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fredd [130]
4 years ago
12

Neglecting air resistance, which of the following statements is true regarding an object in freefall?Lector inmersivo A. An obje

ct’s acceleration increases at a constant rate. B. An object maintains a constant velocity. C. An object falls an equal distance in between each second. D. An object’s velocity and acceleration both increase at a constant rate. E. An object’s velocity changes at a constant rate, and its acceleration remains constant.
Physics
1 answer:
Dafna1 [17]4 years ago
3 0

Answer:

E. An object’s velocity changes at a constant rate, and its acceleration remains constant.

Explanation:

When an object is in freefall, it implies that the object is falling freely under gravity. If it falls towards the earth surface, the fall is in the direction of the Earth's gravitational force.

At the point of release of the object, its initial velocity is zero because it is at rest. But when released, its velocity increases at a constant rate until it is acted upon by an external force. But its acceleration remains constant, acceleration due to gravity.

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A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow
trapecia [35]

Answer:N=13.53 rpm

Explanation:

Given

radius r=7.80 m

Centripetal acceleration a_c=1.6 g

and centripetal acceleration a_c=\omega ^2r

where \omega =angular\ velocity

1.6\times 9.8=\omega ^2\times 7.8

\omega ^2=2.0102

\omega =1.417 rad/s

and  \omega =\frac{2\pi \cdot N}{60}

1.417\times 60=2\pi \cdot N

N=13.53 rpm

8 0
3 years ago
Scientific ideas about the solar system have changed over time. Which of them
Usimov [2.4K]

Answer:

Explanation:

If i'm not wrong and late it might be F

7 0
3 years ago
a paper airplane gliding down towards the ground will experience the force of air resistance pushing up. the weight of the paper
Oduvanchick [21]

The net force acting on the airplane is 25N.

Forces acting on the paper airplane when it is in the air:

  • The forward force generated by the engine, propeller, or rotor is called thrust. It resists or defeats the drag force. It operates generally perpendicular to the longitudinal axis. However, as will be discussed later, this is not always the case.
  • Drag is an airflow disruption generated by the wing, rotor, fuselage, and other projecting surfaces that causes a backward, decelerating force. Drag acts backward and perpendicular to the relative wind, opposing thrust.
  • Weight is the total load carried by airplane, including the weight of the crew, fuel, and any cargo or baggage. Due to the influence of gravity, weight pulls the airplane downward.
  • Lift—acts perpendicular to the flight path through the center of lift and opposes the weight's downward force. It is produced by the air's dynamic influence on the airfoil.

Given.

Weight of the paper airplane, F1 = 16N

The force of air resistance, F2 = 9N

Net force = F1 + F2

Net force = 25N

Thus, the net force acting on the airplane is 25N.

Learn more about the net force here:

brainly.com/question/18109210

#SPJ1

3 0
1 year ago
A faulty thermometer reads 2°C when dipped in ice at 0°C and 95°C when dipped in steam at 100°C. What would this thermometer rea
kvasek [131]

Answer:

The read will be 20.9[C]x=\frac{(y-y_{1} )}{(y_{2} -y_{1} )}*(x_{2}-x_{1})+y_{1}\\  x=\frac{(18-0 )}{(95 -0 )}*(100-2)+2\\\\x= 20.9[C]

Explanation:

This is a problem related to linear interpolation, Linear interpolation consists of tracing a line through two known points y = r (x) and calculating the intermediate values according to this line.

The equation of a known line two points (x1, y1)and (x2, y2) = (2,0) (100,95) is:

\frac{(y-y_{1} )}{(y_{2} -y_{1} )}=\frac{(x-y_{1} )}{(x_{2}-x_{1}  )}

If we clear y from the equation we have:

[tex]y=\frac{(x-y_{1})}{(x_{2}-x_{1} )}*(y_{2} -y_{1} )+y_{1}  \\replacing

4 0
3 years ago
Two forces are acting on an object. The first force has magnitude F1=33.4 N and is pointing at an angle of θ1=23.8 clockwise fro
lesya [120]

Answer:

a)Fnx = -26.92 N

b)Fny= 8.4 N

c)Fex = 26.92 N

d)Fey = - 8.4 N

e)β = 18.34° clockwise from the positive x axis

Explanation:

Look at the attached graphic

a) What is the x component of the net force acting on the object?

Fnx:x component of the net force acting on the object

Fnx= F₁x+ F₂x

F₁x=  33.4*sin 23.8° = 13.48 N

F₂x= - 46.1*cos 28.8°= - 40.4 N

Fnx =  13.48 N- 40.4 N= -26.92 N

b) What is the y component of the net force acting on the object?

Fny= F₁y+ F₂y

F₁y=  33.4*cos 23.8° =30.6 N

F₂y= - 46.1*sin 28.8°= -22.2 N

Fny=30.6 N-22.2 N = 8.4 N

c) What is the x component of the equilibrant?

Fex: component of the equilibrant

Fex = - Fnx

Fex = - ( -26.92 N)

Fex = 26.92 N

d) What is the y component of the equilibrant?

Fey = - Fnx

Fey = - ( 8.4 N)

Fey = - 8.4 N

e) What is the magnitude of the equilibrant?

F_{e} :  equilibrant force

F_{e} = \sqrt{(F_{e}x)^{2}+{(F_{e}y)^{2} }

F_{e} = \sqrt{(26.92)^{2}+{(-8.4)^{2} }

F_{e} = 28.2 N

f) What is the angle the equilibrant makes with the x axis?

\beta =tan^{-1} (\frac{-8.94}{26.96} )

β = -18.34° or β = 18.34° clockwise from the positive x axis

5 0
3 years ago
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