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3 years ago

12

According to the scientific method, how does a physicist formulate and objectively test hypotheses?

1 answer:

AlladinOne [14]3 years ago

4 0

Answer:

According to the scientific method, physicist formulate and objectively test hypotheses through experiments.

Explanation:

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Question 13(Multiple Choice Worth 1 points)

Blizzard [7]

Probably negotiating a treaty with a foreign country because many people need to be involved in the process.

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3 years ago

"A toy airplane flies clockwise at a constant speed in a horizontal circle of radius 8.0 meters. The diagram shows the path of t

mixas84 [53]

When the acceleration is south, the plane is moving east.

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3 years ago

Read 2 more answers

What's the mass show the work?

Alja [10]

You do this one just like the other one that I just solved for you.

For this one ...

The density of the object is 2.5 gm/cm³.
We know that every cm³ of it we have contains 2.5 gm of mass.
We have to find out how many cm³ we have.

The question tells us: We have 2.0 cm³.

Each cm³ of space that the object occupies contains 2.5 gm of mass.

So the 2.0 cm³ that we have contains (2 x 2.5 gm) = 5 gms.
That's the mass of our object.

6 0

4 years ago

Please help! I’ll give brainliest! :)

jeka94

Answer:

C.) Gravity

Explanation:

The projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile.

6 0

3 years ago

A bicyclist starts from rest and accelerates at a rate of 2.3 m/s^2 until it reaches a speed of 23 m/s. It then slows down at a

Rama09 [41]

Answer:

33 seconds.

Explanation:

The equation for speed with constant acceleration at time t its:

$V(t) \ = \ V_0 \ + \ a \ t$

where $V_0$ is the initial speed, and a its the acceleration.

<h3>First half of the problem</h3>

Starting at rest, the initial speed will be zero, so

$V_0 = 0$

the final speed is

$V(t_{f1}) = 23 \frac{m}{s}$

and the acceleration is

$a = 2.3 \frac{m}{s^2}$ .

Taking all this together, we got

$V(t_{f1}) = 23 \frac{m}{s} = 0 + 2.3 \ \frac{m}{s^2} t_{f1}$

$23 \frac{m}{s} = 2.3 \ \frac{m}{s^2} t_{f1}$

$\frac{23 \frac{m}{s}}{2.3 \ \frac{m}{s^2}} = t_{f1}$

$10 s = t_{f1}$

So, for the first half of the problem we got a time of 10 seconds.

<h3>Second half of the problem</h3>

Now, the initial speed will be

$V_0 = 23 \frac{m}{s}$ ,

the acceleration

$a=-1.0 \frac{m}{s^2}$ ,

with a minus sign cause its slowing down, the final speed will be

$V(t_{f2}) = 0$

Taking all together:

$V(t_{f2}) = 0 = 23 \frac{m}{s} - 1.0 \frac{m}{s^2} t_{f2}$

$23 \frac{m}{s} = 1.0 \frac{m}{s^2} t_{f2}$

$\frac{23 \frac{m}{s}}{1.0 \frac{m}{s^2}} = t_{f2}$

$23 s = t_{f2}$

So, for the first half of the problem we got a time of 23 seconds.

<h3>Total time</h3>

$t_total = t_{f1} + t_{f2} = 33 \ s$

5 0

3 years ago