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3 years ago

What forms when acid reacts with a base? please help

Physics

1 answer:

BARSIC [14]3 years ago

7 0

Answer:

salt. (I think)

Explanation:

they neutralize each other and often this results in salt being formed and leftover

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A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal

AysviL [449]

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

$v=\sqrt{v_x^2+v_y^2}$ (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

$v_y^2=v_{oy}^2+2gh$ (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

$v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}$

vx is calculated by using the information about the horizontal range of the ball:

$R=v_o\sqrt{\frac{2h}{g}}$ (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

$v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}$

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

$v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}$

The velocity of the ball just before it touches the ground is 46.99 m/s

5 0

4 years ago

Wilbur ran 1-kilometer. Then he ran 500 meters. How many meters did Wilbur run all together?

Blababa [14]

The answer is B 1,500 meters since 1 kilometer to meters is 1,000 and u add the 500

8 0

3 years ago

A car of mass 1500 kg is negotiating a flat circular curve of radius 50 m with a speed of 20 m/s.

Lilit [14]

Answer:

Explanation:

a. The source of centripetal force on the car is (3) the static friction force.

b. ac = v²/R = (20²)/50 = 8 m/s²

c. Fc = m(ac) = 1500(8) = 12 kN

d. μ = Fc/N = Fc/mg = 12000 / 1500(9.8) = 0.8163... ≈ 0.82

6 0

3 years ago

Two concrete spans of a 380 m long bridge are

Mazyrski [523]

Answer:

4.163 m

Explanation:

Since the length of the bridge is

L = 380 m

And the bridge consists of 2 spans, the initial length of each span is

$L_i = \frac{L}{2}=\frac{380}{2}=190 m$

Due to the increase in temperature, the length of each span increases according to:

$L_f = L_i(1+ \alpha \Delta T)$

where

$L_i = 190 m$ is the initial length of one span

$\alpha =1.2\cdot 10^{-5} ^{\circ}C^{-1}$ is the temperature coefficient of thermal expansion

$\Delta T=20^{\circ}C$ is the increase in temperature

Substituting,

$L_f=(190)(1+(1.2\cdot 10^{-5})(20))=190.0456 m$

By using Pythagorean's theorem, we can find by how much the height of each span rises due to this thermal expansion (in fact, the new length corresponds to the hypothenuse of a right triangle, in which the base is the original length of the spand, and the rise in heigth is the other side); so we find:

$h=\sqrt{L_f^2-L_i^2}=\sqrt{(190.0456)^2-(190)^2}=4.163 m$