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3 years ago

HI PLEASE HELP, I ONLY HAVE 10 MINUTES

Physics

1 answer:

Whitepunk [10]3 years ago

5 0

Answer:

um ok

?????

i dont like you get rejected

jk u gud

Explanation:

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A girl picks up a bag of apples that are at rest on the floor. How dose the force the girl applies compare to the force of gravi

AysviL [449]

Answer:

When the bag of apples are on the floor, the force of gravity is equal to the force applied in the upward direction.

But when the girl picks up the bag of apples, she applies a force on the bag of apples in the upward direction greater than that of the gravitational force.

Explanation:

Gravitational force for an object is always constant at a specific point. When an object is in the state of rest, there is an equal force opposite to the direction of Gravitational Force.

Now, to lift an object upwards, an external force must be applied that is greater than the force of gravitation.

So, when the girl picks up the bag of apples,

Gravitational Force < Force applied by the girl to lift the bag.

6 0

3 years ago

What's the answers for all of the questions ?///

Maru [420]

More energy, colour and I think less bright

8 0

2 years ago

How much time does it take light from a flash camera to reach a subject 6.0 meters across a room in scientific notation?

wariber [46]

The time it takes light from a flash camera to reach a subject 6.0 meters across a room in scientific notation is 2.0 *10^-8 s.

Explanation:

Given

t=?

d=6m

v=3*10^8 m/s

we have, v=d/t

here t=d/v

t=6m/3*10^8 m/s

v=2*10^-8 m/s

The time it takes light from a flash camera to reach a subject 6.0 meters across a room in scientific notation is 2.0 *10^-8 s.

6 0

2 years ago

What situation would give you a mechanical advantage? *

const2013 [10]

D. Using a fixed-pulley system

3 0

2 years ago

A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below th

zhannawk [14.2K]

Answer: $V_{f}=2.96m/s$

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight $w$ of the block has an x-component and y-component:

$w_{x}=wsin(\theta)$ (1)

$w_{y}=wcos(\theta)$ (2)

As well as the Normal Force $N$ :

$N_{x}=Nsin(\theta)$ (3)

$N_{y}=Ncos(\theta)$ (4)

In addition, we know $N=w$ , then $\sum F_{y}=0$

In the X-component:

$\sum F_{x}=m.a$

$m.a=w_{x}$ (5)

Substituting (1) in (5):

$wsin(\theta)=m.a$ (6)

In addition, we know $w=m.g$ , where $m$ is the mass of the block and $g$ the gravity acceleration, which is equal to $9.8m/{s}^{2}$

So:

$m.g.sin(\theta)=m.a$ (7)

$a=g.sin(\theta)$ (8)

$a=5.88m/{s}^{2}$ (9) >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a relation between the distance $d$ with the acceleration $a$ and time $t$ :

$d=\frac{1}{2}a{t}^{2}$ (10)

We already know the value of $d$ and calculated $a$ , we have to find $t$ :

$t=\sqrt{\frac{2d}{a}}$ (11)

$t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}$ (12)

$t=0.50s$ (13) >>>This is the time it takes to the block to go from the initial velocity $V_{o}$ to its final velocity $V_{f}$

If the acceleration is the variation of the velocity in time, we can use the following equation to find $V_{f}$ :

$V_{f}-V_{o}=a.t$ (13)

If $V_{o}=0$

$V_{f}=a.t$ (14)

$V_{f}=(5.88m/{s}^{2})(0.50s)$ (15)

Finally we get the value of the Final Velocity of the block:

$V_{f}=2.96m/s$

6 0

3 years ago