I'm not sure what equation to use.

State pascal is law.<br>please answer me

Angelina_Jolie [31]

Answer:

Pascal's law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container. The pressure at any point in the fluid is equal in all directions

5 0

3 years ago

A Hooke's law spring is compressed 12.0 cm from equilibrium, and the potential energy stored is 72.0 J. What compression (as mea

Anuta_ua [19.1K]

Answer:14.14 cm

Explanation:

Given

Spring Compression $x=12 cm$

Potential energy Stored in spring $=72 J$

Suppose k is the spring constant of spring

Potential Energy of spring is given by $=\frac{kx^2}{2}$

$\frac{k(0.12)^2}{2}=72$

$k(0.12)^2=144$

$k=10,000 N/m$

$k=10 kN/m$

for 100 J energy

$\frac{k(x_0)^2}{2}=100$

$10\times 10^3\cdot (x_0)^2=200$

$(x_0)^2=2\times 10^{-2}$

$x_0=0.1414$

$x_0=14.14 cm$

6 0

3 years ago

A rifle is aimed horizontally at a target 50.0 m away. The bullet hits the target 2.90 cm below the aim point. . . Whats the bul

Marina CMI [18]

The problem ask to calculate the bullet's flight time and the bullet's speed as it left the barrel. So base on the problem, the answer would be that the flight time is 0.076 seconds and the speed of the bullet is 657.9 m/s. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications.

5 0

3 years ago

Which is the best example of the use of imagery in a sentence?

Talja [164]

Answer:

Common Examples of Imagery

Taste: The familiar tang of his grandmother's cranberry sauce reminded him of his youth. Sound: The concert was so loud that her ears rang for days afterward. Sight: The sunset was the most gorgeous they'd ever seen; the clouds were edged with pink and gold.

I hope it's helpful!

6 0

3 years ago

A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M a

aleksklad [387]

Answer:

a) y = 0.98 t², t=1s y= 0.98 m,

b) he two blocks must move the same distance

c) v = 1.96 m / s, d) a = -1.96 m / s², e) x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

N-W = 0

N = M g

X axis

T- fr = Ma

the friction force has the expression

fr = μ N

fr = μ Mg

small block

w- T = m a

we write the system of equations

T - fr = M a

mg - T = m a

we add and resolved

mg- μ Mg = (M + m) a

a = $g \ \frac{m - \mu M}{m+M}$

a = $9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}$

a = 9.8 (6/30)

a = 1.96 m / s²

a) now we can use the kinematic relations

y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

y = ½ a t²

y = ½ 1.96 t²

y = 0.98 t²

for t = 1s y = 0.98 m

t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

v = vo + a t

v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

-fr = M a

- μ M g = M a

a = - μ g

a = - 0.2 9.8

a = -1.96 m / s²

e) the distance to stop the block is

v² = vo² - 2 a x

0 = vo² - 2a x

x = vo² / 2a

x = 1.96² / 2 1.96

x = 0.98 m

the time to travel this distance is

v = vo - a t

t = vo / a

t = 1.96 /1.96

t = 1 s

3 0

3 years ago