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jekas [21]
3 years ago
10

PLS ANSWER FAST WILL GIVE BRAINLEST!!

Physics
2 answers:
Maurinko [17]3 years ago
5 0

\huge\bf{\pink{\underline{\underline{\mathcal{AnSwer࿐}}}}}

<h2>Given:- </h2>

acceleration = 1000 m/s²

force = 5000 N

<h2>To find:- </h2>

Mass of the cannonball

<h2>Formula to be used:- </h2>

\longrightarrow \underline{\boxed{\sf mass = \dfrac{force}{acceleration}}}

<h2>Solution:- </h2>

:\implies \sf mass = \dfrac{5000}{1000}

:\implies \sf mass = 5 kg

hence, the required answer is 5kg.

______________________________

egoroff_w [7]3 years ago
3 0

Answer:

from

force =mass x acceleration

mass = force/acceleration

m = f/a

m = 7.5/15

m=0.5kg

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Simora [160]

The basic sciences are defined as the scientific disciplines of mathematics, physics, chemistry, and biology. They are called basic sciences because they provide a fundamental understanding of natural phenomena and the processes by which natural resources are transformed.

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2 years ago
What is the lowest temperature that a substance can reach
irina1246 [14]
It's known as Absolute Zero. On the Kelvin scale, 0 is the lowest that anything can reach in temperature. It's supposedly impossible to reach, but it's the known limit.
4 0
3 years ago
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A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of
koban [17]

<u>Answer:</u>

 Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

  Let the east point towards positive X-axis and north point towards positive Y-axis.

  Speed of truck = 25 m/s north = 25 j m/s

  Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

                          = (1.43 i + 1.00 j) m/s

    Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

    Magnitude of velocity = 26.04 m/s

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4 0
3 years ago
If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you
Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, Q=121\ nC=121\times 10^{-9}\ C

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

Q=ne

e is the charge on electron

n=\dfrac{Q}{e}

n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}

n=7.5625\times 10^{11}

or

n = 756.25 giga electrons

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6 0
3 years ago
An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of th
jeyben [28]

Answer:

the shooting angle ia 18.4º

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For resolution of this exercise we use projectile launch expressions, let's see the scope

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To know how for the arrow the tree branch we calculate the height of the arrow at this point

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8 0
3 years ago
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