A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at a rate of 2ft/sec. Find the velocity of the top of the ladder at time t=1.

This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.

The minimum horizontal velocity required is "2.6 m/s".

First, we will analyze the vertical motion of the stunt person. We will use the second equation of motion in the vertical direction to find the time interval for the motion.

[tex]h=v_it+\frac{1}{2}gt^2[/tex]

where,

h = height = 3 m

vi = initial vertical speed = 0 m/s

t = time interval = ?

g = acceleration due to gravity = 9.81 m/s²

therefore,

[tex]3\ m = (0\ m/s)(t) + \frac{1}{2}(9.81\ m/s^2)t^2\\\\t^2 = \frac{(3\ m)(2)}{9.81\ m/s^2}\\\\t = \sqrt{0.611\ s^2}[/tex]

t = 0.78 s

Now, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here:

[tex]s = vt\\\\v = \frac{s}{t}[/tex]

where,

s = horizontal distance = 2 m

t =0.78 s

v = minimum horizontal velocity = ?

Therefore,

[tex]v = \frac{2\ m}{0.78\ s}[/tex]

v = 2.6 m/s

Learn more about equations of motion here:  

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

Answer:

I think its d

Explanation:

I'm not sure I'm sorry if I'm wrong

Answer:

If light enters any substance with a higher refractive index (such as from air into glass) it slows down. The light bends towards the normal line. If light travels enters into a substance with a lower refractive index (such as from water into air) it speeds up. The light bends away from the normal line.

Answer:

2

Explanation:

Answer:

[tex]0.886[/tex] N buoyant force is exerted by air

Explanation:

My weight is [tex]75[/tex] Kg

Weight = mass * gravity

As we know

Buoyant Force is equal to the product of density * acceleration due to gravity and volume of the body

Assuming weight density is a bit less than that of water or equal to water i.e [tex]997.77[/tex] kg/m3

Volume is equal to mass / density

[tex]= 75[/tex] Kg * g/[tex]997.777[/tex]

[tex]= 0.0751[/tex] * g

Buoyant Force

= Volume * g * density

[tex]= 0.0751 * 9.8 * 1.2041[/tex]kg/m3

[tex]= 0.886[/tex] N

The final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.

The final angular velocity of the carnival ride is determined by applying third kinematic equation as shown below;

ωf = ωi + 2αθ

where;

ωf = 0 + 2(2.0) x 6.3

ωf = 25.2 rad/s

Thus, the final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.

​

Learn more about angular velocity here: https://brainly.com/question/6860269

Answer: 5.0 rad/s

Explanation: Because that’s what khan said so try it out.

Answer:

question #1 is A

Question #2 is C

Explanation:

Answer:

U=30cm

Explanation:

All you have to do is to put

Mirror formula , 1/f=1/u + 1/v

You should be careful in sign convention .

Virtual image is negative

we take focal length of convex lens negative even if its not given and so on...

Answer:

The moon does not stay at the same distance of the earth because the ortbit of the moon is slightly elliptical. If earth is not tilted at an angle of 66.5°, there will be no change in the season and the earth will have equal length of days and night.

Explanation:

mark me brainlest

Answer:

T

beacuse:

Energy can be transferred from one object to another by doing work. ... When work is done, energy is transferred from the agent to the object, which results in a change in the object's motion (more specifically, a change in the object's kinetic energy).

Answer:

475.7 m/s

Explanation:

Given,

Frequency ( f ) = 67 Hz

Wavelength ( λ ) = 7.1 m

To find : Speed ( v ) = ?

Formula : -

v = f λ

v

= 67 x 7.1

= 475.7 m/s

Therefore,

the speed of the wave is 475.7 m/s.

Answer:

The answer is "15.56 cm".

Explanation:

[tex]v= 2.5 \ cm\\\\f= 2.154 \ cm[/tex]

Calculating object of length is x so:

[tex]u= -x[/tex]

Using formula:

[tex]\to \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\\\to \frac{1}{2.5}-\frac{1}{-x}=\frac{1}{2.154}\\\\\to \frac{1}{x}=\frac{1}{2.154}-\frac{1}{2.5}\\\\\to \frac{1}{2.5}-\frac{1}{-x}=\frac{1}{2.154}\\\\\to x= 15.56 \ cm[/tex]

Answer:

W = 1.875 J

Explanation:

For this exercise let's use the relationship between work and kinetic energy

          W = ΔK

The kinetic energy of rotational motion is

         K₀ = ½ I w²

we can assume that the box is small, so it can be treated as a point object, with moment of inertia

          I = m rₐ²

angular and linear velocity are related

          v = w r

          w = v / r

we substitute in the equation, for point A

         K₀ = ½ (m rₐ²) (v / rₐ)²

         K₀ = ½ m v²

For the final point B, as the system is isolated the angular momentum is conserved

initial        L₀ = Io wo

final          L_f = I_f w_f

                L₀ = L_f

                 I₀ w₀ = I_f w_f

                (m rₐ²) w₀ = (m  [tex]r_{b} ^2[/tex]) w_f

                 w_f = (rₐ/r_b)² w₀

with this value we find the final kinetic energy

         K_f = ½ I_f w_f²

         K_f = ½ (m [tex]r_{b}^2[/tex]) ( (rₐ / r_b)²  w₀) ²

         K_f = ½ m [tex]\frac{r_a^4}{r_b^2} \ w_o^2[/tex]

we substitute in the realcion of work

          W = K_f - K₀

          W = ½ m  [tex]( \( \frac {r_a^2 }{r_b} )^2[/tex] w₀² - ½ m v²

          W = ½ m  [tex]\frac{r_a^4}{r_b^2} ( \frac{v}{r_a} ) ^2[/tex] - ½ m v²

           W = ½ m [tex]\frac{r_a^2}{r_b^2} \ v^2[/tex] - ½ m v2

          W = ½ m v² (([tex]( \ (\frac{r_a}{r_b})^2 -1)[/tex]

let's calculate

           W = ½ ( [tex]\frac{8}{32}[/tex] ) 5 ((2/1)² -1)

           W = 0.625 (3)

           W = 1.875 J

The two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.

