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valentinak56 [21]

3 years ago

15

thr koppean climate map of arizona and three different pictures of cities in arizona describe how ari pressure, temperature and

density have impacted the westher of each other these cities

1 answer:

Sonbull [250]3 years ago

3 0

Answer:

the tempature and the map and climate

Explanation

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Can I have some help please :)

Anastasy [175]

Answer:

didya know this is chrating. everyone has a breain and so do u .

Explanation:

use and u will pass

5 0

3 years ago

A man in a gym is holding an 8.0-kg weight at arm's length, a distance of 0.55 m from his shoulder joint. What is the torque abo

kipiarov [429]

Answer:

$\tau =37.34\ N m$

Explanation:

given,

mass of the weight = 8 Kg

distance = 0.55 m

angle below horizontal = 30°

torque about shoulder

$\tau = \vec{r} \times \vec{F}$

$\tau = r \times F cos \theta$

$\tau = 0.55 \times 8 \times 9.8 \times cos 30^0$

$\tau =37.34\ N m$

torque about his shoulder join is equal to $\tau =37.34\ N m$

5 0

4 years ago

A 10 g bullet moving horizontally with a speed of 2000 m/s strikes and passes through a 4.0 kg block moving with a speed of 4.2

Masteriza [31]

Answer:

The answer is "512 J".

Explanation:

bullet mass $m_1 = 10 g= 10^{-2} \ kg\\\\$

initial speed $u_1 = 2\ \frac{Km}{s}= 2000\ \frac{m}{s}\\\\$

block mass $m_2 = 4\ Kg$

initial speed $v_2 =-4.2 \frac{m}{s}$

final speed $v_2= 0$

Let $v_1$ will be the bullet speed after collision:

throughout the consevation the linear moemuntum

$\to M_1V_1+m_2v_2=M_1U_1+m_2u_2\\\\\to (10^{-2} kg) V_1 +0 = (10^{-2} kg)(2000 \frac{m}{s}) + (4 \ kg)(-4.2 \frac{m}{s}) \\\\\ \to 10^{-2} v_1 = 20 -16.8\\\\$

$= 320 \frac{m}{s}$

The kinetic energy of the bullet in its emerges from the block

$k=\frac{1}{2} m_1 v_1^2$

$=\frac{1}{2} \times 10^{-2} \times 320\\\\=512 \ J$

3 0

3 years ago

Please Please Please Can Help Me On This Question!!!!! I Give Thanks!!!! Please Do 1-4!!!!

ololo11 [35]

1.add the amount of the diagram which is M+Y then dived the answer you get.

5 0

3 years ago

Read 2 more answers

How long (in seconds) does it take a car accelerating at 3.1 m/s2 to go from rest<br> to 51 m/s?

Nesterboy [21]

Answer:

t = 16.5s

Explanation:

Given parameters:

Acceleration = 3.1m/s²

Initial velocity = 0m/s

Final velocity = 51m/s

Unknown:

Time taken = ?

Solution:

To solve this problem we need to reiterate that acceleration is the rate of change of velocity with time.

So;

Acceleration = $\frac{v - u }{t}$

v is the final velocity

u is the initial velocity

t is the time taken

So;

3.1 = $\frac{51 - 0}{t}$

3.1t = 51

t = 16.5s

8 0

3 years ago