All categories

2 years ago

2 kg Use 10 m/s2 for g

Physics

1 answer:

Iteru [2.4K]2 years ago

6 0

Answer:

See below

Explanation:

At point A the PE = mgh = 2 * 10 * 1 = 20 J

at point B, all of the PE , 20 J , is converted to Kinetic Energy

KE = 1/2 m v^2

20 = 1/2 (2)(v^2 )

20 = v^2 v = sqrt 20 = 4.47 m/s

for the friction part

vf = vo t + 1/2 a t^2 vf = final velocity = 0 (stopped)

vo = original velocity = 4.47 m/s

a = -1 m/s^2

0 = 4.47 t + 1/2 (-1) t^2

- .5t^2 + 4.47 t = 0

t ( -.5t+ 4.47) = 0 shows t = 4.47/.5 = 8.9 seconds

You might be interested in

a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. A 19,600 N car is parked 8 meters from one end, whe

Elden [556K]

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

$Mg(r - L/2) = mg(L/2 - 8)$

here we know that

mg = 19600 N

Mg = 400,000 N

L = 20 m

from above equation we have

$400,000(r - 10) = 19,600 (10 - 8)$

$r - 10 = 0.098$

$r = 10.098 m$

so pillar is at distance 10.098 m from one end of the slab

3 0

3 years ago

The ________________ is the measure of how far the pendulum is offset (or pulled back) from a vertical position when it is relea

Dahasolnce [82]

B. Amplitude

It is the maximum distance from the equilibrium point of the pendulum.

6 0

4 years ago

Mr Smith is working in a muddy garden. When he picks up a paving stone his feet sink deeper into the mud. Explain why his feet s

Anni [7]

because he is carrying more mass and as the ground is muddy his feet goes in due to the pull of gravity

3 0

3 years ago

An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.

Pachacha [2.7K]

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

$u -v= d$

$u=71+26=97\ cm$

We need to calculate the focal length

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}$

$\dfrac{1}{f}=-\dfrac{71}{2522}$

$f=-35.52\ cm$

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value in to the formula

$\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}$

$\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}$

$\dfrac{1}{v}=-\dfrac{124}{2755}$

$v=-22.21\ cm$

We need to calculate the magnification

Using formula of magnification

$m=\dfrac{v}{u}$

$m=\dfrac{22.21}{95}$

$m=0.233$

The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.

4 0

3 years ago

GIVING BRAINLIEST PLEASE HELP!!

OLga [1]

A. because I had this question yesterday.

3 0

3 years ago