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Liono4ka [1.6K]

2 years ago

11

After you cross the finish line, you slow down and come to a stop, breathing hard. It takes you 10 seconds to come to a stop. Wh

at is your acceleration?

1 answer:

Otrada [13]2 years ago

4 0

Acceleration = (change in speed) / (time for the change)

<em>Acceleration = -- (speed as you crossed the finish line) / (10 seconds)</em>

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What is the magnitude of electrical force of attraction between an copper nucleus (29 protons) and its innermost electron if the

Agata [3.3K]

The charge of the copper nucleus is 29 times the charge of one proton:
$Q=29 q= 29 \cdot 1.6 \cdot 10^{-19}C=4.64 \cdot 10^{-18}C$
the charge of the electron is
$e=-1.6 \cdot 10^{-19}C$
and their separation is
$r=1.0 \cdot 10^{-12} m$

The magnitude of the electrostatic force between them is given by:
$F=k_e \frac{Qe}{r^2}$
where $k_e$ is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)
$F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(4.64 \cdot 10^{-18}C)(1.6 \cdot 10^{-19}C)}{(1.0 \cdot 10^{-12} m)^2}=6.68 \cdot 10^{-3} N$

3 0

3 years ago

At one instant, a 17.0-kg sled is moving over a horizontal surface of snow at 4.10 m/s. After 6.15 s has elapsed, the sled stops

jasenka [17]

Answer:

force = 11.33 $kg-m/s^{2}$

Explanation:

given data:

sled mass = 17.0 kg

inital velocity (U) = 4.10 m/s

elapsed time (T) 6.15 s

final velocity (V) = 0

final momentum P2 = 0

Initial momentum of sledge is

$P_{1}=mU$

$P_{1}= 17.0 * 4.10 = 69.7 kg- m/s$

from newton second law of motion

$F=\frac{\Delta P}{\Delta t}$

$F = \frac{P_{1}-P_{2}}{T}$

Kgm/s^2

$F = \frac{69.7-0}{6.15}= 11.33[tex]kg-m/s^{2}$ [/tex]

8 0

3 years ago

An object is thrown upwards with the speed of 40.0 km/hr in 5 second, calculate the distance in millimeters (mm).

sertanlavr [38]

V=(40km/hr)(hr/3600s)(1000000mm/km)
v=11111.1mm/s

v=d/t
d=vt
d=(11111.1mm/s)(5s)
d=55555mm
d=5.56x10^4mm

4 0

2 years ago

An automobile with an initial speed of 4.92 m/s accelerates uniformly at the rate of 3.2 m/s2 . Find the final speed of the car

Rudik [331]

Answer:19.32 m/s

Explanation:

Given

initial speed of car(u)=4.92 m/s

acceleration(a)= $3.2 m/s^2$

Speed of car after 4.5 s

using equation of motion

v=u+at

$v=4.92+3.2\times 4.5=4.92+14.4$

v=19.32 m/s

Displacement of the car after 4.5 s

$v^2-u^2=2as$

$19.32^2-4.92^2=2\times 3.2\times s$

$349.05=2\times 3.2\times s$

s=54.54 m

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2 years ago

Hi guys <br> I suggest to make group to exchange the informations if they need add their name here

sveticcg [70]

??

what’s the question

8 0

2 years ago

Read 2 more answers