What effort force will be required to lift the 20 N object using the pulley above?

A shopper walks westward 2.8 meters in the first 5 minutes and then eastward 9.2 meters in the next 10 minutes. What is the shop

Alecsey [184]

Answer:

12m

Explanation:

Given parameters:

Distance walked westward = 2.8m

Time of travel = 5min

Distance walked eastward = 9.2m

Time of travel = 10min

Unknown:

The total shopper's travel distance = ?

Solution:

Total distance traveled is the sum of the length of path covered by a body. It is a scalar quantity.

Total distance = distance walked westward + distance walked eastward

Total distance = 2.8m + 9.2m = 12m

6 0

2 years ago

What is your opinion on Moonman? ( Do not delete this saying it is "racist". it is simply a character from a commercial called M

Leni [432]

Answer:

I think hes cool

Explanation:

8 0

2 years ago

A ball is dropped from rest at point O. After falling for some time, it passes by a window of height 3.3 m and it does so in 0.2

stiv31 [10]

Answer:

Speed at which the ball passes the window’s top = 10.89 m/s

Explanation:

Height of window = 3.3 m

Time took to cover window = 0.27 s

Initial velocity, u = 0m/s

We have equation of motion s = ut + 0.5at²

For the top of window (position A)

$s_A=0\times t_A+0.5\times 9.81t_A^2\\\\s_A=4.905t_A^2$

For the bottom of window (position B)

$s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2$

$\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673$

We also have

$t_B-t_A=0.27$

Solving

$t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s$

So after 1.11 seconds ball reaches at top of window,

We have equation of motion v = u + at

$v_A=0+9.81\times 1.11=10.89m/s$

Speed at which the ball passes the window’s top = 10.89 m/s

7 0

2 years ago

A fire engine approaches a wall at 5 m/s while the siren emits a tone of 500 Hz frequency. At the time, the speed of sound in ai

Svetach [21]

Answer:

The values is $f_b =14.9 \ beats/s$

Explanation:

From the question we are told that

The speed of the fire engine is $v = 5\ m/s$

The frequency of the tone is $f = 500 \ Hz$

The speed of sound in air is $v_s = 340 \ m/s$

The beat frequency is mathematically represented as

$f_b = f_a - f$

Where $f_a$ is the frequency of sound heard by the people in the fire engine and is is mathematically evaluated as

$f_a = [\frac{v_s + v }{v_s -v} ]* f$

substituting values

$f_a = [\frac{340 + 5 }{340 -5} ]* 500$

$f_a = 514.9 \ Hz$

Thus

$f_b =514.9 - 500$

$f_b =14.9 \ beats/s$

8 0

2 years ago

Block B has mass 6.00 kg and sits at rest on a horizontal, frictionless surface. Block A has mass 2.50 kg and sits at rest on to

zzz [600]

Answer:

Explanation:

Block A sits on block B and force is applied on block A . Block A will experience two forces 1) force P and 2 ) friction force in opposite direction of motion . Block B will experience one force that is force of friction in the direction of motion .

Let force on block A be P . friction force on it will be equal to kinetic friction, that is μ mg , where μ is coefficient of friction and m is mass of block A

friction force = .4 x 2.5 x 9.8

= 9.8 N

net force on block A = P - 9.8

acceleration = ( P - 9.8 ) / 2.5

force on block B = 9.8

acceleration = force / mass

= 9.8 / 6

for common acceleration

( P - 9.8 ) / 2.5 = 9.8 / 6

( P - 9.8 ) / 2.5 = 1.63333

P = 13.88 N .

4 0

3 years ago