A double replacement reaction have two ionic compounds that are exchanging anions or cations.

From the given options, we can choose the following based on their exchange of anions or cations.

Thus, the two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.

Learn more about double replacement reaction here: https://brainly.com/question/14281077

#SPJ2

Answer:

true

Explanation:

Answer:

Distance travel by go-cart = 500 meter

Explanation:

Given:

Speed of go cart = 25 m/s

Time travel = 20 seconds

Find:

Distance travel by go-cart

Computation:

Distance = Speed x time

Distance travel by go-cart = Speed of go cart x Time travel

Distance travel by go-cart = 25 x 20

Distance travel by go-cart = 500 meter

Answer:

The second bulb will have thicker filament

Explanation:

Given;

First electric bulb: Power, P₁ = 100 W and Voltage, V₁ = 220 V

Second electric bulb: Power, P₂ = 200 W and Voltage, V₂ = 200 V

Resistivity of tungsten, ρ = 4.9 x 10⁻⁸ ohm. m

Resistance of the first bulb:

[tex]P = IV = \frac{V}{R} .V = \frac{V^2}{R} \\\\R = \frac{V^2}{P} \\\\R_1 = \frac{V_1^2}{P_1} = \frac{(220)^2}{100} = 484 \ ohms[/tex]

Resistance of the second bulb:

[tex]R_2 = \frac{V_2^2}{P_2} = \frac{(200)^2}{200} = 200 \ ohms[/tex]

Resistivity of the tungsten filament is given by the following equation;

[tex]\rho = \frac{RA}{L}[/tex]

where;

L is the length of the filament

R is resistance of each filament

A is area of each filament

[tex]A = \pi r^2[/tex]

where;

r is the thickness of each filament

[tex]\rho = \frac{R (\pi r^2)}{L} \\\\\frac{\rho L}{\pi} = Rr^2 \\\\Recall ,\ \frac{\rho L}{\pi} \ is \ constant \ for \ both \ filaments\\\\R_1r_1^2 = R_2r_2^2\\\\(\frac{r_1}{r_2} )^2 = \frac{R_2}{R_1} \\\\\frac{r_1}{r_2} = \sqrt{\frac{R_2}{R_1} } \\\\\frac{r_1}{r_2} = \sqrt{\frac{200}{484} } \\\\\frac{r_1}{r_2} = 0.64\\\\r_1 = 0.64 \ r_2\\\\r_2 = 1.56 \ r_1[/tex]

Therefore, the second bulb will have thicker filament

Answer:

W = 1/2 K x^2

x^2 = 2 * W / K = 2 * 49 J / (N/m)  = .218 / m^2

x =  .467 m

Answer:

B i think

Explanation:

...

Answer:

The final speed will be "[tex]1.185\times 10^7 \ m/sec[/tex]".

Explanation:

The given values are:

Potential difference,

Δv = 400 v

Radius,

r = 0.5580 cm

As we know,

⇒  [tex]W=e \Delta v[/tex]

and,

⇒  [tex]\frac{1}{2}mv^2=e \Delta v[/tex]

then,

⇒  [tex]v=\sqrt{\frac{2e \Delta v}{m} }[/tex]

On substituting the values, we get

⇒     [tex]=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 400}{9.11\times 10^{-31}} }[/tex]

⇒     [tex]=\sqrt{\frac{1.6\times 10^{-19}\times 800}{9.11\times 10^{-31}}}[/tex]

⇒     [tex]=1.185\times 10^7 \ m/sec[/tex]

Answer:

speed: 35m/s

direction: left

Explanation:

Assuming the right side is the positive direction:

before explosion:

P = mv = 0

after explosion:

P' = 15P + 5P

(Set the velocity of the 15kg piece after explosion as v1' and the velocity of the 5kg piece after explosion as v2')

P' = 0.75mv1' + 0.25mv2'

P' = (15kg)v' + (5kg)(105m/s)

P' = 525kg/m/s + (15kg)v1'

P = P'

525kg/m/s + (15kg)v1' = 0

(15kg)v1' = -525kg/m/s

v1' = -35m/s

speed = |-35| = 35m/s

direction is to the left since the right side is the positive direction.

Answer:

1)   P₁ = -2 D,   2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

          [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

          [tex]\frac{1}{f_1} = \frac{1}{q}[/tex]

           q = f₁

2) for an object located at p = 25 cm

            [tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}[/tex]

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

2)  

        [tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}[/tex]

        [tex]\frac{1}{f_2}[/tex] = 0.06

         f₂ = 16.67 cm

the expression for the power of the lenses is

          P = [tex]\frac{1}{f}[/tex]

where the focal length is in meters

1)       P₁ = 1/0.50

        P₁ = -2 D

2)     P₂ = 1 /0.16667

        P₂ = 6 D

Answer:

B. it increases

Explanation:

As shown in the table provided, the speed of sound in water (1493 m/s) is greater than the speed of sound in air (346 m/s).

Answer:

B is the correct answer.

Explanation:

Answer:

the length of the wire is 134.62 m.

Explanation:

Given;

resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm

cross-sectional area of the wire, A  = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²

resistance of the wire, R = 10Ω

The length of the wire is calculated as follows;

[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]

Therefore, the length of the wire is 134.62 m.