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katrin2010 [14]

3 years ago

13

a races car travels on a circular track at an average rate of 125mi/h the radius of fhe track is 0.320 miles what is that centri

petal accelerration of the car

1 answer:

Sholpan [36]3 years ago

6 0

390.625
I hope this helps

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Dr. Smith is conducting an experiment to determine if paintings of landscapes produce more peaceful feelings than abstract paint

skad [1K]

The option that could be the hypothesis for this experiment is that D. Abstract paintings may elicit feelings of stress as students try to determine the meanings of the paintings; therefore, abstract paintings may cause college students to report lower feelings of peace.

<h3>What is a hypothesis?</h3>

It should be noted that a hypothesis simply means the proposed explanation made on the basis of limited evidence.

From the information, Smith is conducting an experiment to determine if paintings of landscapes produce more peaceful feelings than abstract paintings.

Therefore, the option that could be the hypothesis for this experiment is that abstract paintings may elicit feelings of stress as students try to determine the meanings of the paintings; therefore, abstract paintings may cause college students to report lower feelings of peace.

Learn more about hypothesis on:

brainly.com/question/11555274

#SPJ1

3 0

2 years ago

What effect does Earth’s spherical shape have on the amount of sunlight that reaches the equator?

Natasha_Volkova [10]

Because the Earth<span> is a sphere, the surface gets much more intense </span>sunlight<span>, hence heat, at </span>the equator<span>than at the poles.</span>

6 0

3 years ago

Read 2 more answers

A ball is dropped from rest at the top of a 6.10 m

natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

$\frac{v_1}{v}$ = e ( coefficient of restitution ) = $\frac{1}{\sqrt{10} }$

and

$\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }$

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

$e = \sqrt{\frac{h_1}{6.1} }$

$e = \sqrt{\frac{h_2}{h_1} }$

So on

$e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }$

$(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}$ = .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0

3 years ago

This problem has been solved!

lana66690 [7]

Answer:

Charge on each metal sphere will be $8\times 10^{8}C$

Explanation:

We have given number of electron added to metal sphere A $n=10^{12}electron$

As both the spheres are connected by rod so half -half electron will be distributed on both the spheres.

So electron on both the spheres $=\frac{10^{12}}{2}=5\times 10^{11}electron$

We know that charge on each electron $e=1.6\times 10^{-19}C$

So charge on both the spheres will be equal to $q=1.6\times 10^{-19}\times 5\times 10^{11}=8\times 10^{8}C$

So charge on each metal sphere will be equal to $8\times 10^{8}C$

6 0

3 years ago

suppose that 273 g of one of the substances listed above displaces 26 mL of water. What is the substance?

guajiro [1.7K]

<span>The unknown substance is silver. I don't see a list of available substances, but let's see if there's something reasonable available that will match. First, let's calculate the density of the unknown substance. Density is mass per volume, so 273 g / 26 mL = 10.5 g/mL Looking up a list of elements sorted by density, I see the following: 10.07 Actinium 10.22 Molybdenum 10.5 Silver 11.35 Lead And silver at 10.5 g/ml is a very nice match for the unknown substances' density of 10.5 g/ml.</span>

6 0

3 years ago

Read 2 more answers