in a certain right triangle, the two sides that are perpendicular to each other are 5.9 and 5.1 m long. what is the tangent of the angle for which 5.9 m is the opposite side?

1. Area of Circular End: [tex]A = π x (1.5)^2 = 7.07 sq ft.[/tex]

2. Weight of Gasoline: W = 133.5π lbs.

3. Fluid Force: F = 2043.45π lbs on half-full tank end.

let's break down the problem step by step.

1. Identify Relevant Parameters:

  - Diameter of the tank (D) = 3 feet

  - Radius of the tank (r) = D/2 = 1.5 feet

  - Density of gasoline (ρ) = 42 pounds per cubic foot

  - Depth of the gasoline (h) when the tank is half full = 1.5 feet

2. Calculate the Area of the Circular End:

  - The area (A) of a circle is given by the formula:[tex]A = π x r^2[/tex]

  - Substituting the value of radius (r = 1.5 feet), we get:

  - [tex]A = πx (1.5)^2 = π x 2.25 = 7.07[/tex] square feet

3. Determine the Weight of the Gasoline:

  - The weight (W) of the gasoline can be calculated using the formula: W = ρx V

  - Where V is the volume of the gasoline.

  - Since the tank is half full, the volume of the gasoline is half the volume of the cylinder.

  - The volume (V) of the cylinder is given by the formula: [tex]V = πxr^2 x h[/tex]

  - Substituting the values of radius (r = 1.5 feet) and height (h = 1.5 feet), we get:

[tex]- V = π x (1.5)^2 x 1.5 = π x 2.25x 1.5 = 3.375π cubic feet[/tex]

  - So, the weight of the gasoline (W) is: W = 42x3.375π = 133.5π pounds

4. Calculate the Fluid Force:

  - The fluid force (F) on a circular end of the tank can be calculated using the formula: F = ρ x g x V x h

  - Where g is the acceleration due to gravity.

  - Since we're working in pounds and feet, we'll use the value of acceleration due to gravity, [tex]g ≈ 32.2 ft/s^2.[/tex]

  - Substituting the values, we get:

  - F = 42x 32.2x3.375π x 1.5 = 2043.45π pounds

5.Final Calculation:

  - Therefore, the fluid force on a circular end of the tank when the tank is half full is approximately 2043.45π pounds.

So, the complete calculation and explanation show that the fluid force on a circular end of the tank when it is half full is about 2043.45π pounds.

Answer:

(a) 328 Nm

(b) 79.35 Nm

Explanation:

N = =150, side = 17.5 cm = 0.175 m, i = 42 A, B = 1.7 T

A = side^2 = 0.175^2 = 0.030625 m^2

(a) Torque = N x i x A x B x Sinθ

For maximum torque, θ = 90 degree

Torque = 150 x 42 x 0.030625 x 1.7 x Sin 90

Torque = 328 Nm

(b) θ = 14 degree

Torque =  150 x 42 x 0.030625 x 1.7 x Sin 14

Torque = 79.35 Nm

The torque on a current-carrying loop in a magnetic field is calculated using the formula T = NIAB sin ϴ. Part A finds the maximum torque by setting ϴ to 90 degrees, and Part B calculates the torque at an angle of 14 degrees by applying the sine of that angle.

To calculate the torque on a current-carrying loop in a magnetic field, we use the formula T = NIAB sin ϴ, where T is the torque, N is the number of turns in the coil, I is the current, A is the area, B is the magnetic field strength, and ϴ is the angle between the magnetic field and the normal to the plane of the loop.

Part A: Maximum Torque

To find the maximum torque, we set ϴ to 90°, as sin(90°) = 1. Thus, Tmax = NIAB. Using the given values: N = 150 turns, I = 42 A, A (area of the square loop) = (0.175 m)2 and B = 1.7 T, we calculate the maximum torque.

Part B: Torque at 14°

To find the torque when ϴ is 14°, sin(14°) is used in the formula. We thus get T = NIAB sin(14°).

Answer:

0.695 m

Explanation:

α = angular acceleration of the fan blade = 14.1 rad/s²

a = tangential acceleration at the point concerned = acceleration due to gravity = 9.8 m/s²

r = distance of the point from axis of rotation at which tangential acceleration is same as acceleration due to gravity

We know the relation between

a = r α

Inserting the values

9.8 = 14.1 r

r = 0.695 m

Answer:

The time taken for ATP to diffuse across an average cell is 0.66 seconds

Explanation:

It is given that,

Diffusion constant of ATP is, [tex]D=3\times 10^{-10}\ m^2s^{-1}[/tex]

Distance to be diffused across is, [tex]x=20\ \mu m=20\times 10^{-6}\ m[/tex]

We need to find the time taken for ATP to diffuse across an average cell. It is given by :

[tex]t=\dfrac{x^2}{2D}[/tex]

[tex]t=\dfrac{(20\times 10^{-6}\ m)^2}{2\times 3\times 10^{-10}\ m^2s^{-1}}[/tex]

t = 0.66 seconds

So, the time taken for ATP to diffuse across an average cell is 0.66 seconds. Hence, this is the required solution.

Answer:

The distance is 5 m.

Explanation:

Given that,

Horizontal velocity = 20.0 m/s

Time t = 0.250 s

We need to calculate the horizontal distance

The distance traveled by the baseball with horizontal velocity in time t is given by

Using formula for horizontal distance

[tex]d = v\times t[/tex]

[tex]d = 20.0\times0.250[/tex]

[tex]d=5\ m[/tex]

Hence, The distance is 5 m.

Answer:

0.091 sec

Explanation:

f = frequency of oscillation of the mass attached to the spring = 11 Hz

T = Time period of oscillation of the mass attached to the spring = ?

Time period and frequency of oscillation of the mass attached to the end of spring are related as

[tex]T = \frac{1}{f}[/tex]

Inserting the values

[tex]T = \frac{1}{11}[/tex]

T = 0.091 sec

The net work done by friction on the body of a snake slithering in a complete circle can be found by calculating the work done by the friction force. The friction force is equal to the coefficient of friction multiplied by the normal force. The net work done by friction is -2πr × friction force = -2π(1.04 m) × 19.5 N = -129.77 J.

The net work done by friction on the body of a snake slithering in a complete circle can be found by calculating the work done by the friction force. The friction force is equal to the coefficient of friction multiplied by the normal force. In this case, the friction force is opposing the motion of the snake, so the work done by friction is negative.

Considering a complete circle of radius 1.04 m, the distance traveled by the snake is equal to the circumference of the circle, which is 2πr. The normal force is equal to the weight of the snake, which is 78.0 N. The friction force is equal to the coefficient of friction (0.25) multiplied by the normal force (78.0 N), so the friction force is 19.5 N.

The net work done by friction is then calculated by multiplying the friction force by the distance traveled and taking into account the negative sign to indicate the opposing direction of friction. Therefore, the net work done by friction on the body of the snake is -2πr × friction force = -2π(1.04 m) × 19.5 N = -129.77 J.

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Answer:

Spring constant, k = 2.304 N/m

Explanation:

It is given that,

Mass of spring, m = 0.7 kg

Angular frequency, [tex]\omega=2.4\ rad/s[/tex]

We need to find the spring constant of the spring. It is a case of SHM. Its angular frequency is given by :

[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]

[tex]k=\omega^2\times m[/tex]

[tex]k=(2.4)^2\times 0.4\ kg[/tex]

k = 2.304 N/m

So, the spring constant of the spring is 2.304 N/m. Hence, this is the required solution.

Answer:

option (A)

Explanation:

Angle between the plane of coil and the magnetic field = 60 degree

Angle between the normal of coil and the magnetic field = 90 - 60 = 30 degree

N = 7, B = 0.4 T, T = 0.06 s , w = 2 pi / T, A = 900 cm^2 = 0.09 m^2

peak emf, e0 = N x B x A x (2 x 3.14 / T) x Sin 30

e0 = 7 x 0.4 x 0.09 x 6.28 x 0.5  / 0.06 = 13.18 Volt = 13 volt

To change the wave speed on a guitar string from 328 m/s to 363 m/s, the tension must be increased by 22.04%, as calculated through the wave speed formula for a string, taking into account the ratio of the squares of the wave speeds.

