Ask question

Ask question

All categories

2 years ago

10

A characteristic in an experiment, such as how much food, how much space, how hot or how cold, is termed a ____________. (spelli

ng counts, so be careful!)

2 answers:

KATRIN_1 [288]2 years ago

6 0

I think the term is a variable?

yuradex [85]2 years ago

3 0

Variables are what we change in an experiment to test their effects on whatever it is we are measuring. Hope that helps!

You might be interested in

An object attached to one end of a spring makes 20 vibrations in 10 seconds. Its angular frequency is: 0. 79 rad/s 1. 57 rad/s 2

morpeh [17]

Angular frequency in radian per second for 20 vibrations in 10 seconds is 12.6 rad/s

<h3>What is Angular frequency?</h3>

Angular frequency is the number of vibrations in radian per second.

The total number of vibrations n is 20 and the time taken for these vibrations is 10 s

The frequency of the vibrations will be

f = 20 / 10 = 2 Hz

Angular frequency is related to the frequency as

ω = 2πf

ω=2π × 2

ω = 12.6 rad/s

Thus, the angular frequency is 12.6 rad/s.

Learn more about Angular frequency.

brainly.com/question/14244057

#SPJ

5 0

1 year ago

24. An elevator is moving vertically up with an acceleration a. The force exerted on the floor by a passenger of mass m is

Wewaii [24]

<u>Given </u><u>:</u><u>-</u>

An elevator is moving vertically up with an acceleration a.

<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>

The force exerted on the floor by a passenger of mass m .

<u>Solution</u><u> </u><u>:</u><u>-</u>

As the man is in a accelerated frame that is <u>non </u><u>inertial</u><u> frame</u><u> </u>, we would have to think of a pseudo force .

The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .

For the FBD refer to the attachment . From that ,

$\implies Weight_{apparent}= mg + ma$

<u>Hence</u><u> </u><u>option</u><u> </u><u>d </u><u>is </u><u>correct</u><u> </u><u>choice </u><u>.</u>

<em>I </em><em>hope</em><em> this</em><em> helps</em><em> </em><em>.</em>

3 0

1 year ago

The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second if i

Sergio [31]

Answer with Explanation:

We are given that

Diameter of fighter plane=2.3 m

Radius= $r=\frac{d}{2}=\frac{2.3}{2}=1.15 m$

a.We have to find the angular velocity in radians per second if it spins=1200 rev/min

Frequency= $\frac{1200}{60}=20 Hz$

1 minute=60 seconds

Angular velocity= $\omega=2\pi f$

Angular velocity= $2\times \frac{22}{7}\times 20=125.7 rad/s$

b.We have to find the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac.

$v=r\omega=1.15\times 125.7=144.56 m/s$

c.Centripetal acceleration= $\omega^2 r=(125.7)^2(1.15)=18170.56 m/s^2$

Centripetal acceleration== $\frac{18170.56\times g}{9.81}=1852.25 g m/s^2$

7 0

2 years ago

It is correct to say that impulse is equal toA) momentum.B) the change in momentum.C) the force multiplied by the distance the f

goldenfox [79]

Answer:

B) the change in momentum

Explanation:

Impulse is defined as the product between the force exerted on an object (F) and the contact time ( $\Delta t$ )

$I=F \Delta t$

Using Newton's second law (F = ma), we can rewrite the force as product of mass (m) and acceleration (a):

$I=(ma) \Delta t$

However, the acceleration is the ratio between the change in velocity ( $\Delta v$ ) and the contact time ( $\Delta t$ ): $a=\frac{\Delta v}{\Delta t}$ , so the previous equation becomes

$I=m \frac{\Delta v}{\Delta t}\Delta t$

And by simplifying $\Delta t$ ,

$I=m \Delta v$

which corresponds to the change in momentum of the object.

8 0

2 years ago

Determine the magnitude of force F so that the resultant FR of the three forces is as small as possible.

Kamila [148]

Answer: FR=2.330kN

Explanation:

Write down x and y components.

Fx= FSin30°

Fy= FCos30°

Choose the forces acting up and right as positive.

∑(FR) =∑(Fx )

(FR) x= 5-Fsin30°= 5-0.5F

(FR) y= Fcos30°-4= 0.8660-F

Use Pythagoras theorem

F2R= √F2-11.93F+41

Differentiate both sides

2FRdFR/dF= 2F- 11.93

Set dFR/dF to 0

2F= 11.93

F= 5.964kN

Substitute value back into FR

FR= √F2(F square) - 11.93F + 41

FR=√(5.964)(5.964)-11.93(5.964)+41

FR= 2.330kN

The minimum force is 2.330kN

8 0

2 years ago