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Natasha2012 [34]
3 years ago
15

Which atoms have the lowest electronegativity?

Physics
2 answers:
Kobotan [32]3 years ago
8 0

Answer:

the element Francium

Explanation:

AlexFokin [52]3 years ago
7 0

Francium(sorry if I'm wrong)

Explanation:

which has an electronegativity of 0.7

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An archer pulls her bowstring back 0.376 m by exerting a force that increases uniformly from zero to 251 N. (a) What is the equi
Amanda [17]

Answer:

(A) 667.5 N/m

(B)

Explanation:

(A) Let the spring constant be k.

Using the formula F = kx

k = 251 / 0.376

K = 667.5 N/m

(B)

Work done

W = 0.5 × kx^2

W = 0.5 × 667.5 × 0.376 × 0.376

W = 47.2 J

8 0
3 years ago
cindy threw a ball with Kinetic energy of 3.5 ana a velocity of 17 m/s. What is the mass of the ball?
abruzzese [7]
  • K.E=3.5J
  • Velocity=17m/s
  • Mass be m

\\ \sf\Rrightarrow K.E=\dfrac{1}{2}mv^2

\\ \sf\Rrightarrow 3.5=\dfrac{1}{2}m(17)^2

\\ \sf\Rrightarrow 7=289m

\\ \sf\Rrightarrow m=\dfrac{7}{289}

\\ \sf\Rrightarrow m=0.024kg

7 0
3 years ago
HELP DUE TODAY
vichka [17]

Answer:

The correct options here are C and D

Explanation:

DC here means "direct current". Direct current is characterized by it's flow of current been <u>unidirectional</u> (meaning it's current flows in just <u>one direction</u>). Thus, to get the answer to the question one must look for options that only move in a single direction. Options A and B move in a back and forth direction meaning they are not unidirectional.

A longitudinal wave moves in the same direction OR opposite direction to the direction of the propagating wave. This means longitudinal wave is also unidirectional. <u>Option C is correct</u>.

1st graders, in a line, going to the cafeteria also move in one direction (direction of the cafeteria). Thus, <u>option D is also correct.</u>

4 0
3 years ago
Scientists estimate the age of the universe to be (1 point)
Nesterboy [21]

12-15 billion years i think

4 0
4 years ago
Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 27.0°below the horizontal. If it st
andreev551 [17]

a) The time of flight is 3.78 s

b) The initial speed is 17.0 m/s

c) The speed at impact is 46.4 m/s at 70.3^{\circ} below the horizontal

Explanation:

The picture of the previous problem (and some data) is missing: find it in attachment.

a)

The motion of the ball is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

We start by considering the vertical motion, to find the time of flight of the ball. We do it by using the following suvat equation: for the y-displacement:

y=u_y t+\frac{1}{2}at^2

where we have:

y = -45.0 m is the vertical displacement of the ball (the height of the building)

u_y=u sin \theta is the initial vertical velocity, with u being the initial velocity (unknown) and \theta=-27.0^{\circ} the angle of projection

t is the time of the fall

a=g=-9.8 m/s^2 is the acceleration of gravity

Along the x-direction, the equation of motion is instead

x=(u cos \theta)t

where ucos \theta is the horizontal component of the velocity. Rewriting this equation as

t=\frac{x}{ucos \theta}

And substituting into the previous equation, we get

y=xtan \theta + \frac{1}{2}gt^2

And using the fact that the horizontal range is

x = 59.0 m

And solving for t, we find the time of flight:

t=\sqrt{\frac{y-x tan \theta}{g}}=\sqrt{\frac{-45-(59.0)(tan(-27^{\circ}))}{-9.8}}=3.78 s

b)

We can now find the initial speed, u, by using the equation of motion along the x-direction

x=u cos \theta t

where we know that:

x = 59.0 m is the horizontal range

\theta=-23^{\circ} is the angle of projection

t=3.78 s is the time of flight

Solving for u, we find the initial speed:

u=\frac{x}{cos \theta t}=\frac{59.0}{(cos (-23^{\circ}))(3.78)}=17.0 m/s

c)

First of all, we notice that the horizontal component of the velocity remains constant during the motion, and it is equal to

v_x = u cos \theta = (17.0)(cos (-23^{\circ})=15.6 m/s

The vertical velocity instead changes according to the equation

v_y = u sin \theta + gt

Substituting all the values and t = 3.78 s, the time of flight, we find the vertical velocity at the time of impact:

v_y = (17.0)(sin (-23^{\circ}))+(-9.8)(3.78)=-43.7 m/s

Where the negative sign means it is downward.

Therefore, the speed at impact is

v=\sqrt{v_x^2+v_y^2}=\sqrt{(15.6)^2+(-43.7)^2}=46.4 m/s

while the direction is given by

\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-43.7}{15.6})=-70.3^{\circ}

So, 70.3^{\circ} below the horizontal.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0
3 years ago
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