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17. Suppose you were standing on a scale in an elevator in free fall. What would the scale read?

lianna [129]

17

What would the scale read? zero

18 In free fall you are being pulled by a gravity. "Truly" weightless presumably happens in deep space where there is nothing to pull you.

19 coasters accelerate down to simulate weight loss/zeroised. As do NASA planes,

Roller coasters are for fun seekers. NASA is for science

5 0

3 years ago

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A student makes a model of the sun-Earth system by swinging a ball around her head. Using this model, the student is trying to e

DedPeter [7]

Answer:

c. gravitational attraction between the sun and earth

8 0

3 years ago

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A crate of 45.2-kg tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and observe that th

iragen [17]

Answer:

μ = 0.725

Explanation:

This problem refers to Newton's second law.

F = ma

Let's write the equations on each axis

Y Axis

N-W = 0

N = W

N = mg

X axis

F-fr = ma

With the body not started moving its acceleration is zero

F-fr = 0

F = fr

The friction force equation is

fr = μ N

fr = μ m g

Let's replace and calculate

F = μ m g

μ = F / mg

μ = 321 /45.2 9.8

μ = 0.725

8 0

4 years ago

Projectile Motion: A hobbyist launches a projectile from ground level on a horizontal plain. It reaches a maximum height of 72.3

marin [14]

Answer:

The angle of launch from the horizontal direction is 20.99° .

Explanation:

Let u and θ be the initial speed and angle of projection from the horizontal axis of the object respectively.

The equations for projectile motion are :

H = ( u² sin²θ)/ 2g ......(1)

Here H is maximum height of the projectile motion and g is acceleration due to gravity.

R = ( u² sin2θ)/g .......(2)

Here R is the maximum horizontal displacement of the object.

Rearrange equation (1) in terms of u².

u² = (2gH)/sin²θ

Substitute this equation in equation (2).

R = (2gH sin2θ) / (sin²θ x g)

R = (2H sin2θ)/sin²θ

Using trigonometry property, sin2θ = 2 cosθ sinθ

So, above equation becomes,

R = (2H x 2 cosθ sinθ)/sin²θ

R = (4H cosθ)/sinθ

tanθ = R/4H

θ = tan⁻¹(R/4H)

Substitute 111 m for R and 72.3 m for H in the above equation.

θ = tan⁻¹( 111/ 4 x 72.3 )

θ = tan⁻¹(0.38)

θ = 20.99°

3 0

3 years ago

In one contest at the county fair, a spring-loaded plunger launches a ball at a speed of 3.2m/s from one corner of a smooth, fla

lara31 [8.8K]

Answer:

Explanation:

Given

Speed of ball $u=3.2\ m/s$

Plane is inclined at an angle $20^{\circ}$

To win the Game we need to hit the target at $x=2.4\ m$ away

Launch angle of ball $\theta$

Motion of ball can be considered in two planes i.e. Vertical to the plane and horizontal to the plane

So Net acceleration in vertical plane is $g\sin 20$

Range of Projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

for $R=2.4\ m$

$2.4=\frac{3.2^2\times sin 2\theta }{g\sin 20}$

$\sin 2\theta =\frac{2.4\times 9.8\times \sin 20}{3.2^2}$

$\sin 2\theta =0.7855$

$2\theta =51.77$

$\theta =25.88^{\circ}$

so ball must be launched at an angle of $25.88^{\circ}$

4 0

3 years ago