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2 years ago

Which of the following statements is false?

Physics

2 answers:

Allushta [10]2 years ago

7 0

Answer:

Explanation:

The weight of an object is 1/6 of the earth's weight in the moon

For example if your weight is 54 kg on earth,

Weight on moon=1/6 of 54=1/6 x 54 = 9kg

olasank [31]2 years ago

7 0

Answer:

C. The weight of an object is the same on the Earth and the moon

Explanation:

The mass of an object is ⅙ of earth's weight on moon. So, option (C) is the false statement.

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)If a force of 5.00 N is needed to open a 90.0 cm wide door when applied at the edge opposite the hinges, what force must be app

masha68 [24]

Answer:

A force of 12.857 newtons must be applied to open the door.

Explanation:

In this case, a force is exerted on the door, a moment is performed and the door is opened. If moment remains constant, the force is inversely proportional to distance respect to axis of rotation passing through hinges. That is:

$F \propto \frac{1}{r}$

$F = \frac{k}{r}$ (Eq. 1)

Where:

$F$ - Force, measured in newtons.

$k$ - Proportionality ratio, measured in newton-meters.

$r$ - Distance respect to axis of rotation passing through hinges, measured in meters.

From (Eq. 1) we get the following relationship and clear the final force within:

$F_{A}\cdot r_{A} = F_{B}\cdot r_{B}$

$F_{B}=\left(\frac{r_{A}}{r_{B}} \right)\cdot F_{A}$ (Eq. 2)

Where:

$F_{A}$ , $F_{B}$ - Initial and final forces, measured in newtons.

$r_{A}$ , $r_{B}$ - Initial and final distances, measured in meters.

If we know that $F_{A} = 5\,N$ , $r_{A} = 0.9\,m$ and $r_{B} = 0.35\,m$ , then final force is:

$F_{B}= \left(\frac{0.9\,m}{0.35\,m} \right)\cdot (5\,N)$

$F_{B} = 12.857\,N$

A force of 12.857 newtons must be applied to open the door.

3 0

2 years ago

I KNOW YALL SEE THIS I NEED HELP GOD WILL GIVE U MANY BLESSING HELP A POOR SOUL OUT THE RECENT QUESTIONSSSSSSSSSSSSS. ://///

photoshop1234 [79]

Answer:

Explanation:

will do

4 0

2 years ago

Read 2 more answers

A flat uniform circular disk (radius = 2.30 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the

Vadim26 [7]

The resulting angular speed = 0.6 rad / s.

<u>Explanation:</u>

Here there is no external torque acting on the system thus we can apply the law of conservation of angular momentum

Angular momentum of the man = Iω

Where I = Inertia of the man about the axis of rotation

or I = M r 2

I = 50 * 1.25*1.25 = 78.125

w = Angular velocity of the man, that can be calculated as follows

Tangential velocity of man = v = 2m/s

So time taken to describe this circle is t = (2*pi* r) / v

Now angle described in 1 revolution θ = 2*pi radians

This angle is subtended in time t = (2*pi* r) / v

Thus angular speed = w = θ/t = 2*pi* ( v/ 2π r) = v/r = 2.70 / 1.25 = 2.16 rad/s

So angular momentum of man = Iw = 78.125 * 2.16 = 168.75.

To conserve the angular momentum before and after,

Angular momentum of disk = angular momentum of the man

i.e. Iw of disk = 168.75

disk of I = (disk of M*R^2) / 2

= (1.00 * 102 * 2.30 * 2.30) / 2

= 269.79

Thus 269.79 of disk of w = 168.75

Resulting angular speed of disk = 168.75 / 269.79 = 0.6 ras / s

7 0

3 years ago

A food packet is dropped from a helicopter during a flood-relief operation from a height of 750 meters. Assuming no drag (air fr

Leviafan [203]

1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration $a=g=9.8 m/s^2$ directed downward, initial vertical position $d=750 m$ , and initial vertical velocity $v_0 = 0$ . We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

$v_f^2 -v_i^2 =2ad$

substituting, we find

$v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s$

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

$v_f=\frac{121.2 m/s}{2}=60.6 m/s$

In order to do that, we use again the same SUVAT equation substituting $v_f$ with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity: