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ziro4ka [17]
3 years ago
15

Please select all that would be categorized as vectors

Physics
1 answer:
blsea [12.9K]3 years ago
6 0

Answer:

1. 10 ft/s² straight up

2. 4 Km north-west

3. 20 m/s South

Explanation:

To successfully categorize the quantities given above as vectors, we must understand what vector quantity is.

A vector quantity is a quantity that has both magnitude and direction.

Considering the question given above, the vector quantities are:

1. 10 ft/s² straight up

2. 4 Km north-west

3. 20 m/s South

The remaining quantities in the question are not considered as vectors because they only have magnitude but no direction. They are therefore referred to as scalar quantities.

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A human being can be electrocuted if a current as small as 55 mA passes near the heart. An electrician working with sweaty hands
evablogger [386]

Answer:

The lethal voltage for the electrician under those conditions is 126.5 V.

Explanation:

To discover what is the lethal voltage to the electrician we need to find out what is the voltage that produces 55 mA = 0.055 A when across a resistance of 2300 Ohms (Electrician's body resistancy). For that we'll use Ohm's Law wich is expressed by the following equation:

V = i*R

Where V is the voltage we want to find out, i is the current wich is lethal to the electrician and R is his body resistance. By applying the given values we have:

V = 0.055*2300 = 126.5 V.

The lethal voltage for the electrician under those conditions is 126.5 V.

4 0
3 years ago
Read 2 more answers
In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-l
Ksenya-84 [330]

Answer:

10.4 m/s

Explanation:

The problem can be solved by using the following SUVAT equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the diver in the problem, we have:

u=+6.3 m/s is the initial velocity (positive because it is upward)

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

v=+6.3 + (-9.8)(1.7)=-10.4 m/s

And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.

3 0
4 years ago
Read 2 more answers
What do we call the small changes that
Lelu [443]

Answer:

The "butterfly Effect"

Explanation:

The "butterfly effect" will probably have big changes in the future.

6 0
3 years ago
If I accidentally touch a stove and burn myself, which type of heat transfer is occurring?
allochka39001 [22]

Answer:

radiation is the correct answer

5 0
4 years ago
The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.
S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }

Where:

\omega - Angular frequency, measured in radians per second.

m - Mass of the physical pendulum, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

d - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

I_{O} - Moment of inertia with respect to pivot point, measured in kg\cdot m^{2}.

In addition, frequency and angular frequency are both related by the following formula:

\omega =2\pi\cdot f

Where:

f - Frequency, measured in hertz.

If f = 0.658\,hz, then angular frequency of the physical pendulum is:

\omega = 2\pi \cdot (0.658\,hz)

\omega = 4.134\,\frac{rad}{s}

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}

I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}

Given that m = 1.15\,kg, g = 9.807\,\frac{m}{s^{2}}, d = 0.425\,m and \omega = 4.134\,\frac{rad}{s}, the moment of inertia associated with the physical pendulum is:

I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}

I_{o} = 0.280\,kg\cdot m^{2}

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

8 0
3 years ago
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