Ask question

Ask question

All categories

3 years ago

5

The talk test can be used to measure the __________. A. knowledge of an activity B. intensity of an activity C. length of an act

ivity D. cost of an activity

2 answers:

gladu [14]3 years ago

4 0

The answer would be A

Alexxx [7]3 years ago

4 0

I’m certain the answer is a

You might be interested in

If the coefficient of friction is 0.3900 and the cylinder has a radius of 2.700 m, what is the minimum angular speed of the cyli

Aleks04 [339]

Answer:

w=3.05 rad/s or 29.88rpm

Explanation:

k = coefficient of friction = 0.3900

R = radius of the cylinder = 2.7m

V = linear speed of rotation of the cylinder

w = angular speed = V/R or to rewrite V = w*R

N = normal force to cylinder

N= $=\frac{m(V)^{2}}{R}=m*(w)^2*R$

$Friction force\\Ff = k*N\\Ff= k*m*w^2*R$

$Gravitational force \\Fg = m*g$

These must be balanced (the net force on the people will be 0) so set them equal to each other.

$Fg = Ff$

$m*g = k*m*w^2*R$

$g=k*w^{2}*R$

$w^2 =\frac{g}{k*R}$

$w=\sqrt{\frac{g}{k*R}} \\w =\sqrt{\frac{9.8\frac{m}{s^{2}}}{0.3900*2.7m}}\\ w=\sqrt{9.306}=3.05 \frac{rad}{s}$

There are 2*pi radians in 1 revolution so:

$RPM=\frac{w}{2\pi }*60\\RPM=\frac{3.05\frac{rad}{s}}{2\pi}*60\\RPM= 0.498*60\\RPM=29.88$

So you need about 30 RPM to keep people from falling out the bottom

7 0

3 years ago

A leaky faucet drips once each second. at this rate, it takes 1 hour to fill a 1-liter graduated cylinder to the 200-cm3 mark. w

UNO [17]

Define
v = volume of a drop per second, cm³/s

The time taken to fill 200 cm³ is 1 hour.
Let V = 200 cm³, the filled volume.
Let t = 1 h = 3600 s, the time required to fill the volume.

Therefore,
$\frac{200 \, cm^{3}}{v \, cm^{3}/s} = 3600 \, s \\ v = \frac{200}{3600} =0.0556 \, cm^{3}$

The average volume of a single drop is approximately 0.0556 cm³.

Answer: 0.0556 cm³

8 0

3 years ago

My Notes Ask Your Teacher 6. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclin

Tresset [83]

Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

where

E₀: Mechanical energy at the beginning of the motion (top of the incline)

E₁: Mechanical energy at the end (bottom of the incline)

then

K₀ + U₀ = K₁ + U₁

If v₀ = 0 ⇒ K₀

and h₁ = 0 ⇒ U₁ = 0

we get

U₀ = K₁

U₀ = m*g*h₀ = K₁

we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

finally, we can say that

Kf > Ka = Kb > Kc > Kd > Ke

8 0

3 years ago

Is an electrical current which comes from a battery

Oxana [17]

Hrdudikdodidbshshsjjsksks

6 0

3 years ago

. A huge pile of leaves was wrapped in a tarp in the middle of a lawn. The wrapped leaves weigh 580 newtons. The coefficient of

Rina8888 [55]

The force required is 319 N

Explanation:

The force of static friction is a force that acts an object on a surface, when this object is pushed by another force to put it in motion. The direction of the force of friction is opposite to the direction of the force of push, and its value increases as the force of push increases, up to a maximum value given by:

$F_f = \mu W$

where

$\mu$ is the coefficient of friction

W is the weight of the object

Therefore, in order to put the object in motion, the force applied must be greater than this value.

For the pile of leaves in this problem, we have:

$\mu = 0.55$ (coefficient of friction)

$W=580 N$ (weight of the leaves)

Substituting, we find:

$F=(0.55)(580)=319 N$

Learn more about force of friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

7 0

3 years ago