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3 years ago

12

. At what velocity will the box in Problem 6 be traveling when it hits the ground? Use formula: Va = Vf - Vi / 2

1 answer:

VladimirAG [237]3 years ago

5 0

Answer:

How much time does his victim on the ground below have to move out of harm's way? At what velocity will the safe hit the ground? sownt d-200m. n a = 100/2. Vi-o.

Explanation:

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Explain why you can only change one variable in an experiment

ohaa [14]

The purpose of an experiment is to LEARN the EFFECT of something.

The way you do that is to CHANGE the thing and see what happens.

You can change as many things as you want to. But If you change
TWO things and observe the result, then you don't know which one
of them caused the effect you see.

Or maybe BOTH of them working together caused it. You don't know.

So your experiment is not really much good. You need to do it again.

5 0

3 years ago

Atoms in a molecule are bonded together by sharing gaining or losing_

Alex

It is by gaining! I just learned this in school. Hope this helps:)

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3 years ago

Read 2 more answers

Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, in

olya-2409 [2.1K]

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

$I=\Delta p = m\Delta v$

where

m = 62.0 kg is the mass of the passenger

$\Delta v$ is the change in velocity of the car (and the passenger), which is

$\Delta v = 5.30 m/s - 0 = 5.30 m/s$

So, the linear impulse experienced by the passenger is

$I=(62.0 kg)(5.30 m/s)=328.6 kg m/s$

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

$I=F \Delta t$

where in this case

$I=328.6 kg m/s$ is the linear impulse

$\Delta t = 0.812 s$ is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

$F=\frac{I}{\Delta t}=\frac{328.6 kg m/s}{0.812 s}=404.7 N$

7 0

4 years ago

A ship leaves a harbor and sails at 23.5 ∘ to the north of due west. After traveling 575 km, how far west is the ship from the h

solniwko [45]

Answer:

D = 527.31 Km

Explanation:

given,

angle of ship, θ = 23.5° N of W

distance travel in the direction = 575 Km

Distance of ship in west from harbor = ?

now,

Distance of the ship in the west direction

D = d cos θ

d = 575 Km

θ = 23.5°

inserting all the values

D = 575 x cos 23.5°

D = 575 x 0.91706

D = 527.31 Km

Hence, the distance travel by the ship in west from harbor is equal to D = 527.31 Km

6 0

3 years ago

One knife is longer from the cutting edge to the back of the blade. It's easier to push the longer knife through a pear and more

AlladinOne [14]

By the formula of torque balance we know that

$\tau = r\times F$

here we know that

r = distance of force

F = applied force

so here we know that as we will take large edge knife then the distance of force will increase

due to this the torque will also increase in it so it is easy to pear it

So here we can say

$F_{in}r_1 = F_{out}r_2$

$F_{out} = \frac{r_1}{r_2} F_{in}$

so output force will increase in this case

So here the correct answer would be

<em>b. the longer the knife, the stronger the output force</em>

6 0

4 years ago