How can a dictatorship best be classified

A Person whose weight is 5.20 x 10^2 N is being pulledup

Gnoma [55]

Answer:

Explanation:

Given

Weight of Person $W=5.20\times 10^{2} N$

Cave is $h=35.1 m$ deep

Breaking stress $T=569 N$

Net Force on Person

$F_{net}=569-520=49 N$

$a_{net}=\frac{F_{net}}{\frac{W}{g}}$

$a_{net}=\frac{49}{\frac{520}{9.8}}$

$a_{net}=0.923 m/s^2$

The shortest time such that the person can be taken out of cave

$h=ut+\frac{1}{2}at^2$

where

h=distance moved

t=time

a=acceleration

$35.1=0+\frac{1}{2}(0.923)(t)^2$

$t^2=76.05$

$t=\sqrt{76.05}$

$t=8.72\ s$

6 0

3 years ago

If a 5 N force pushes a 20 kg mass on a frictionless surface, how fast is the mass going in 5 seconds.

stira [4]

Explanation:

Let me know if you have questions

3 0

3 years ago

You push on a cart (18.0kg) at a 30 degree below horizontal angle. The coefficient of kinetic friction between the chair and the

podryga [215]

Given that force is applied at an angle of 30 degree below the horizontal

So let say force applied if F

now its two components are given as

$F_x = Fcos30$

$F_y = Fsin30$

Now the normal force on the block is given as

$N = Fsin30 + mg$

$N = 0.5F + (18\times 9.8)$

$N = 0.5F + 176.4$

now the friction force on the cart is given as

$F_f = \mu N$

$F_f = 0.625(0.5F + 176.4)$

$F_f = 110.25 + 0.3125F$

now if cart moves with constant speed then net force on cart must be zero

so now we have

$F_f + F_x = 0$

$Fcos30 - (110.25 + 0.3125F) = 0$

$0.866F - 0.3125F = 110.25$

$F = 199.2 N$

so the force must be 199.2 N

8 0

3 years ago

A force of 40 N is required to hold a spring that has been stretched from its natural length of 10 cm to a length of 15 cm. How

Sati [7]

0.36 J of work is done in stretching the spring from 15 cm to 18 cm.

To find the correct answer, we need to know about the work done to strech a string.

<h3>What is the work required to strech a string?</h3>

Mathematically, the work done to strech a string is given as 1/2 ×K×x².
K is the spring constant.

<h3>What will be the spring constant, if 40N force is required to hold a 10 cm to 15 cm streched spring?</h3>

The force experienced by a streched spring is given as Kx. x is the length of the spring streched from its natural length.
Then K = Force / x.
Here x = 15 - 10 = 5 cm = 0.05 m
K = 40/0.05 = 800N/m.

<h3>What will be the work required to strech that spring from 15 cm to 18 cm?</h3>

Work done = 1/2×k×x²
Here x= 18-15=3cm or 0.03 m
So, W= 1/2×800×0.03² = 0.36 J.

Thus, we can conclude that the work done is 0.36 J.

Learn more about the spring force here:

brainly.com/question/14970750

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6 0

2 years ago

a college student rests a backpack upon his shoulder. the pack is suspended motionless by one strap from one shoulder.

Minchanka [31]

The student's shoulder supports the weight of the bag.

<h3>What is the free body diagram?</h3>

Free-body diagrams are utilized to display the relative direction and strength of all forces that are being applied to an item in a certain scenario. A unique illustration of the geometric diagrams that were covered in a previous lesson is the free-body diagram. We will make use of these graphics throughout the entire study of physics.

A university student is carrying a backpack. One strap is hanging the rucksack immobile from one shoulder.

The weight of the backpack is balanced by the shoulder of the student.

The free-body diagram is attached below.

More about the free body diagram link is given below.

brainly.com/question/24087893

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4 0

1 year ago