Changed velocity in a circuit

Describe what is happening to the speed during the period (I). 0s - 10s __________________________________________________ (II).

aleksley [76]

Answer:

- There was a constant acceleration at 0 to 10s

- There was a zero acceleration at 10 to 25s

- There was a constant deceleration at 25 to 30s

Explanation:

See attachment for complete question.

Solving (a): What happens at 0s to 10s

There was a constant acceleration and this is proven below.

At time 0, velocity = 15

At time 10, velocity = 30

This is represented as:

$(t_1,v_1) = (0,15)$

$(t_2,v_2) = (10,30)$

Acceleration (A) is the rate of change of velocity against time.

So:

$A = \frac{v_2 - v_1}{t_2-t_1}$

$A = \frac{30-15}{10 - 0}$

$A = \frac{15}{10}$

$A = 1.5$

Since the acceleration is positive, then it shows a constant acceleration.

Solving (b): What happens at 10s to 25s

There was a zero acceleration and this is because the velocity do not change.

See proof below

At time 10, velocity = 30

At time 25, velocity = 30

This is represented as:

$(t_1,v_1) = (10,30)$

$(t_2,v_2) = (25,30)$

Acceleration (A) is the rate of change of velocity against time.

So:

$A = \frac{v_2 - v_1}{t_2-t_1}$

$A = \frac{30-30}{25 - 10}$

$A = \frac{0}{15}$

$A = 0$

Solving (c): What happens at 25s to 30s

There was a constant deceleration and this is proven below.

At time 25, velocity = 30

At time 30, velocity = 0

This is represented as:

$(t_1,v_1) = (25,30)$

$(t_2,v_2) = (30,0)$

Acceleration (A) is the rate of change of velocity against time.

So:

$A = \frac{v_2 - v_1}{t_2-t_1}$

$A = \frac{0-30}{30-25}$

$A = \frac{-30}{5}$

$A = -6$

Since the acceleration is negative, then it shows a constant deceleration

4 0

3 years ago

A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predete

Alex17521 [72]

Answer:

0.0366 m

Explanation:

We are given;

Current density; J = 540 A/cm² = 540 × 10⁴ m

Current; I = 0.57 A

Now, formula for current density is;

J = I/A

Where A is area = πr²

Thus;

J = I/(πr²)

r = √(I/(Jπ))

r = √(0.57/(540π))

r = 0.0183 m

Diameter = 2 × radius

Diameter = 2 × 0.0183

Diameter = 0.0366 m

7 0

3 years ago

The left fielder throws the baseball home from 60 m away. The ball's horizontal velocity is 30 m/s, and its vertical velocity is

Vikentia [17]

Answer:

The base runner

Explanation:

To know the correct answer to the question, we shall determine the time taken for the baseball and the base runner to get to the home plate. This is illustrated below:

For the baseball:

Horizontal velocity (u) = 30 m/s

Horizontal distance (s) = 60 m

Time (t) =?

s = ut

60 = 30 × t

Divide both side by 30

t = 60 / 30

t = 2 s

Thus, it will take the baseball 2 s to get to the home plate.

For the base runner:

Horizontal velocity (u) = 5 m/s

Horizontal distance (s) = 5 m

Time (t) =?

s = ut

5 = 5 × t

Divide both side by 5

t = 5 / 5

t = 1 s

Thus, it will take the base runner 1 s to get to the home plate.

SUMMARY:

Time taken for the baseball to get to the home plate = 2 s

Time taken for the base runner to get to the home plate = 1 s

From the calculations made above, we can conclude that the base runner will arrive at the home plate first because it took him 1 s to get to the home plate whereas the baseball took 2 s to get there.

7 0

3 years ago

A tightly wound 1000-turn toroid has an inner radius 1.00 cm and an outer radius 2.00 cm, and carries a current of 1.50 A. The t

V125BC [204]

Therefore, the magnitude of magnetic field at a distance 1.10cm from the origin is 27.3mT

Explanation:

Given;

Number of turns, N = 1000

Inner radius, r₁ = 1cm

Outer radius, r₂ = 2cm

Current, I = 1.5A

Magnetic field strength, B = ?

The magnetic field inside a tightly wound toroid is given by B = μ₀ NI / 2πr

where,

a < r < b and a and b are the inner and outer radii of the toroid.

The magnetic field of toroid is

$B = \frac{u_oNI}{2\pi r}$

Substituting the values in the formula:

$B (1.10cm) = \frac{(4\pi X 10^-^7 ) ( 1000)(1.5)}{2\pi (1.10) } \\\\$

$B (1.10cm) = 27.3mT$

Therefore, the magnitude of magnetic field at a distance 1.10cm from the origin is 27.3mT

7 0

3 years ago

11) Arthur and Betty start walking toward each other when they are 100 m apart. Arthur has a speed of 3.0 m/s and Betty has a sp

sleet_krkn [62]

Answer:

100 m

Explanation:

Arthur and Betty should be walking the same amount of time if they start walking at the same time and stop when they meet = t

Speed of Arthur = 3 m/s

Speed of Betty = 2 m/s

Distance = Speed × time

Distance covered by Arthur = 3t

Distance covered by Betty = 2t

The distance covered by both of them will be 100 m

$3t+2t=100\\\Rightarrow 5t=100\\\Rightarrow t=\frac{100}{5}\\\Rightarrow t=20\ s$

The speed of dog is 5 m/s

Spot is running back and forth at 5 m/s for 20 seconds

In 20 seconds the distance covered by the dog is

$\mathbf{5\times 20}=\mathbf{100\ m}$

5 0

4 years ago