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2 years ago

A community calendar allows nonprofit organizations to add events and their intended purposes for community members to see.

Physics

1 answer:

lbvjy [14]2 years ago

8 0

Out of all given choices, activities and donor list can be found on the given kind of informative website.

Answer: Option A and B

<u>Explanation:</u>

Non-profit need to do the examination and for sake activities that simply aren't successful. And afterwards, they have to look to a portion of these auxiliary changes that to discover increasingly productive approaches to make a feasible money related model for their social change work.

When occasions are crucial, allowed to visit, and concentrated on developing as well as managing present or potential significant donors (people, establishments, corporate pioneers, they can bode well. Yet, only in the event that you catch up with participants on a one-on-one premise to additionally put them in the association and in the long run request that they contribute or reestablish their commitments.

On the off chance, you don't charge those to visit with the goal that can approach them for a greater, and progressively important blessing not far off.

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A 5.0-kilogram box is sliding across a level floor. The box is acted upon by a force of 27 newtons east and a frictional force o

marshall27 [118]

Answer:

The magnitude of the acceleration of the box is 2 m/s².

Explanation:

Given:

Mass of the box, $m=5.0$ kg

Force acting towards east, $F=27$ N

Frictional force acting towards west, $f=17$ N

Let the acceleration be $a$ m/s².

Now, net force acting on the box towards east is given as:

$F_{net}=F-f=27-17=10\textrm{ N}$

From Newton's second law of motion,

$F_{net}=ma\\10=5.0\times a\\a=\frac{10}{5.0}=2\textrm{ }m/s^2$

Therefore, the magnitude of the acceleration of the box is 2 m/s².

6 0

2 years ago

Read 2 more answers

Ugonna stands at the top of an incline and pushes a 100−kg crate to get it started sliding down the incline. The crate slows to

Anna [14]

Answer:(a)891.64 N

(b)0.7

Explanation:

Mass of crate $m=100 kg$

Crate slows down in $s=1.5 m$

initial speed $u=1.77 m/s$

inclination $\theta =30^{\circ}$

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

$W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2$

$W_{gravity}=mg(0-h)=mgs\sin \theta$

$W_{gravity}=-mgs\sin \theta$

$W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N$

change in kinetic energy $=\frac{1}{2}\times 100\times 1.77^2=156.64 J$

$W_{friction}=156.64+735=891.645$

(b)Coefficient of sliding friction

$f_r\cdot s=W_{friciton}$

$891.645=f_r\times 1.5$

$f_r=594.43 N$

and $f_r=\mu mg\cos \theta$

$\mu 100\times 9.8\times \cos 30=594.43$

$\mu =0.7$

5 0

2 years ago

You are standing at the midpoint between two speakers, a distance D away from each. The speakers are playing the exact same soun

Rzqust [24]

Answer:

Explanation:

wave length of sound waves = velocity / frequency

= 340 / 170

λ = 2 m.

When the position of man is exactly at the meddle point between the speakers , sound waves from the speakers reaching man are in same phases ( path difference is zero. ) so intensity of sound is maximum .

Now , the man starts moving towards one of the speakers , his distance from one speaker becomes closer than the other creating path difference for the sound waves reaching his ears.

If he walks a distance of .5 m towards one speaker , path difference created

= .5 x 2 = 1 m

So , path difference = λ /2 ,

there will be destructive interference so minimum sound will be heard there.

When he walks a distance of 1 m , path difference created = 2m

path difference = λ

so there is constructive interference and maximum sound will be heard there.

Again we he walks a distance of 1.5 m , path difference created = 3 m

path difference = 3 λ /2

So there will be destructive interference so minimum sound will be heard there.

In this way we see that man starts from a point of maximum sound intensity , reaches a point of minimum sound intensity , then reaches a point of maximum sound intensity . At last he reaches a position of minimum sound intensity.

3 0

2 years ago

Suppose you push a hockey puck of mass m across frictionless ice for a time 1.0 s, starting from rest, giving the puck speed v a

EleoNora [17]

Newton's second law ...Force = momentum change/time.momentum change = Forcextme.also, F=ma -> a=F/m - the more familiar form of Newton's second law
using one of the kinematic equations for m ... V=u+at; u=0; a=F/m -> V=(F/m)xt.-> t=mV/F using one of the kinematic equations for 2m ... V=u+at; u=0; a=F/2m -> V=(F/2m)xt. -> t=2mV/F (twice as long, maybe ?)
I think I've made a mistake somewhere below, but I think that the principle is right ...using one of the kinematic equations for m ... s=ut + (1/2)at^2); s=d;u=0;a=F/m; t=1; -> d=(1/2)(F/m)=F/2musing one of the kinematic equations for 2m ... s=ut + (1/2)at^2); s=d;u=0;a=F/2m; t=1; -> d=(1/2)(F/2m)=F/4m (half as far ????? WHAT ???)

3 0

2 years ago

Read 2 more answers

You toss a rock of mass m vertically upward. Air resistance can be neglected. The rock reaches a maximum height h above your han

VladimirAG [237]

Answer with Explanation:

We are given that

Mass of rock=m

Maximum height=h

a.At maximum height, velocity,v=0

We know that

$v^2=u^2-2gh$