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2 years ago

A nerve impulse travels along a myelinated neuron at 90.1 m/s. What is this speed in mi/h?

Physics

1 answer:

algol [13]2 years ago

7 0

Answer:

201.5537 mph

Explanation:

Given the following data;

Speed = 90.1 m/s

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the formula;

Speed = distance/time

To convert this value into miles per hour;

Conversion;

1 meter = 0.000621 mile

90.1 meters = 90.1 * 0.000621 = 0.05595 miles

1 metre per second = 2.237 miles per hour

90.1 meters per seconds = 90.1 * 2.237 = 201.5537 miles per hour

90.1 m/s = 201.5537 mph

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A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 6.3 m/s2. For what value of the coe

Levart [38]

Answer:

Coefficient of static friction will be equal to 0.642

Explanation:

We have given acceleration $a=6.3m/sec^2$

Acceleration due to gravity $g=9.8m/sec^2$

We have to find the coefficient of static friction between truck and a cabinet will

We know that acceleration is equal to $a=\mu g$ , here $\mu$ is coefficient of static friction and g is acceleration due to gravity

So $\mu =\frac{a}{g}=\frac{6.3}{9.8}=0.642$

So coefficient of static friction will be equal to 0.642

3 0

3 years ago

A car traveled north 25 km for 1 hour. The car’s SPEED is 25 km/hr north.

kramer

Answer:

hgk

Explanation:

6 0

3 years ago

Part A

7nadin3 [17]

Answer:

2.5 m/s²

Explanation:

Using the formula, v = u + at ( v = Final velocity; u = Initial velocity; t = Time; a = Acceleration)

25 = 0 + 10a

a = 25/10 = 2.5 m/s²

8 0

3 years ago

find the potential energy of an aircraft weighing 10000 bs at 5000 ft true altitude and 125 kts true air speed

Cloud [144]

Answer:

$U=5*10^7ft.Ib$

Explanation:

From the question we are told that

Weight $W= 10000bs$

Altitude $H=5000ft$

Speed $V=125kts\\1kts=0.514m/s\\V=125*0.514=>64.25m/s$

Generally the equation for Potential energy ids mathematically given as

$Potential\ Energy\ U=mgh$

$U=Wh$

$U=10000*5000$

$U=5*10^7ft.Ib$

6 0

3 years ago

According to the law of reflection, the angle of incidence is equal to the angle of

Darina [25.2K]

Before going to answer this question first we have to understand reflection and laws of reflection.

Reflection is the optical phenomenon in which light will bounce back to the same medium from which it had originated .

Whenever a light ray will incident on a mirror or any reflecting surface, it will be reflected. The ray which falls on the reflecting surface is called incident ray and the ray which is reflected is called reflected ray.

Let us consider a normal to the point of incidence.The angle made by incident ray with the normal is called angle of incidence.Let it be denoted as[ i ]

The angle made by the reflected ray with the normal is called angle of incidence.Let it be denoted as [r]

There are two types of reflection.One is called regular and other one is called as irregular.The laws of reflection is valid for both the types of reflection.

There are two laws of reflection.

FIRST LAW -It states that the incident ray,reflected ray and the normal to the point of incidence,all lie in one plane.

SECOND LAW- It states that that the angle of incidence is equal to the angle of reflection irrespective of the type of reflection.i.e i =r

Hence the correct answer will be angle of reflection.

3 0

3 years ago

Read 2 more answers