To find by what percent the tension in a guitar string must increase to change the wave speed on the string from 328 m/s to 363 m/s, we can use the formula for wave speed on a string, which is [tex]v = (\sqrt{T/mu}\)[/tex], where v is the wave speed in meters per second, T is the tension in Newtons, and [tex](mu)[/tex] is the linear mass density in kilograms per meter.

Starting with the initial tension (let's call it T1) and the initial wave speed (328 m/s), we have:

[tex]v1 = \(\sqrt{T1/\mu}\)[/tex]

For the final wave speed (363 m/s), we have:

[tex]v2 = \(\sqrt{T2/mu}\)[/tex]

Using the ratio of the squares of the wave speeds:

[tex]\((v2/v1)^2 = (T2/T1)[/tex]

[tex]\((363/328)^2 = (T2/T1)[/tex]

[tex]\(T2/T1 = 1.2204[/tex]

To find the percentage increase in tension:

Percent increase = [tex]\((T2/T1 - 1) \times 100\%)[/tex]

Percent increase = [tex]\((1.2204 - 1) \times 100\%\)[/tex]

Percent increase = 22.04%

Therefore, the tension in the guitar string must be increased by 22.04% to achieve a wave speed change from 328 m/s to 363 m/s.

Answer:

45.3°C

Explanation:

Heat gained = mass × specific heat × increase in temperature

q = mC (T − T₀)

Given C = 0.128 J/g/°C, m = 94.0 g, q = 305 J, and T₀ = 20.0°C:

305 J = (94.0 g) (0.128 J/g/°C) (T − 20.0°C)

T = 45.3°C

The final temperature of the metal, after adding 305 J of heat to 94.0 g of metal initially at 20.0 °C, is 45.34 °C. The calculation uses the specific heat capacity formula and involves solving for the change in temperature.

To find the final temperature, we can use the formula:

where:

First, solve for ΔT:

Next, find the final temperature:

Therefore, the final temperature of the metal is 45.34 °C.

Answer:

23.4 m/s

Explanation:

f = actual frequency of the wave = 6.2 x 10⁹ Hz

[tex]f_{app}[/tex] = frequency observed as the ball approach the radar

[tex]f_{rec}[/tex] = frequency observed as the ball recede away from the radar

V = speed of light

[tex]v[/tex] = speed of ball

B = beat frequency = 969 Hz

frequency observed as the ball approach the radar is given as

[tex]f_{app}=\frac{f(V+v)}{V}[/tex]                                 eq-1

frequency observed as the ball recede the radar is given as

[tex]f_{rec}=\frac{f(V-v)}{V}[/tex]                                  eq-2

Beat frequency is given as

[tex]B = f_{app} - f_{rec}[/tex]

Using eq-2 and eq-1

[tex]B = \frac{f(V+v)}{V}- \frac{f(V-v)}{V}[/tex]

inserting the values

[tex]969 = \frac{(6.2\times 10^{9})((3\times 10^{8})+v)}{(3\times 10^{8})}- \frac{(6.2\times 10^{9})((3\times 10^{8})-v)}{(3\times 10^{8})}[/tex]

[tex]v[/tex] = 23.4 m/s

Answers:

(a) [tex]0.0073kg[/tex]

(b) Volume gold: [tex]3.79(10)^{-7}m^{3}[/tex], Volume cupper: [tex]7.6(10)^{-8}m^{3}[/tex]

(c) [tex]17633.554kg/m^{3}[/tex]

Explanation:

We are told the total mass [tex]M[/tex] of the coin, which is an alloy  of gold and copper is:

[tex]M=m_{gold}+m_{copper}=7.988g=0.007988kg[/tex]   (1)

Where  [tex]m_{gold}[/tex] is the mass of gold and [tex]m_{copper}[/tex] is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats [tex]K[/tex] and mass is:

[tex]K=24\frac{m_{gold}}{M}[/tex]   (2)

Finding [tex]{m_{gold}[/tex]:

[tex]m_{gold}=\frac{22}{24}M[/tex]   (3)

