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7nadin3 [17]
3 years ago
15

A load of 500N is carried by 200N effort in a simple machine having load distance 3m Calculate effort distance.​

Physics
1 answer:
vlada-n [284]3 years ago
5 0

Answer:

<h2>2.5 m</h2>

Explanation:

Load ( L ) = 500 N

Effort ( E ) = 200 N

Load distance ( LD ) = 3 m

Effort distance ( ED ) = ?

Now, Let's find the Effort distance ( ED )

We know that,

Output work = Input work

i.e L × LD = E × ED

plug the values

500 \times 3 = 200 \times ED

multiply the numbers

1500 = 200 \times ED

Swipe the sides of the equation

200 \: ED \:  = 500

Divide both sides of the equation by 200

\frac{200 \: ED}{200}  =  \frac{500}{200}

Calculate

ED\:  = 2.5 \: m

Hope this helps..

best regards!!

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LenaWriter [7]

Answer:

Explanation:

Applying conservation of momentum

mA v0 + 0 = mA x 12v0 + mB x 92v0

11 mA = - 92mB .

mA / mB = 92 / 11

mA : mB = 92 : 11.

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3 years ago
How much does a 45 kg rock weigh
bazaltina [42]

Answer:

441.45N

Explanation:

Weight = mass(kg) x acceleration (m/s^2)

Weight = 45kg x 9.81m/s^2=441.45N

Recall that weight =mass of the rock x gravitational force acting on the rock

Therefore, the rock will weigh = 441.45N

7 0
3 years ago
A parachutist bails out and freely falls 63 m. Then the parachute opens, and thereafter she decelerates at 1.5 m/s2. She reaches
yaroslaw [1]

Answer:

(a) The parachutist spent 24.84 secs in air

(b) The height the fall begins is 472 m

Explanation:

Here is the complete question:

A parachutist bails out and freely falls 63 m. Then the parachute opens, and thereafter she decelerates at 1.5 m/s2. She reaches the ground with a speed of 3.3 m/s. (a) How long is the parachutist in the air  (b) At what height does the fall begin?

Explanation:

From one of the equations of kinematics for free fall

H = ut - \frac{1}{2}gt^{2}

Where H is the height

u is the initial velocity

t is the time

and g is the acceleration due to gravity (Take g = 9.8 m/s2)

Now, we can find the time spent before the parachute opens.

u = 0 m/s (we assume the parachutist starts from rest)

H = - 63 m

∴-63 = 0(t) - \frac{1}{2}(0.98) t^{2}  \\-63 = -4.9 t\\t^{2} = \frac{63}{4.9} \\t^{2} = 12.86\\t = \sqrt{12.86} \\t = 3.59 s

This the time spent before the parachute opens

Also from one of the equations of kinematics for free fall

v = u - gt

where v is the final velocity

We can determine the final velocity before the parachute opens and she starts to decelerate

∴v = - 9.8(3.59)\\v = - 35.18m/s

Now, we will calculate the time spent after the parachute opens

From one of the equation of kinematics for linear motion,

v = u + at

Here, the initial velocity will be the final velocity just before the parachute opens, that is

u = - 35.18 m/s

From the question,

v = - 3.3 m/s

a = 1.5 m/s^{2}

We then get

-3.3 = - 35.18 + (1.5)t

-3.3 + 35.18 =  1.5t\\31.88 = 1.5t

t = \frac{31.88}{1.5}

t = 21.25 secs

(a) To determine how long the parachutist is in the air,

That is sum of the time used when falling freely and the time used after the parachute opens

= 3.59 secs + 21.25 secs

24.84 secs

Hence, the parachutist spent 24.84 secs in air

(b) To determine what height the fall begins

First, we will calculate the height from which the parachute opens

From one of the equation of kinematics for linear motion,

x = ut + \frac{1}{2}at^{2}  \\

x = -35.18(21.25) + \frac{1}{2}(1.5)(21.25)^{2}  \\x = -747.58 + 338.67\\x = -408.91m\\

x ≅ - 409m

Hence, the height the fall begins is 63m + 409m

= 472 m

3 0
3 years ago
What term refers to the product of an object's mass and velocity?
aleksandr82 [10.1K]

Momentum (symbol p) and the equation of momentum is:

p = m*v

5 0
4 years ago
Read 2 more answers
1) Halving the distance (i.s., decreasing by a factor of two) between two charged objects will cause the electrical force betwee
Vedmedyk [2.9K]

Answer:

Explanation:

For an electric force, F the formula:

F = kQq/r^2

Given:

r2 = 1/2 × r1

F1 × r1 = k

F1 × r1 = F2 × r2

F2 = (F1 × r1^2)/(0.5 × r1)^2

= (F1 × r1^2)/0.25r1^2

= 4 × F1.

7 0
4 years ago
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