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2 years ago

Will give 45 points to who ever solves this

Physics

2 answers:

kotegsom [21]2 years ago

8 0

Answer:

1)m=230 kg

w= 2254( approximately 2300) newtons

g= 9.8 m/s^2(approximately 10)

2)m= 90.5( approximately 88.7) kg

w= 887 N

g= 9.8 m/s^2(approximately 10)

3) m= 420 kg

w= 4,116 (approximately 4220

g= 9.8 m/s^2(approximately 10)

4) m=15kg

w=55.5N

g=3.7 m/s^2

Explanation:

The wanswers given in brackets is to signify the results of using 10m/s^2 as the gravity if u have any misunderstanding text back

Lubov Fominskaja [6]2 years ago

7 0

Answer:

1. 230 kg...

W = m * g

W= 230 kg * 9.81 m/s^2

W= 2256,3 N

m= 230kg , W = 2256,3 N , g= 9.81 m/s^2

2. 887 N

W= m * g

887 N = m * 9.81 m/s^2

m= 90,42 kg

m= 90,42 kg, W = 887 N , g= 9.81 m/s^2

3. 420 kg

W= m * a

w= 420 kg * 9.81 m/s^2

w= 4120,2 N

m= 420 kg , W = 4120,2 N , g= 9.81 m/s^2

4. Determine the gravity on Pluto where a 15 kg object weighs 55.5N.

w = m * g

55.5 N = 15 kg * g

a= 3,7 m/s^2

m = 15 kg , W= 55.5 N , g= 3,7 m/s^2

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3 years ago

A projectile is thrown with an initial speed vo = 25.0 m/s at an initial angle with the horizontal ao = 45.0. a . Find the time

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Answer:

See the explanation below

Explanation:

To solve this problem we must decompose the initial speeds into x & y.

$v_{o}_{x}=25*cos(45)=17.67[m/s]\\v_{o}_{y}=25*sin(45)=17.67[m/s]\\$

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Note: Acceleration is taken as negative as it is directed downwards.

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$x=(v_{x})_{o}*t\\x=17.67*1.8\\x=31.82[m]$

The position in the y component can be found using the following kinematic equation

$y =(v_{y})_{o}*t+\frac{1}{2} *g*t^{2} \\y=17.67*1.8-0.5*9.81*(1.8)^{2}\\ y=15.91[m]$

Since the motion on the X-axis is at constant speed, there is no acceleration, so the only acceleration that exists is due to gravity

In the attached image we can see, the projectile with the vectors of acceleration and velocity.

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3 years ago