[tex]m_{gold}=\frac{22}{24}(0.007988kg)[/tex]   (4)

[tex]m_{gold}=0.0073kg[/tex]   (5)  This is the mass of gold in the coin

The density [tex]\rho[/tex] of an object is given by:

[tex]\rho=\frac{mass}{volume}[/tex]

If we want to find the volume, this expression changes to: [tex]volume=\frac{mass}{\rho}[/tex]

For gold, its volume [tex]V_{gold}[/tex] will be a relation between its mass [tex]m_{gold}[/tex]  (found in (5)) and its density [tex]\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}[/tex]:

[tex]V_{gold}=\frac{m_{gold}}{\rho_{gold}}[/tex]   (6)

[tex]V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}[/tex]   (7)

[tex]V_{gold}=3.79(10)^{-7}m^{3}[/tex]   (8)  Volume of gold in the coin

For copper, its volume [tex]V_{copper}[/tex] will be a relation between its mass [tex]m_{copper}[/tex]  and its density [tex]\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}[/tex]:

[tex]V_{copper}=\frac{m_{copper}}{\rho_{copper}}[/tex]   (9)

The mass of copper can be found by isolating [tex]m_{copper}[/tex] from (1):

[tex]M=m_{gold}+m_{copper}[/tex]  

[tex]m_{copper}=M-m_{gold}[/tex]  (10)

Knowing the mass of gold found in (5):

[tex]m_{copper}=0.007988kg-0.0073kg=0.000688kg[/tex]  (11)

Now we can find the volume of copper:

[tex]V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}[/tex]   (12)

[tex]V_{copper}=7.6(10)^{-8}m^{3}[/tex]   (13)  Volume of copper in the coin

Remembering density is a relation between mass and volume, in the case of the coin the density [tex]\rho_{coin[/tex] will be a relation between its total mass [tex]M[/tex] and its total volume [tex]V[/tex]:

[tex]\rho_{coin}=\frac{M}{V}[/tex] (14)

Knowing the total volume of the coin is:

[tex]V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}[/tex] (15)

[tex]\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}[/tex] (16)

Finally:

[tex]\rho_{coin}=17633.554kg/m^{3}}[/tex] (17)  This is the total density of the British sovereign coin

Answer:

option d)

Explanation:

The amount of heat required to convert the 1 g of ice at 0 degree C to 1 g water at 0 degree C is called latent heta of fusion.

H = m Lf

where,, h is the heat required, m is the mass and Lf is the latent heat of fusion

1140 = m x 79.7

m = 14.3 g

Answer:

(a) 3.82 x 10⁷ m/s

(b) 4.5 MV/m

Explanation:

(a)

ΔV = change in the electric potential as the proton moves = 7.60 x 10⁶ Volts

q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C

v = speed gained by the proton

m = mass of proton = 1.67 x 10⁻²⁷ kg

Using conservation of energy

Kinetic energy gained by proton = Electric potential energy

(0.5) m v² = q ΔV

inserting the values

(0.5) (1.67 x 10⁻²⁷) v² = (1.6 x 10⁻¹⁹) (7.60 x 10⁶)

v = 3.82 x 10⁷ m/s

(b)

d = distance over which the potential change = 1.70 m

Electric field is given as

E = ΔV/d

E = 7.60 x 10⁶/1.70

E = 4.5 x 10⁶ V/m

E = 4.5 MV/m

Answer:

It takes to a spacecraft 100,837.13 years to cross the galaxy if could travels at 99% the speed of light.

Explanation:

d= 9.461 *10²⁰m

V= 297 *10⁶ m/s

t= d/V

t= 3.18 * 10¹² seconds = 100837.13 years

The required tension to produce the transverse waves with a speed of 50.0 m/s on a 0.06 kg, 5.00 m long string is approximately 30.0 N. If the string has a tension of 8.00 N, the wave speed will be roughly 26.0 m/s.

This question is about the physics of waves on strings and involves the concepts of wave speed, tension, and linear mass density. Let's handle this in two parts.

