What type of matter is absorbent

In the diagram, each charge is

Jlenok [28]

Answer:

Quick maths

First you find the fafarick and the lalickc and the caprisum and the joinnt

3 0

3 years ago

Give the number of significant figures: 832.000 km

Galina-37 [17]

It has three significant figure

8 0

3 years ago

45 V =

Valentin [98]

Explanation:

https://educationalghana.news.blog/2021/08/09/geography-human-physical-and-practical-for-wassce-novdec-candidates/

8 0

2 years ago

A ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally aw

astraxan [27]

26.54 m/s is the magnitude of its velocity just before it strikes the ground

h=100m,v=20m/s,g=9.8m/s

time it takes to reach the ground,

$[t=\sqrt2h/g],[=\sqrt2*100/9.8=4.51s]$

x= 120m

t= 4.52

v= x/t

v= 120/4.52

v= 26.54 m/s

The "speed at which an object changes its location" can be expressed using a vector number called velocity. Consider a person who moves swiftly while taking two steps forward and two steps back while remaining in one location. Velocity is a vector quantity. Therefore, velocity is cognizant of direction. The direction must be taken into account when determining an object's velocity. A speed of 55 mph is not enough information. The direction must be used to appropriately depict the item's velocity. Simply said, the direction of the velocity vector indicates the direction of motion of an object.

To know more about velocity visit : brainly.com/question/16379705

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5 0

1 year ago

Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, th

alex41 [277]

Answer: 321 J

Explanation:

Given

Mass of the box $m=3\ kg$

Force applied is $F=25\ N$

Displacement of the box is $s=15\ m$

Velocity acquired by the box is $v=6\ m/s$

acceleration associated with it is $a=\dfrac{F}{m}$

$\Rightarrow a=\dfrac{25}{3}\ m/s^2$

Work done by force is $W=F\cdot s$

$W=25\times 15\\W=375\ J$

change in kinetic energy is $\Delta K$

$\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J$

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy

$\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J$

Therefore, the magnitude of work done by friction is $321\ J$

3 0

3 years ago