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3 years ago

Ill give brainliest! 1. Which item below describes a quick change to

Physics

1 answer:

TiliK225 [7]3 years ago

3 0

Answer:

hmmmmm

Explanation:

I am gonna say D.

if I am right plz give me a brainliest

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3. Order the following lengths from shortest to longest.

horrorfan [7]

Answer: $d < a < c < b$

$3.3 cm < 400 mm < 170 m < 22 km$

Explanation:

Firstly, let's convert these length to meters $m$ , in order to work with the same units:

a. 400 millimeters

$400 mm \frac{1m}{1000 mm}=0.4 m$

b. 22 kilometers

$22 km \frac{1000m}{1 km}=22000 m$

c. 170 meters

Here the lenght is already in meters

d. 3.3 centimeters

$3.3 cm \frac{1m}{100 cm}=0.033 m$

Now that we have all the lengths in meters, we can order them from shortest to longest:

$0.033 m < 0.4 m < 170 m < 22000 m$

$d < a < c < b$

7 0

4 years ago

A . Nitrogen and oxygen

drek231 [11]

Answer:

Option B is correct

Because they lie in same group

4 0

3 years ago

Read 2 more answers

Two boys with masses of 40 kg and 60 kg stand on a horizontal frictionless surface holding the ends of a light 10-m long rod. Th

zlopas [31]

When they meet the 40-kg boy would have moved a distance of 6 m.

<h3>Distance moved by the 40 kg boy</h3>

Apply the principle of center mass;

Take the 40 kg mass as the reference point;

X(40 kg) = (40kg x 0 + 60kg x 10 m)/(40 kg + 60 kg)

X(40 kg) = (600)/(100)

X(40 kg) = 6 m

Thus, when they meet the 40-kg boy would have moved a distance of 6 m.

Learn more about distance here: brainly.com/question/2854969

#SPJ1

4 0

2 years ago

A boy walks a distance of 100 meters to the right at a steady speed of 1.00 m/s. Then he stops for 30 seconds. Then returns back

Mamont248 [21]

Explanation:

$speed=\frac{Distance}{time}$

In the figure attached:

A boy travels AB distance of 100 m, with speed of 1.00 m/s

AB = 100 m

$1.00 m/s=\frac{100 m}{t_1}$

$t_1=100 s$

After reaching at point B he stops there at for 30 seconds.

$t_2= 30 s$

After , 30 seconds he he comes back to his initial position that is A with steady speed of 1.00 m/s.

Distance covered from B to A= 100 m

Time taken by him during coming back= $t_3$

$1.00=\frac{100 m}{t_3}$

$t_3=100 s$

After returning to to point A he turns left and moves towards point C with speed of 1.5 m/s for 2 minutes.

Distance of AC = ?

$t_4=2 min= 120 s$

$1.5 m/s=\frac{AC}{120 s}$

AC = 180 m

The total time of his round trip is:T

$T=t_1+t_2+t_3+t_4=100 s+30 s(stop)+ 100 s(returning)+120 s=350 s$

The total distance: D

D = AB + BA (returning) + AC=100 m + 100 m + 180 m = 380 m[/tex]

The total displacement of boy:

Displacement is the shortest distance between the initial point and final point.

He first walked to point B from A and then came back to A . And after that walking to point C from A.So, the final displacement (d) is from A to C.

d = AC = 180 m

The total displacement of boy is 180 m.

The average speed of the boy is given by:

$=\frac{AB+BA+AC}{t_1+t_2+t_3+t_4}=\frac{D}{T}$

$\frac{D}{T}=\frac{380 m}{350 s}=1.085 m/s$

The average velocity of the boy is given by:

$=\frac{AC}{t_1+t_2+t_3+t_4}=\frac{d}{T}$

$\frac{d}{T}=\frac{180 m}{350 s}=0.5142 m/s$

5 0

4 years ago

Explain how a magnetic compass can be used to find directions on Earth’s surface

irakobra [83]

The north and the south of the globe are magnitized and the needles of a compass are magnetized. Since positives attract negatives. The negative end of the earth attarcts the positive point of the needle to point towards it. The same happens with positive end of the earth and the negative point of the needle.

6 0

3 years ago