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3 years ago

An Object moving 10 m/s undergoes acceleration for 6 seconds. If it moves 8 m to the left, find the acceleration

1 answer:

4 0

$\Delta x= V_0t+ \frac{1}{2}at^{2} \\ \\ 8=10(6)+ \frac{1}{2}a(6)^{2} \\ \\ 8=60+ \frac{1}{2}(36)a \\ \\ 8=60+13a \\ \\ -52=13a \\ \\ \frac{-52}{13}=a \\ \\ \boxed{a=-4m/s^{2}}$

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A metal block has a mass of 5 grams and a volume of 1 cm3. What is the density of a piece of the same identical metal but with a

Fynjy0 [20]

Answer:

ρ = Mass / Volume definition of density

ρ = 5 g / 1cm^3 = 5 g / cm^3

Since the other object is made of the same metal its density is the same:

ρ = 5 g/cm^3

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2 years ago

What is the word that goes with the definition “ a disturbance in space that causes a force”

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Gravity

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Gravity is the force that causes a "disturbance" in space between planets.

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3 years ago

The record for the highest speed achieved in alaboratory for a

Ira Lisetskai [31]

Answer:

0.000234 seconds

Explanation:

Since the row is 0.15m, its radius of rotation must be 0.15 / 2 = 0.075 m

We can start by calculating the angular speed of the rod:

$\omega = \frac{v}{r} = \frac{2010}{0.075} = 26800 rad/s$

Since one revolution equals to 2π rad. The speed in revolution per second must be

26800 / 2π = 4265 revolution/s

The number of seconds per revolution, or period, is the inverse:

1/4265 = 0.000234 seconds

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2 years ago

The blood pressure of human body is greater at the feet than at the brain. why?give reason

SOVA2 [1]

Answer:

Gravity stronger closer to the core of the earth

Explanation:

The gravity applied to the feet is stronger than to the upper part of the body for example the brain because of the distance it is between the body part and the core of the earth where the gravity force is pulling towards. Even though the difference between the gravity force between the brain and the feet is minimal it is still a greater force at the feet than at the brain

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3 years ago

What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?

Daniel [21]

To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as

$F=\frac{kq_{1}q_{2} }{r^{2} }$

Here, $q_{1}$ and $q_{2}$ are the charges on the pith balls, r is the separation between the charges and k is constant and its value is $8.99\times 10^9 N m^2/C^2$ .

Given $q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C$ and $r=8 cm=8\times10^{-2} m$ .

Substituting these values in above formula we get,

$F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9} N\\\\F=8.8\times10^{-4}N$

Thus, the repulsive force between two pith balls is $8.8\times10^{-4}N$ .

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3 years ago

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