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3 years ago

What allows for the investigation of a variable? Can someone help please.

Physics

2 answers:

ladessa [460]3 years ago

5 0

The answer is <span>C: Experiment </span>

nevsk [136]3 years ago

3 0

Answer: Experiment

Explanation: If you want to study a variable you should do small changes and observe how the variable changes.

This process is called an experiment, where the investigator does small changes and tries to understand how these changes affect the variable. Trying to create a model that describes the variable.

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Traveling at an average rateaverage rate of between 50 and 60 miles per hour for 4 to 7 hours select the best estimate for the d

borishaifa [10]

Answer:

best close estimate distance is 315 miles

Explanation:

given data

speed v1 = 50 miles per hour

speed v2 = 60 miles per hour

time t1 = 4 hr

time t2 = 7 hr

to find out

best estimate for the distance traveled

solution

we know lower end distance

distance = v1 × t1

distance = 50 × 4 = 200 miles

and

upper end distance

distance = v2 × t2

distance = 60 × 7 = 420 miles

so middle value distance is

v = 55 miles per hour and t = 5.5 hours

distance = v × t

distance = 55 × 5.5

distance = 302.5 miles

so best close estimate distance is 315 miles

3 0

3 years ago

5. Graph A below plots a race car's speed for 5 seconds. The car's rate of acceleration is 6 m/s^2

Georgia [21]

Answer:

The answer is below

Explanation:

We are to check if the statement is true of false. If it is false, we correct the statement.

Solution:

Acceleration is the time rate of change of velocity. It is the ratio of the change in velocity to the change in time. The acceleration can be gotten from a velocity time graph by finding the slope of the graph.

The x coordinate represent the time and the y coordinate velocity.

5) Graph A passes through the point (0, 0) and (4, 24). Therefore the acceleration (slope) is:

Acceleration = $\frac{24-0}{4-0}=6\ m/s^2$

This is correct.

6) Graph B is a straight line of 12 m/s. It passes through (0, 12) and (4, 12). Hence:

Acceleration = $\frac{12-12}{4-0}=0\ m/s^2$

This is false.

Therefore the acceleration of graph B is 0 m/s².

8 0

3 years ago

A car traveling at 20 m/s when the driver sees a child standing in the road. He takes 0.80 s to react, then steps on the brakes

mr Goodwill [35]

When driver see the child standing on road his speed is 20 m/s

So here at that instant his reaction time is 0.80 s

He will cover a total distance given by product of speed and time

$d_1 = v* t$

$d_1 = 20 * 0.8$

$d_1 = 16 m$

now after this he will apply brakes with acceleration a = 7 m/s^2

so the distance covered before it stop is given by

$v_f^2 - v_i^2 = 2 a d$

$0 - 20^2 = 2*(-7)*d_2$

$d_2 = 28.6 m$

so the total distance covered by it

$d = d_1 + d_2$

$d = 16 + 28.6 = 44.6 m$

<em>so it will cover a total distance of 44.6 m</em>

3 0

3 years ago

Superman does an exhibition run at a track meet. When he runs the 200 m

Ber [7]

Answer:

6.32s

Explanation:

Given parameters:

Length of track and distance covered = 200m

Acceleration = 10m/s²

Unknown:

Time taken to cover the track = ?

Solution:

To solve this problem, we apply one of the motion equations as shown below:

S = ut + $\frac{1}{2}$ at²

S is the distance covered

t is the time taken

a the acceleration

u is the initial velocity

The initial velocity of Superman is 0;

So;

S = $\frac{1}{2}$ at²

200 = $\frac{1}{2}$ x 10 x t²

200 = 5t²

t² = 40

t = 6.32s

5 0

3 years ago

An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _

Viefleur [7K]

Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
where
$f=5 cm$ is the focal length
$p=10 cm$ is the distance of the object from the mirror
$q$ is the distance of the image from the mirror.

Rearranging the equation, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}$
so, the distance of the image from the mirror is $q=10 cm$ .

1b) The image height is given by the magnification equation:
$\frac{h_i}{h_o}=- \frac{p}{q}$
where $h_i$ is the heigth of the image and $h_o=1 cm$ is the height of the object. By rearranging the equation and using p and q, we find
$h_i=-h_o \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm$
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

3) As before, we find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

And now we can use the magnification equation to find the image height:
$\frac{h_i}{h_o}=- \frac{p}{q}$
Rearranging it, we find
$h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm$
and the negative sign means the image is inverted.

5 0

3 years ago