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3 years ago

Can someone help me with any of these problems or all of them? It will get you an easy ten points per question

Physics

1 answer:

MrRa [10]3 years ago

6 0

Answer: For number 9, the children would not be able to close the box.

Explanation: The combined force of the children is 55 N. The box spring overpowers them with 60 N. I hope this helps!

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If you walk at an average speed of 5 km/h for 30 minutes, how

Inessa05 [86]

The distance that would be accumulated during the journey is 2.5 meters

The parameters given in the question are written below;

average speed= 5 km/hr

time = 30 minutes

convert 30 minutes to hours

= 30/60

= 0.5 hours

Distance-= speed × time

= 5 × 0.5

= 2.5 meters

Hence the distance of the entire journey is 2.5 meters

Please see the link below for more information

brainly.com/question/24268730?referrer=searchResults

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2 years ago

What is the resultant of a and b if a = 3i 3j and b = 3i − 3j?

ss7ja [257]

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3 years ago

Read 2 more answers

A machine gun fires 50-g bullets at the rate of 4 bullets per second. The bullets leave the gun at a speed of 1000 m/s. What is

TEA [102]

Answer:

Average recoil force experienced by machine will be 200 N

Explanation:

We have give mass of each bullet m = 50 gram = 0.05 kg

There are 4 bullets

So mass of 4 bullets = 4×0.05 = 0.2 kg

Initial speed of the bullet u = 0 m/sec

And final speed of the bullet v = 1000 m/sec

So change in momentum $P=m(v-u)=0.2\times (1000-0)=200kgm/sec$

Time is given per second so t = 1 sec

We know that force is equal to rate of change of momentum

So force will be equal to $F=\frac{200}{1}=200N$

So average recoil force experienced by machine will be 200 N

5 0

3 years ago

Mấy bạn việt nam giúp mình với. cần gấp quá

Basile [38]

Saying english so we can help u

6 0

2 years ago

A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back

elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity= $9.8 m/s^2$

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates= $\mu mg=0.202\times 9.8\times 35$

Force by which box accelerates= $62.286 N$

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

$ma=\mu mg$

$a=\mu g=9.8\times 0.202=1.9796 m/s^2$

Hence, the truck can have maximum acceleration before the box slide= $1.9796 m/s^2$

3 0

3 years ago