(a) To determine the required tension to produce the transverse waves with a speed of 50.0 m/s, we first need to calculate the string's linear mass density, which is the mass of the string divided by its length. So, the linear mass density (μ) would be 0.06 kg / 5.00 m = 0.012 kg/m. Now there is a formula that determines the wave speed (v) on a string: v = sqrt(FT/μ) where FT is the tension in the string. Rearranging to solve for FT, we get FT = μv^2. Substituting the values we have,

(b) If the tension is 8.00 N, we can use the same formula to calculate the wave speed. This time, rearranging for v, we get v = sqrt(FT/μ). Substituting the values we have, v = sqrt((8.00 N)/(0.012 kg/m)) which gives us approximately 26.0 m/s. Therefore, the wave speed with a tension of 8.00 N is roughly 26.0 m/s.

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The linear mass density of the string is calculated first, which is 0.012 kg/m. Using this in the wave speed equation, the required tension to produce a wave speed of 50.0 m/s is 30.0 N. If the tension is 8.00 N, then the resulting wave speed would be around 25.82 m/s.

The student is attempting to produce transverse waves on a string. To determine the required tension to achieve a wave speed of 50.0 m/s, we should first calculate the linear mass density of the string, using the formula μ = m/L, where m is the mass of the string and L is its length. So, for a string 5.0 m long and a mass of 0.060 kg, μ = 0.060 kg / 5.00 m = 0.012 kg/m.

For (a), the formula for wave speed is v = √(FT/μ), where FT is the tension in the string and μ is the linear mass density. We need to rearrange it to calculate the required tension: FT = μ * v^2 = 0.012 kg/m * (50.0 m/s)^2 = 30.0 N.

For (b), using the same formula and given new tension value, the wave speed is v = √(FT/μ) = √(8.00 N / 0.012 kg/m) = approx. 25.82 m/s.

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Answer:

68.445 cm³/s

Explanation:

Given:

Volume, V = [tex]\frac{4}{3}\pi r^3[/tex]

radius = 5.85 cm

Growth rate of radius = 0.5 cm/week

now

differentiating the volume with respect to time 't', we get:

[tex]\frac{dV}{dt}=\frac{d(\frac{4}{3}\pi r^3)}{dt}[/tex]

or

[tex]\frac{dV}{dt}=(\frac{4}{3}\pi )3r^2\frac{dr}{dt}[/tex]

now, substituting the value of r (i.e at r = 5.85cm) in the above equation, we get:

[tex]\frac{dV}{dt}=4\pi 5.85^2\times 0.5[/tex]

or

[tex]\frac{dV}{dt}=68.445cm^3/s[/tex]

hence, the rate of change of volume at r = 5.85cm is 68.445 cm³/s

Answer:

[tex]a = 0.53 m/s^2[/tex]

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi ( \frac{20}{60})[/tex]

[tex]\omega = 2.10 rad/s[/tex]

so final tangential speed is given as

[tex]v = r\omega[/tex]

[tex]v = 1.71 (2.10) = 3.58 m/s[/tex]

now average acceleration of the girl is given as

[tex]a = \frac{v_f - v_i}{\Delta t}[/tex]

[tex]a = \frac{3.58 - 0}{6.73}[/tex]

[tex]a = 0.53 m/s^2[/tex]

The speed at which the satellite travels in its circular orbit around the Earth is 3,000 m/s.

In order to find the speed at which the satellite travels, we can use the formula:

v = 2πr / T

where v is the velocity of the satellite, r is the radius of the orbit, and T is the period of the satellite. Substituting the given values into the formula, we have:

v = (2π × 6,370,000) / (1.26 × 10^4) = 3,000 m/s

Therefore, the speed at which the satellite travels is 3,000 m/s.

Answer:

Momentum, p = 5 kg-m/s

Explanation:

The magnitude of the momentum of an object is the product of its mass m and speed v i.e.

p = m v

Mass, m = 3 kg

Velocity, v = 1.5 m/s

So, momentum of this object is given by :

[tex]p=3\ kg\times 1.5\ m/s[/tex]

p = 4.5 kg-m/s

or

p = 5 kg-m/s

So, the magnitude of momentum is 5 kg-m/s. Hence, this is the required solution.

