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Minchanka [31]
3 years ago
15

For an object to be in projectile motion, what force must be acting on it

Physics
1 answer:
Lina20 [59]3 years ago
3 0

Answer:

In projectile motion, only the force of gravity (weight) acts on the object.

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Peter, a 100 kg basketball player, lands on his feet after completing a slam dunk and then immediately jump up again to celebrat
Ludmilka [50]

Answer:

1800 N

Solution:

Impulse = mΔv = m * (u - v) .

 here m = 100 kg

          u = 4 m/s

          v = -5 m/s

  impulse = 100 x ( 4 - ( -5 ) ) = 900 Kg m/s .

Average reaction Force ( Favg ) = impulse / Δt

Average reaction Force ( Favg )  = 900kg·m/s / 0.5s

Average reaction Force ( Favg )  = 1800 N

6 0
2 years ago
A man applies a force of 100 Newtons to a rock for 60 seconds, but the rock does not not move. What is the amount of work done b
frozen [14]
Well if the rock doesn't move, then there is no amount of work done. There is no work done on an object if a force is applied to the object but it DOES NOT change its position, in this case is the rock.
7 0
3 years ago
Ryan is looking at investment opportunities as a cloud service provider. he wants to invest in a deployment-based cloud model th
Dominik [7]

Answer : a. Community

Allows a system to be accessible by a group of organizations. It can be shared between several organizations. It may be managed by organizations or by the third party.


This should be chosen by Ryan, since this computing model is cost effective and best to share among companies and organizations.



Other options explained:

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5 0
2 years ago
Read 2 more answers
A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. The stone lands 36.0 meters away.
alisha [4.7K]
A) 

     It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:

S=S_{o}+v_{o}t+ \frac{at^2}{2} \\ 20= \frac{10t^2}{2} \\ t=2s
 
     Through Definition of Velocity, comes:

\Delta v=  \frac{\Delta S}{\Delta t}  \\ v_x= \frac{36}{2}  \\ \boxed {v_{x}=18m/s}


B)
 
     Using the Velocity Hourly Equation in vertical direction, we have:

v_{y}=v_{y_{o}}+gt \\ v_{y}=10\times2 \\ \boxed {v_{y}=20m/s}
  
     The angle of impact is given by:

cos(\theta) =\frac{v_{x}}{v_{y}}  \\ cos(\theta) = \frac{18}{20}  \\ cos(\theta) =0.9 \\ arccos(0.9)=\theta \\ \boxed {\theta \approx 25.84}


If you notice any mistake in my english, please let me know, because i am not native.

7 0
2 years ago
Read 2 more answers
g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr
Lorico [155]

To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

In other words the acceleration can be described as

a = \frac{GM}{r^2}

Where

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius of Earth

This equation can be differentiated with respect to the radius of change, that is

\frac{da}{dr} = -2\frac{GM}{r^3}

da = -2\frac{GM}{r^3}dr

At the same time since Newton's second law we know that:

F_w = ma

Where,

m = mass

a =Acceleration

From the previous value given for acceleration we have to

F_W = m (\frac{GM}{r^2} ) = 600N

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:

dF_W = mda

dF_W = m(-2\frac{GM}{r^3}dr)

dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})

dF_W = -2F_W(\frac{dr}{r})

But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})

dF_W = -0.3N

Therefore there is a weight loss of 0.3N every kilometer.

4 0
3 years ago
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