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3 years ago

Describe how you would draw a diagram that shows the electric field between two particles that

Physics

2 answers:

Furkat [3]3 years ago

8 0

i got a 5/5

20 characters

dimaraw [331]3 years ago

6 0

Answer:

draw the two atomic structures and then focus in on the electrons either gaining or losing and draw them with arrows and dots

Explanation:

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Which of the following changes when an unbalanced force acts on an object?

KengaRu [80]

What are the choices? then i will answer

3 0

4 years ago

At the train station, you notice a large horizontal spring at the end of the track where the train comes in. This is a safety de

xxMikexx [17]

Answer:

v₀ = 9,798 ft / s

Explanation:

We can solve this problem with kinematics in one dimension, when the train stops the speed is zero, the acceleration is negative so that the train stops. Let's use the equation

v² = v₀² - 2 a d

v = 0

v₀ = √2 a d

In the problem it indicates that the acceleration is g / 2, we substitute

v₀ = √2 (g / 2) d

Let's calculate

v₀ = √ 2 32/2 3 = √32 3

v₀ = 9,798 ft / s

8 0

4 years ago

Jesse is swinging Miguel in a circle at a tangential speed of 3.50 m/s. If the radius of the circle is 0.600 m and Miguel has a

vovangra [49]

Answer:

224.6 N

Explanation:

We can use first the formula to calculate the centripetal acceleration, given by:

$a_c=\frac{v_t^2}{R}$

where the Vt is the tangential velocity, and R is the radius of the circular motion.

Then, for our case we have:

$a_c=\frac{v_t^2}{R}=\frac{3.5^2}{0.6}\approx 20.42\,\,\frac{m}{s^2}$

And now we multiply this acceleration by Miguel's mass (11 kg) to obtain the centripetal force acting on him:

$F_c=11 \,*\,20.42\,N = 224.6\,\,N$

4 0

3 years ago

Read 2 more answers

Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.

KIM [24]

Answer:

The velocity is $v = 4.76 \ m/s$

Explanation:

From the question we are told that

The first distance is $d_1 = 4.0 \ km = 4000 \ m$

The first speed is $v_1 = 5.0 \ m/s$

The second distance is $d_2 = 1.0 \ km = 1000 \ m$

The second speed is $v_2 = 4.0 \ m/s$

Generally the time taken for first distance is

$t_1 = \frac{d_1 }{v_1 }$

$t_1 = \frac{4000}{5}$

$t_1 = 800 \ s$

The time taken for second distance is

$t_1 = \frac{d_2 }{v_2 }$

$t_1 = \frac{1000}{4}$

$t_1 = 250 \ s$

The total time is mathematically represented as

$t = t_1 + t_2$

=> $t = 800 + 250$

=> $t = 1050 \ s$

Generally the constant velocity that would let her finish at the same time is mathematically represented as

$v = \frac{d_1 + d_2}{t }$

=> $v = \frac{4000 + 1000}{1050 }$

=> $v = 4.76 \ m/s$

7 0

4 years ago

How long would it take you to travel 163 meters at a speed of 33 meters per second?

Leviafan [203]

Speed=

$33m {s}^{ -1}$

Distance=

$163m$

speed= distance/time

So,

Time= distance/speed

Time= 163/33

$\huge\underline\mathtt\colorbox{cyan}{4.9393s}$

7 0

3 years ago