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raketka [301]
3 years ago
8

PLEASE HELP!!! WILL MARK THE BRAINLIEST! HIGH SCHOOL PHYSICAL SCIENCE List the independent, dependent, controlled variables of t

he experiment.
Materials:
- hot plate
- thermometer
- scale
- measuring spoon
- water
- beakers
Physics
2 answers:
podryga [215]3 years ago
8 0

Answer:

Controlled variables are hot plate and measuring spoon

Dependant variables are scale and thermometer

Independant variables are beakers and water

Explanation:I tried my best, this should be right

serious [3.7K]3 years ago
3 0

Answer:

Explanation:

You didn't last any of the variables. You have to list the variables to tell which are which.

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An elevator starts from rest with a constant. upward acceleration andmoves 1m in the first. 1.8 s. A passenger in the elevator i
mina [271]
Newton's second law tells you: 

Sum of forces on an object = ma 

Here, the forces acting on the bundle are the tension in the string and the force of gravity, these two must combine to yield the acceleration of the bundle. 

So we have: 

T-mg = ma 

or T=m(g+a) 

We know m=8.7kg, we need to find a from the information 

starting from rest, an accelerating object covers distance according to: 

<span>dist = 1/2 at^2 </span>

to cover 1m in 1.8s, we have: 

a=2d/t^2 = 2x1/1.8^2 = 0.62 m/s/s 

Thus, the tension in the string is: 

<span>T = m(g+a)
=  8.7</span>kg(9.8m/s/s+0.62m/s/s) 
<span>
<span>T =  90.654 N
</span>
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

</span>
5 0
2 years ago
A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 6 ft from the path and is kept fo
BARSIC [14]

We are given that,

\frac{dx}{dt} = 4ft/s

We need to find \frac{d\theta}{dt} when x=8ft

The equation that relates x and \theta can be written as,

\frac{x}{6} tan\theta

x = 6tan\theta

Differentiating each side with respect to t, we get,

\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}

\frac{dx}{dt} = (6sec^2\theta)\cdot \frac{d\theta}{dt}

\frac{d\theta}{dt} = \frac{1}{6sec^2\theta} \cdot \frac{dx}{dt}

Replacing the value of the velocity

\frac{d\theta}{dt} = \frac{1}{6} cos^2\theta (4)^2

\frac{d\theta}{dt} = \frac{8}{3} cos^2\theta

The value of cos \theta could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to

cos\theta = \frac{6}{10}

\frac{d\theta}{dt} = \frac{8}{3} (\frac{6}{10})^2

\frac{d\theta}{dt} = \frac{24}{25}

Search light is rotating at a rate of 0.96rad/s

4 0
2 years ago
How much watt is called 1 H. P​
agasfer [191]

Answer:

746 watts

Explanation:

7 0
2 years ago
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