Answer:

reeeeeeeeeeeeeeeeeeeeeeee

Answer:

[tex]intensity = 12.98 W/m^2[/tex]

[tex]Energy = 3.93 J[/tex]

Explanation:

As we know that the magnitude of electric field intensity is given as

[tex]E = 98.9 V/m[/tex]

now we know that intensity of the wave is given as the product of energy density and speed of the wave

[tex]intensity = \frac{1}{2}\epsilon_0 E^2 c[/tex]

[tex]intensity = \frac{1}{2}(8.85 \times 10^{-12})(98.9)^2(3\times 10^8)[/tex]

[tex]intensity = 12.98 W/m^2[/tex]

so intensity is the energy flow per unit area per unit of time

so the energy that flows through the area of 0.0259 m^2 in 11.7 s is given as

[tex]Energy = Area \times time \times intensity[/tex]

[tex]Energy = 0.0259(11.7)(12.98)[/tex]

[tex]Energy = 3.93 J[/tex]

Answer:

16.035 revolutions

Explanation:

Part 1:

t = 6 s, f0 = 0 , f = 5 rps,

Let the number of revolutions be n1.

Use first equation of motion for rotational motion

w = w0 + α t

2 x 3.14 x 5 = 0 + α x 6

α = 5.233 rad/s^2

Let θ1 be the angle turned.

Use second equation of motion for rotational motion

θ1 = w0 t + 12 x α x t^2

θ1 = 0 + 0.5 x 5.233 x 6 x 6 = 94.194 rad

n1 = θ1 / 2π = 94.194 / 2 x 3.14 = 15 revolutions

Part 2:

f0 = 5 rps, f = 0, t = 13 s

Let the number of revolutions be n2.

Use first equation of motion for rotational motion

w = w0 + α t

0 = 2 x 3.14 x 5 + α x 13

α = - 2.415 rad/s^2

Let θ2 be the angle turned.

Use third equation of motion for rotational motion

w^2 = w0^2 + 2 x α x θ2

0 = 2 x 3.14 x 5 - 2 x 2.415 x θ2

θ2 = 6.5 rad  

n2 = θ2 / 2π = 6.5 / 2 x 3.14 = 1.035 revolutions

total revolutions n = n1 + n2 = 15 + 1.035 = 16.035 revolutions

The work done on the particle as it moves from x = b to x = a is k(1/a - 1/b).

To calculate the work done on a particle by the attractive force, we need to find the integral of the force function over the distance the particle moves. In this case, the force function is given by F(x) = k/x^2, where k is the constant. The work done when the particle moves from x = b to x = a is given by:

Work = ∫(k/x^2) dx from x = b to x = a

To evaluate this integral, we need to use the power rule of integration. The result will be:

Work = k(1/a - 1/b)

Therefore, the work done on the particle as it moves from x = b to x = a is k(1/a - 1/b).

Answer:

150 fils

Explanation:

Power = 125 Watt

Time 4 hours a day

Energy per day = Power x time per day = 125 x 4 = 500 watt hour

Energy for 30 days = 500 x 30 watt hour = 15000 watt hour = 15 KWH

Cost of 1 KWH = 10 fils

Cost of 15 KWH = 15 x 10 = 150 fils

It takes approximately 2.80 seconds for the second-place cyclist to catch up with the leader.

To find the time it takes for the second-place cyclist to catch up with the leader, we can set up an equation based on their positions.

Let's assume that the time it takes for the second-place cyclist to catch up is "t" seconds.

For the leader:

Distance = Velocity × Time

Distance(leader) = Velocity(leader) × t

For the second-place cyclist:

Distance = Initial Distance + Velocity × Time + 0.5 × Acceleration × Time²

Distance(second) = Initial Distance(second) + Velocity(second) × t + 0.5 × Acceleration(second) × t²

Given:

Velocity(leader) = 11.9 m/s

Initial Distance(second) = 10.6 m

Velocity(second) = 9.80 m/s

Acceleration(second) = 1.20 m/s²

Equating the two distances:

Distance(leader) = Distance(second)

Velocity(leader) × t = Initial Distance(second) + Velocity(second) × t + 0.5 × Acceleration(second) × t²

Now, plug in the values and solve for "t":

11.9t = 10.6 + 9.80t + 0.5 × 1.20 × t²

Simplify the equation:

0.6t² + 2.1t - 10.6 = 0

solving the equation we get,

t = 2.80292945

t = -6.2898

Since time can not be negative so considering t = 2.80292945.

