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3 years ago

Suppose your walking speed is 2m/s in a period of 1 s . What is your acceleration?

Physics

1 answer:

Veronika [31]3 years ago

6 0

Answer:

Explanation:

The acceleration of an object given it's velocity and time taken can be found by using the formula

$a = \frac{v}{t} \\$

v is the velocity

t is the time taken

From the question we have

$a = \frac{2}{1} \\$

We have the final answer as

Hope this helps you

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Neon has 10 protons and 10 electrons. How many additional electrons would it take to fill the outer (second) electron shell?

Yakvenalex [24]

None because Neon is already stable

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3 years ago

A helium nucleus is composed of two protons (positively charged) and two neutrons (charge neutral). The distance between the two

lora16 [44]

Answer:

Using the given values

F = K q^2 / r^2 = 9 * 10E9 * (1.6 * E-19)^2 / (5.18 * E-15)^2 N

E = 9 * 1.6^2 / 5.18^2 * 10 = 8.5 N

5 0

3 years ago

so classmate states that continental drift could not be possible because it would take far too much force to move tectonic plate

Juliette [100K]

Any student who states that tectonic plates are too massive to move doesn't understand how much heat exists in the center of the earth. This heat is capable of moving the most massive of plates.

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3 years ago

A baby carriage is rolling down a hill. Take into consideration, the carriage has 90J of kinetic energy and the mass of the carr

Sphinxa [80]

Answer:

YES. 6√10

Explanation:

apply E=1/2mv^2 and find the velocity of carriage which is 6√10 ms-1

as our velocity is higher than that of cart. we can catch the carriage

7 0

3 years ago

Two remote control cars with masses of 1.16 kilograms and 1.98 kilograms travel toward each other at speeds of 8.64 meters per s

Black_prince [1.1K]

The initial momentum of the system can be expressed as,

$p_i=m_1u_1+m_{2_{}}u_2$

The final momentum of the system can be given as,

$p_f=m_1v_1+m_{2_{}}v_2$

According to conservation of momentum,

$p_i=p_f$

Plug in the known expressions,

$\begin{gathered} m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ m_2v_2=m_1u_1+m_2u_2-m_1v_1 \\ v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2} \end{gathered}$

Initially, the second mass move towards the first mass therefore the initial speed of second mass will be taken as negative and the recoil velocity of first mass is also taken as negative.

Plug in the known values,

$\begin{gathered} v_2=\frac{(1.16\text{ kg)(8.64 m/s)+(1.98 kg)(-3.34 m/s)-(1.16 kg)(-2.16 m/s)}}{1.98\text{ kg}} \\ =\frac{10.02\text{ kgm/s-}6.61\text{ kgm/s+}2.51\text{ kgm/s}}{1.98\text{ kg}} \\ =\frac{5.92\text{ kgm/s}}{1.98\text{ kg}} \\ \approx2.99\text{ m/s} \end{gathered}$

Thus, the final velocity of second mass is 2.99 m/s.

3 0

1 year ago