #1: The rest of them are used mainley by farmers, and crops are used by common citizens in the world.

Question 1: Crops.

Question 2: Diagnostic Services.

Question 3: A cable company needs to lay new fiber optic cable to reach its customers across a large lake.

Question 4: A bachelor's degree in energy research.

Question 5: Environmental Resources.

If any of these answers are incorrect, please tell me, so I can fix my mistake. Thank you.

Answer:

a

Explanation:

I think don't get mad if I'm wrong

Answer:

Option 2

Explanation:

As per the relation between the distance of the galaxy and shifting of the light of the galaxy towards any specific wavelength of the electromagnetic spectrum, a galaxy at great distance shifts more towards the red spectra that has the highest wavelength.

Thus, this observation give details about the distance of the galaxy from earth.

Answer:

b

Explanation:

Answer: 22.1 m/s

Explanation:

The velocity of Car traveling 30 m/s towards the north

In vector form it is

[tex]v_x=30\hat{j}[/tex]

The velocity of car Y w.r.t X is

[tex]\Rightarrow v_{yx}=15[-\cos 45^{\circ}\hat{i}-\sin 45^{\circ}\hat{j}][/tex]

Solving this

[tex]\Rightarrow v_{yx}=v_y-v_x\\\Rightarrow v_y=v_{yx}+v_x[/tex]

putting values

[tex]\Rightarrow v_y=15[-\cos 45^{\circ}\hat{i}-\sin 45^{\circ}\hat{j}]+30\hat{j}[/tex]

[tex]\Rightarrow v_y=-10.606\hat{i}+19.39\hat{j}[/tex]

absolute velocity relative to ground is

[tex]\left | v_y\right |=\sqrt{(-10.606)^2+(19.39)^2}\\\left | v_y\right |=22.101\ m/s[/tex]

Answer:

BaF2

Explanation:

since you got the valence numbers just do the scissors move where you:

give the F the 2 and give the Ba the 1 so it be like

BaF2 here is the chemical compound

Answer:

See Explanation

Explanation:

Given

Steps: a - f

Table

[tex]\begin{array}{ccccccc}{cm} & {9} & {10} & {13} & {18} & {50} & {83} \ \\ {Age} & {10} & {20} & {30} & {40} & {50} & {60} \ \end{array}[/tex]

Note that: The question is a practical question and the result may differ base on individuals and environment.

So, I will pick up the question from how to determine the age of the eye after the distance between the eyes and the pencil has been established

In my case, the measurement is:

[tex]Length= 10.4[/tex]

Approximate

[tex]Length= 10[/tex]

From the above table, the corresponding age to 10cm is:

[tex]Age = 20cm[/tex]

If in your measurement, the length is approximately (for example):

[tex]Length = 9cm[/tex]

The age will be:

[tex]Age = 10[/tex]

Answer:

The mass of water boiled off is [tex]0.0 \overline{185}[/tex] kg

Explanation:

The given percentage by weight of protein solids in raspberries = 10 weight%  

The ratio of sugar to raspberries in ja-m = 45:55

The mass of the mixture after boiling = 0.4 weight fraction water

Let 's' represent the mass of sugar in the mixture, and let 'r' represent the mass of raspberry

The mass of raspberry, r = 1 kg

The percentage by weight of water in raspberry = 90 weight %

The mass of water in 1 kg of raspberry =  90/100 × 1 kg = 0.9 kg

The ratio of the mass of sugar to the mass of raspberry in jam = r/s = 45/55

∴ s = 1 kg × 55/45 = 11/9 kg

The mass of the mixture before boiling = 1 kg + 11/9 kg = 20/9 kg

The weight fraction of water in the remaining mixture after boiling = 0.4 weight fraction

Let 'w' represent the mass of water boiled off, we have;

(0.9 - w)/(20/9 - w) = 0.4

(0.9 - w) = 0.4 × (20/9 - w)

0.9 - w = 8/9 - 0.4·w

9/10 - 8/9 = w - 0.4·w = 0.6·w = (6/10)·w

(81 - 80)/(90) = (6/10)·w

1/90 = (6/10)·w

w = ((10/6) × 1/90) = 1/54

w = 1/54  

The mass of water boiled off, w = (1/54) kg = [tex]0.0 \overline{185}[/tex] kg

Answer:

speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Explanation:

Given:

mass of truck M = 1370 kg

speed of truck = 12.0 m/s

mass of car m = 593 kg

collision is elastic therefore,

Applying law of momentum conservation we have

momentum before collision = momentum after collision

1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2               ....(i)

Also for a collision to be elastic,

velocity of approach = velocity of separation

12 -0 = v2-v1                  ....(ii)

using (i) and (ii) we have

So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Answer:

I hypothesis that the motion involving the balls in the experiment were moving to create data.

Explanation:

I hope this helps!

Answer:

the answer, the correct one is C

Explanation:

Let's propose the solution of this problem to know which explanation is correct, when the concha stick with the disc is an impulse exercise

             I = ΔP

            ∫ F dt = pf-po

             ∫ F dt = m v_f - m v₀

Therefore, during the time that the contact lasts, a force is applied to the disk, which causes that if the amount of movement increases and therefore its speed increases, when the constant ceases the forces are reduced to zero and the disk no longer changes its momentum following with constant velocity.

When reviewing the answer, the correct one is C

Answer: barometer.

A cyclone is a storm or system of winds that rotates around a center of low atmospheric pressure. An anticyclone is a system of winds that rotates around a center of high atmospheric pressure.

Answer:

A)  a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²

Explanation:

Part A) The relation of the test tube is centripetal

               a_c = v² / r

the angular and linear variables are related

              v = w r

we substitute

               a_c = w² r

let's reduce the magnitudes to the SI system

              w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s

               r = 1 cm (1 m / 100 cm) = 0.10 m

let's calculate

              a_c = 418.88² 0.1

               a_c = 1.75 10⁴ m / s²

part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor

as part of rest the initial velocity is zero and on the floor the height is zero

                v² = v₀² - 2g (y- y₀)

                v² = 0 - 2 9.8 (0 + 1)

                v =√19.6

                v = -4.427 m / s

now let's look for the applied steel to stop the test tube

                v_f = v + a t

                0 = v + at

                a = -v / t

                a = 4.427 / 0.001

                a = 4.43 10³ m / s²

Answer:

0.6

Explanation:

Given that :

Radius, R = 7m

Period, T = 6.9s

The Coefficient of static friction, μs can be obtained using the relation :

μs = v² / 2gR

Recall, v = 2πR/T

μs becomes ;

μs = (2πR/T)² / 2gR

μs = (4π²R² / T²) ÷ 2gR

μs = (4π²R² / T²) * 1/ 2gR

μs = 4π²R / T²g

μs = 4π²*7 / 6.9^2 * 9.8

μs = 28π² / 466.578

μs = 276.34892 / 466.578

μs = 0.5922887

μs = 0.6

Answer:

Distance S = 11.77 m (Approx.)

Explanation:

Given:

Time t = 1.55 Second

Gravity acceleration = 9.8 m/s²

Find:

Distance S

Computation:

S = ut + (1/2)(g)(t)²

S = (0)(1.55) + (1/2)(9.8)(1.55)²

S = (0)(1.55) + (1/2)(9.8)(1.55)²

Distance S = 11.77 m (Approx.)

Answer:

false I think

Explanation:

hope that help

so it's not divided in 3 regions