So, it takes approximately 2.80 seconds for the second-place cyclist to catch up with the leader.

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To determine the time elapsed before the second-place cyclist catches the leader in a bicycle race, we use the equations of motion to calculate when their positions will be equal, accounting for the leader's constant velocity and the second-place cyclist's acceleration.

The problem at hand involves the second-place cyclist catching up to the leader in a bicycle race. Given that the leader has a constant velocity of 11.9 m/s and the gap between them is 10.6 m, while the second-place cyclist has a velocity of 9.80 m/s and an acceleration of 1.20 m/s2, we can find out the time it takes for the second-place cyclist to catch up by using the equations of motion.

The equation for the leader's position as a function of time will be:
x1(t) = 11.9t + 10.6 ,
and for the second-place cyclist, considering initial velocity (u), acceleration (a), and the initial position being 0, the position as a function of time is:
x2(t) = 0 + 9.80t + 0.5(1.20)t2.

To find the time when the second-place cyclist catches up to the leader, we need to set x1(t) = x2(t) and solve for t:

11.9t + 10.6 = 9.80t + 0.5(1.20)t2

Solving this quadratic equation will give us the time elapsed before the second-place cyclist catches the leader.

Answer:

The velocity of the 4.00 kg object after the collision is 12 m/s.

Explanation:

Given that,

Mass of object [tex]m_{1} = 4.00\ kg[/tex]

Velocity of object [tex]v_{1} = 20.0\ m/s[/tex]

Mass of another object [tex]m_{2} = 6\ kg[/tex]

Velocity of another object [tex]v_{2}= 12.0\ m/s[/tex]

We need to calculate the relative velocity

[tex]v_{r}=v_{1}-v_{2}[/tex]

[tex]v_{r}=20-12=8\ m/s[/tex]

The relative velocity is 8 m/s in west before collision.

We know that,

In one dimensional elastic collision, the relative velocity before collision equals after collision but with opposite sign.

So, The relative velocity after collision must be 8 m/s in east.

So, The object of 6.00 kg is going 20 m/s and the object of 4.00 kg is slows down to 12 m/s.

Hence, The velocity of the 4.00 kg object after the collision is 12 m/s.

The 4.00-kg object is moving east at 2.00 m/s after the perfectly elastic collision with the 6.00-kg object.

This question is tackled using the concept of Conservation of Momentum. Since the collision is perfectly elastic, both the momentum and kinetic energy of the system are conserved.

In the east-west direction, momentum before is equal to momentum after the collision. Let's denote the velocity of the 4.00-kg object after the collision as v1. Thus, P(before) = P(after) gives us:
4.00kg * 20.00m/s (west) + 6.00kg * 12.00m/s (east) = 4.00kg * v1 (west) + 6.00kg * -12.00m/s (west).

Solving the above equation we find that v1 = -2.00 m/s, where the negative sign indicates it’s moving towards the east. Therefore, the 4.00-kg object is moving east at 2.00 m/s after the collision.

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Answer:

a)frequency of the oscillation = 1/(2*pi)*square root (k/x) =1/(2*pi)*square root (450/(15*10^-2))=8.72 cycle/second

b)spring potential energy = 0.5*k*(x)^2

=0.5*450*(15*10^-2)^2 =5.0625 joule

maximum kinetic energy =spring potential energy

c)

maximum kinetic energy=5.0625

kinetic energy=0.5*m*v^2

v=square toot ((5.0625/(0.5*2.5)) =2 m/s

Answer:

c

Explanation:

When a satellite is orbiting the earth , a constant force is being applied on it which means it must has acceleration. Also the direction of satellite is always being changed when it is orbitting to there is always change in the velocity vector which means acceleration.

You can view in the attached diagram to understand how the velocity is being changed.