One of alcohol's effects is to _____.

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:

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

Explanation:

I hope this helps!

Answer:

extreme heat, because no physical damage can demagnetize a magnet

Explanation:

Answer:

the 3rd one

Explanation:

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: 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:

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:

[tex]\boxed{\text{\sf \Large 42 m}}[/tex]

Explanation:

Use height formula

[tex]\displaystyle \sf H=\frac{u^2 sin(\theta)^2}{2g}[/tex]

u is initial velocity

θ = 90° (fired vertically upward)

g is acceleration of gravity

[tex]\displaystyle \sf H=\frac{29^2 \times sin(90 )^2}{2 \times 10}=42.05[/tex]

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:

Explanation:

Maximum emf produced = nωAB. where n is no of turns , A is area , B is magnetic field and ω is angular velocity .

n = 100 , A = .035 m²

B = .05 T

ω = 2π f , f is no of revolution per second by coil

= 2 x 3.14 x 6 x 3

= 113.04 rad /s

Maximum emf produced = 100 x 113.04 x .035 x .05

= 19.78  volt .

Answer:

i don't know but i hope you get it right

Explanation

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:

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:

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

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:

Tectonic Plate Movement

Explanation:

Each continent and ocean sits on its own tectonic plate which floats on the Earths upper mantle. They move very little over time.

Answer:

tectonic plates movement

Answer:

false I think

Explanation:

hope that help

so it's not divided in 3 regions

Answer:

The right approach is "0.85 m".

Explanation:

According to the question, the diagram will is provided below.

So that as per the diagram,

The values will be:

My height,

AO = 1.70 m

My eyes at,

AB = 12 cm

i.e.,

     = 0.12 m

As we can see, the point of incidence lies between the feet as well as the eyes, then

BO = 1.58 m

Now,

⇒ [tex]O'D = \frac{1.58}{2}[/tex]

           [tex]=0.79 \ m[/tex]

The point of incidence of the ray will be:

⇒ [tex]CO'=1.70-\frac{0.12}{2}[/tex]

           [tex]=1.70-0.06[/tex]

           [tex]=1.64 \ m[/tex]

Hence,

The smallest length of the mirror will be:

= [tex]CO'-O'D[/tex]

On substituting the values, we get

= [tex]1.64-0.79[/tex]

= [tex]0.85 \ m[/tex]

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:

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:

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:

"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:

 v = 27.28 m /s, θ = 63.9º

Explanation:

For this exercise we can approximate the movement to a projectile launch, let's analyze the situation.

* We must find the horizontal speed, for this we will find the descent time and the horizontal distance

* We look for the vertical speed

At the highest point the speed is horizontal

Let's find the time it takes for the kite to reach the ground

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

             0 =y₀ + 0 -1/2 gt²

             t = [tex]\sqrt{ \frac{2y_o}{g} }[/tex]

             t = √(2 40/32)

             t = 2.5 s

to find the horizontal velocity we must know the horizontal distance, let's use trigonometry

          sin θ = y / l

          θ = sin⁻¹1 y / l

          θ = sin⁻¹ 40/50

          θ = 53.1º

therefore the horizontal distance is

          x = l cos 53.1

          x = 50cos 53.1

          x = 30 m

let's use the equation

          x = v₀ₓ t

          v₀ₓ = x / t

          v₀ₓ = 30 / 2.5

          v₀ₓ = 12 m / s

we look for the vertical component of the velocity

          v_y = v_{oy} - g t

          v_y = 0 - g t

          v_y = - 9.8 2.5

          v_y = -24.5 m / s

the negative sign indicates that the speed is directed downwards, because it is the arrival point, as they indicate that there is no friction, the exit speed is the same, worse with the opposite sign

We already have the two components of the velocity, let's use the Pythagorean theorem to find the modulus

          v = [tex]\sqrt{v_x^2 + v_y^2}[/tex]

          v = [tex]\sqrt{12^2 + 24.5^2}[/tex]

          v = 27.28 m /s

we use trigonometry for the angle

          tan θ = v_y / vₓ

          θ = tan⁻¹ v_y / vₓ

          θ = tan⁻¹ 24.5 / 12

          θ = 63.9º

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:

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:

a). 53.78 m/s

b) 52.38 m/s

c) -75.58 m

Explanation:

See attachment for calculation

In the c part, The negative distance is telling us that the project went below the lunch point.

Answer:

a

Explanation:

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