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3 years ago

AreaAgNO3 + ___Response area_Li →_Response areaLiNO3 +Response areaAg

Physics

1 answer:

Solnce55 [7]3 years ago

6 0

Answer:

1 is 3 -1

Explanation:

yan po ang tamang sagot thank you four gave thise answer thise if iam rong sorry i dont gate thinking

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Un automovil parte del reposo y acelera uniformemente hasta alcanzar una rapidez de 0,255km/h en un tiempo de 3/4 Minutos determ

Elden [556K]

Answer:

a = 1.5*10^-3 m/s^2

x = 0.033m = 3.3cm

Explanation:

To calculate the acceleration and the distance traveled by the car you use the following formulas:

$v=v_o+at$ (1)

$x=v_ot+\frac{1}{2}at^2$ (2)

v: final velocity = 0,255 km/h

vo: initial velocity = 0 m/s

t: time = 3/4 min

a: acceleration = ?

x: distance

In order to use the equations (1) and (2) you first convert the units of the final velocity to m/s, and the time to seconds.

$v=0,255\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}\\\\v=0.07m/s\\\\t=\frac{3}{4}min*\frac{60s}{1min}=45s$

Next, you solve the equation (1) for the acceleration a:

$a=\frac{v}{t}=\frac{0.07m/s}{45s}=1.5*10^{-3}\frac{m}{s^2}$

With this value of a you can calculate the distance traveled by the car, by using the equation (2):

$x=\frac{1}{2}(1.5*10^{-3}m/s^2)(45s)^2=0.033m=3.3cm$

hence, the acceleration of the car is 1.5*10^-3 m/s^2 and the distance traveled in 3/4 min is 0.033m

5 0

4 years ago

The fundamental resolution of an optical instrument is set by __________.

inysia [295]

Answer:C . The wave nature of light.

Explanation: The wave nature of light has been described in the refraction and diffraction of light when it comes in contact with a prism.

Wave nature of light has been described through this equation C= velocity( measured in meters per second square), Wavelength (lambda) which is the distance between successive antinodes or nodes,it is measured in meters. and Frequency is the number of times light waves pass through a point in one second, usually measured in Hertz(Hz)

Below is a

Thin-lens equation (also works for mirrors)

1/f=1/u+1/v

Where f is the focal length, u is distance of object,v is the distance between the optical centre and the image of the object.

6 0

3 years ago

A 24 toy falls from 2 to 1 m. How much does the toys CPE change

pshichka [43]

GPE change:

$\tt =24\times 9.8(2-1)=235.2~J$

4 0

3 years ago

Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 27.0°below the horizontal. If it st

andreev551 [17]

a) The time of flight is 3.78 s

b) The initial speed is 17.0 m/s

c) The speed at impact is 46.4 m/s at $70.3^{\circ}$ below the horizontal

Explanation:

The picture of the previous problem (and some data) is missing: find it in attachment.

The motion of the ball is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

We start by considering the vertical motion, to find the time of flight of the ball. We do it by using the following suvat equation: for the y-displacement:

$y=u_y t+\frac{1}{2}at^2$

where we have:

y = -45.0 m is the vertical displacement of the ball (the height of the building)

$u_y=u sin \theta$ is the initial vertical velocity, with u being the initial velocity (unknown) and $\theta=-27.0^{\circ}$ the angle of projection

t is the time of the fall

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Along the x-direction, the equation of motion is instead

$x=(u cos \theta)t$

where $ucos \theta$ is the horizontal component of the velocity. Rewriting this equation as

$t=\frac{x}{ucos \theta}$

And substituting into the previous equation, we get

$y=xtan \theta + \frac{1}{2}gt^2$

And using the fact that the horizontal range is

x = 59.0 m

And solving for t, we find the time of flight:

$t=\sqrt{\frac{y-x tan \theta}{g}}=\sqrt{\frac{-45-(59.0)(tan(-27^{\circ}))}{-9.8}}=3.78 s$

We can now find the initial speed, u, by using the equation of motion along the x-direction

$x=u cos \theta t$

where we know that:

x = 59.0 m is the horizontal range

$\theta=-23^{\circ}$ is the angle of projection

$t=3.78 s$ is the time of flight

Solving for u, we find the initial speed:

$u=\frac{x}{cos \theta t}=\frac{59.0}{(cos (-23^{\circ}))(3.78)}=17.0 m/s$

First of all, we notice that the horizontal component of the velocity remains constant during the motion, and it is equal to

$v_x = u cos \theta = (17.0)(cos (-23^{\circ})=15.6 m/s$

The vertical velocity instead changes according to the equation

$v_y = u sin \theta + gt$

Substituting all the values and t = 3.78 s, the time of flight, we find the vertical velocity at the time of impact:

$v_y = (17.0)(sin (-23^{\circ}))+(-9.8)(3.78)=-43.7 m/s$

Where the negative sign means it is downward.

Therefore, the speed at impact is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(15.6)^2+(-43.7)^2}=46.4 m/s$

while the direction is given by

$\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-43.7}{15.6})=-70.3^{\circ}$

So, $70.3^{\circ}$ below the horizontal.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0

4 years ago

Which of the following is a true statement about mass and weight? a. Mass is a measure of how much matter an object has, while w

natka813 [3]

Answer:

b option mass will not change based on location while weight will changr based on gravitational pull

bcz in the formula of weight w=mg

so w is directly proportional to g

8 0

4 years ago

AreaAgNO3 + ___Response area_Li →_____Response areaLiNO3 +____Response areaAg

AreaAgNO3 + ___Response area_Li →_Response areaLiNO3 +Response areaAg