What will the weather be like in your area and your friend’s area over the next 24 hours?

A motorcyclist travels around a curved path that has a radius of 350 ft. While traveling around the curved path, the motorcyclis

Ilya [14]

Answer:

A) t = 29.3 s

B) V = 58.56 ft/s

C) a_net = 2.3 ft/s²

Explanation:

A) The formula for radial acceleration is given as;

a_c = V²/R

We are given;

Radius;R = 350 ft

So, a_c = V²/350

Where V is velocity

Tangential acceleration;a_t = 2 ft/s²

Formula for net acceleration is;

a_net = √((a_c)² + (a_t)²)

We are given a_net = 10 ft/s²

Thus;

10 = √(V²/350)² + (2²)

10² = V⁴/350² + 4

100 - 4 = V⁴/350²

96 × 350² = V⁴

V = 58.56 ft/s

Now, formula for angular velocity is;

ω = V/r

ω = 58.56/350

ω = 0.1673 rad/s

Angular acceleration is given by;

α = a_t/r

α = 2/350

α = 0.00571 rad/s²

Time needed will be gotten from the formula;

t = ω/α

t = 0.1673/0.00571

t = 29.3 s

B) we are told total acceleration is 10 ft/s², thus it's the same as velocity gotten earlier which is 58.56 ft/s

C) we are told that the speed is now 20 ft/s

Thus;

a_c = 20²/350

a_c = 1.1429 ft/s²

Since a_net = √((a_c)² + (a_t)²)

We are given a_t = 2 ft/s²

Thua;

a_net = √(1.1429² + 2²)

a_net = √5.30622041

a_net = 2.3 ft/s²

6 0

3 years ago

What are the characteristics of an ideal transformer

Vsevolod [243]

It is an imaginary transformer which has no core loss, no ohmic resistance and no leakage flux. The ideal transformer has the following important characteristic. The resistance of their primary and secondary winding becomes zero. The core of the ideal transformer has infinite permeability.

4 0

4 years ago

What does 760mmHg represents??

malfutka [58]

Explanation:

760 mmHg (millimeters of mercury) is a measure of atmospheric pressure. It represents the height of a column of mercury at which the static pressure at the bottom is equal to the atmospheric pressure.

1 atm = 760 mmHg = 101,325 Pa = 14.7 psi

8 0

3 years ago

The fastest time ever run at Pikes Peak is 7:57. That’s way too long. If we removed all the curves, making a straight line from

anygoal [31]

Answer: 271.4 s

Explanation:

We are told the top speed (maximum speed) $V_{max}$ the car has is:

$V_{max}=203 mph=90.74 m/s$ taking into account $1 mile=1609.34 m$

And the car's average acceleration $a_{ave}$ is:

$a_{ave}=0.091 g=2.93 ft/s^{2}=0.89 m/s^{2}$

Since:

$a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}$ (1)

Where:

$V_{f}=V_{max}=90.74 m/s$ is the car's final speed (top speed)

$V_{o}=0 m/s$ because it starts from rest

$\Delta t$ is the time it takes to reach the top speed

Finding this time:

$\Delta t=\frac{V_{f}-V_{o}}{a_{ave}}$ (2)

$\Delta t=\frac{90.74 m/s - 0 m/s}{0.89 m/s^{2}}$ (3)

$\Delta t=t_{1}=101.95 s$ (4)

Now we have to find the distance $d$ the car traveled at this maximum speed with the following equation:

$V_{f}^{2}=V_{o}^{2} + 2a_{ave} d$ (5)

Isolating $d$ :

$d=\frac{V_{f}^{2}}{2a_{ave}}$ (6)

$d=\frac{(90.74 m/s)^{2}}{2(0.89 m/s^{2})}$ (7)

$d=4625.70 m$ (8)

On the other hand, we know the total distance $D$ traveled by the car is:

$D=12.42 miles = 19988.052 m$

Hence the remaining distance is:

$d_{remain}=D-d=19988.052 m - 4625.70 m$ (9)

$d_{remain}=15362.35 m$ (10)

So, we can calculate the time $t_{2}$ it took to this car to travel this remaining distance $d_{remain}$ at its top speed $V_{max}$ , with the following equation: