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3 years ago

what is the approximate weight of a 20-kg cannonball on the moon if the acceleration due to gravity is 1.6m/s^2

1 answer:

monitta3 years ago

5 0

On Earth, a cannonball with a mass of 20 kg would weigh 196 Newtons.
With the formula F=mg, where F is the weight in Newtons, m is the mass, and g is the acceleration due to gravity on the Earth which is 9.8m/s^2.
F=20kg x 9.8m/s^2= 196 Newtons

BUT on the moon, acceleration due to gravity is 1.6 m/s^2,
so F=mg=20kgx1.6m/s^2= 32 N

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What are the three subatomic particles of an atom?

Anestetic [448]

Protons, neutrons, and electrons are the three main subatomic particles found in an atom. Protons have apositive (+) charge. An easy way to remember this is to remember that both proton and positive start with theletter "P." Neutrons have no electrical charge.

5 0

3 years ago

Read 2 more answers

A field measuring 12 meters by 16 meters is to have a brick paver walkway installed all around it, increasing the total area to

kolezko [41]

Answer:

1.5 m

Explanation:

Length. L = 12 m

Width, W = 16 m

Area, A = 12 x 16 = 192 m^2

Let the width of pavement be d.

The new length, L' = 12 + 2d

the new width, W' = 16 + 2d

New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2

Difference in area = A' - A

285 = 192 + 56 d + 4d^2 - 192

93 = 56 d + 4d^2

4d^2 + 56 d - 93 = 0

$d = \frac{-56\pm \sqrt{56^{2}+4\times 4\times 93}}{8}$

$d=\frac{-56\pm 87.72}{8}$ \

d = 1.5 m

Thus, the width of the pavement is 1.5 m.

6 0

4 years ago

While traveling along a highway a driver slows from 34 m/s to 17 m/s in 6 seconds. What is the automobiles acceleration?

xxTIMURxx [149]

Answer:

-2.83 m/s²

Explanation:

Initial velocity (u) = 34 m/s
Final velocity (v) = 17 m/s
Time taken (t) = 6 seconds

❖ Acceleration is defined as the rate of change in velocity with time.

→ a = (v - u)/t

v denotes final velocity
a denotes acceleration
u denotes initial velocity
t denotes time

→ a = (17 - 34)/6 m/s²

→ a = -17/6 m/s²

<h3>→ Acceleration = -2.83 m/s²</h3>

(Minus sign implies that the velocity is decreasing.)

5 0

3 years ago

A person swings a 0.546-kg tether ball tied to a 4.56-m rope in an approximately horizontal circle. If the maximum tension the r

Murrr4er [49]

Answer:

2.1 rad/s

Explanation:

Given that,

Mass of a tether ball, m = 0.546 kg

Length of a rope, l = 4.56 m

The maximum tension the rope can withstand before breaking is 11.0 N

We need to find the maximum angular speed of the ball. Let v is the linear velocity. The maximum tension is balanced by the centripetal force acting on it. It can be given by :

$F=\dfrac{mv^2}{r}\\\\v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{11\times 4.56}{0.546}} \\\\=9.584\ m/s$

Let $\omega$ is the angular speed of the ball. The relation between the angular speed and angular velocity is given by :

$v=r\omega\\\\\omega=\dfrac{v}{r}\\\\=\dfrac{9.584}{4.56}\\\\=2.1\ rad/s$

So, the maximum angular speed of the ball is 2.1 rad/s.

4 0

3 years ago

I NEED THIS QUICKLY

Yanka [14]

True.

I think that’s the answer.

8 0

4 years ago