A squirrel is running a race where she is on track for her average velocity to be 6.0 m/s. She is distracted by a dummy of an attractive male squirrel and pauses for 3.0 s. As a result her average velocity ends up being 5.0 m/s instead. What is the length of the race? [HINT: construct two equations with the same two unknowns in them and you can solve the system of equations]

Answer:

the derivative with respect to time

Explanation:

This is an exercise in kinematics, where the velocity is defined as a function of the position of a body of the form

            v = dx/dt

where v is the velocity of the body, x is the position that we assume is a continuous and differentiable function.

The function written in the equation is the derivative with respect to time

Answer:

51 mph

Explanation:

Answer:

x(t) = 20t + 12.75e⁻¹•⁶ᵗ + 487.5

t = 24.375 s

Explanation:

The force balance on the object is given as

Net force = W - Drag force

ma = W - 10v

a = (dv/dt)

ma = m(dv/dt) = 200 - 10v

W = mg

200 = m×32

m = 6.25 kg

m(dv/dt) = 200 - 10v

6.25(dv/dt) = 200 - 10v

(dv/dt) = 32 - 1.6v

v' + 1.6v = 32

Solving this differential equation using the integrating factor method

(ve¹•⁶ᵗ) = ∫ (32e¹•⁶ᵗ) dt

ve¹•⁶ᵗ = (20e¹•⁶ᵗ) + c (where c = constant of integration)

v = (20 + ce⁻¹•⁶ᵗ)

At t = 0, v = 0

0 = 20 + c

c = -20

v = (20 - 20e⁻¹•⁶ᵗ)

v = (dx/dt)

(dx/dt) = 20 - 20e⁻¹•⁶ᵗ

dx = (20 - 20e⁻¹•⁶ᵗ) dt

x(t) = 20t + 12.5e⁻¹•⁶ᵗ + c (c is still the constant of integration)

At t = 0, x = - 500

- 500 = 0 + 12.5 + c

c = 512.5

x(t) = 20t + 12.75e⁻¹•⁶ᵗ - 487.5

when the object hits the ground, x = 0

0 = 20t + 12.75e⁻¹•⁶ᵗ - 487.5

20t + 12.75e⁻¹•⁶ᵗ = 487.5

Solving by trial and error,

t = 24.375 s

Hope this Helps!!!

Answer:

It will take 313.376 sec to raise temperature to boiling point

Explanation:

We have given that potential difference V = 120 Volt

Current i = 4.50 A

So resistance

Heat flow in resistor will be equal to

It is given that this heat is used for boiling the water

Mass of the water = 0.525 kg = 525 gram

Specific heat of water 4.186 J/gram/°C

Initial temperature is given as 23°C

Boiling temperature of water = 100°C

So change in temperature = 100-23 = 77°C

Heat required to raise the temperature of water

So

t = 313.376 sec

So it will take 313.376 sec to raise temperature to boiling point

Answer:

Explanation:

Voltage, V = 120 V

Current, i = 4.5 A

mass of water, m = 0.525 kg

initial temperature of water, T1 = 23°C

Final temperature of water, T2 = 100 °C

specific heat of water, c = 4.18 x 1000 J/kg °c

let the time taken is t.

Heat given by the heater = heat gain by the water

V x i x t = m x c x (T2 - T1)

120 x 4.5 x t = 0.525 x 4.18 x 1000 x (100 - 23)

540 t = 47701.5

t = 88.34 s

To lift a 10-kg bundle of shingles with an upward acceleration of 1.5 m/s², a motor with a pulley of radius 0.17 m must provide a torque of at least 19 N.m.

We wish to use a motor to lift a 10-kg (m) bund of shingles with an upward acceleration of 1.5 m/s² (a).

The resulting force (F = m.a) is the difference between the tension (T) of the pulley and the weight (w = m.g) of the shingles.

where,

• g: gravity

Then, we can calculate the minimum torque (τ) that the motor must apply using the following expression.

where,

• r: radius of the pulley

To lift a 10-kg bundle of shingles with an upward acceleration of 1.5 m/s², a motor with a pulley of radius 0.17 m must provide a torque of at least 19 N.m.

Learn more: brainly.com/question/19247046

Answer:

Explanation:

It is given that,

Mass of bundle of shingles, m = 10 kg

Upward acceleration of the shingles,

The radius of the motor of the pulley, r = 0.17 m

Let T is the tension acting on the shingles when it is lifted up. It can be calculated as :

T = 113 N

Let is the minimum torque that the motor must be able to provide. It is given by :

So, the minimum value of torque is 19.21 N-m. Hence, this is the required solution.