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3 years ago

What elements are not a major category of elements on the periodic table

Physics

1 answer:

lana66690 [7]3 years ago

3 0

Answer:

Group and Periods of the Periodic Table of Elements. The three major groups on the Periodic Table are the metals, nonmetals and metalloids. Elements within each group have similar physical and chemical properties.

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What height would a 4 kg book need to be to have a potential energy of

zloy xaker [14]

Answer:

5.99 m = 6 m

Explanation:

PE = m*g*h

235.2 J = (4 kg)(9.81 m/s^2)(h)

h = (235.2 J)/(9.81*4)

h = 5.99 m

h = 6 m

5 0

3 years ago

I need help with the last question- it’s very simple

77julia77 [94]

You calculated the density correctly.

When you drop anything into a fluid . . .

-- If the object is MORE dense than the fluid, it sinks.

-- If the object is LESS dense than the fluid, it floats.

3 0

3 years ago

Dennis throws a volleyball up in the air. It reaches its maximum height 1.1\, \text s1.1s1, point, 1, start text, s, end text la

rewona [7]

Answer:

If max height = 1.1 meters, then initial velocity is 3.28 m/s

If max height is 1.1 feet, then the initial velocity is 5.93 ft/s

Explanation:

Recall the formulas for vertical motion under the acceleration of gravity;

for the vertical velocity of the object we have

$v=v_0-g \,t$

for the object's vertical displacement we have

$y-y_0=v_0\,t - \frac{g}{2} \,t^2$

If the maximum height reached by the object is given in meters, we use the value for g in $m/s^2$ which is: $9.8\,\,m/s^2$

If the maximum height of the object is given in feet, we use the value for g in $ft/s^2$ which is : $32\,\,ft/s^2$

Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:

$v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}$

and now we use this to express the maximum height in the second equation we typed:

$y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g}) - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}$

Then if the max height is 1.1 meters, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s$

If the max height is 1.1 feet, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s$

5 0

3 years ago

Read 2 more answers

1. A bus starting from rest moves

skelet666 [1.2K]

Answer:

Speed acquired (Final velocity) = 12 m/s

Distance travelled = 720 meter

Explanation:

Given:

Uniform acceleration = 0.1 m/s²

Initial velocity = 0 m/s

Time taken = 2 minutes = 2 x 60 = 120 second

Find:

(a) Speed acquired (Final velocity)

(b) Distance travelled

Computation:

v = u + at

v = 0 + 0.1(120)

Speed acquired (Final velocity) = 12 m/s

Distance travelled = ut + (1/2)(a)(t²)

Distance travelled = (0)(120) + (1/2)(0.1)(120²)

Distance travelled = (1/2)(0.1)(14400)

Distance travelled = 720 meter

8 0

3 years ago

A body is thrown up with a velocity of 78.4 m per second.How high will it rise and how much time it will take to return to its p

a_sh-v [17]

Answer:

The maximum height reached by the body is 313.6 m

The time to return to its point of projection is 8 s.

Explanation:

Given;

initial velocity of the body, u = 78.4 m/s

at maximum height (h) the final velocity of the body (v) = 0

The following equation is applied to determine the maximum height reached by the body;

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u²/2g

h = (78.4²) / (2 x 9.8)

h = 313.6 m

The time to return to its point of projection is calculated as follows;

at maximum height, the final velocity becomes the initial velocity = 0

h = v + ¹/₂gt²

h = 0 + ¹/₂gt²

h = ¹/₂gt²

2h = gt²

t² = 2h/g

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 313.6}{9.8} }\\\\t = 8 \ s$

4 0

3 years ago