All categories

3 years ago

The following reaction is an example of a base reacting with water: NH3 + H2O → H3O + NH4+

Physics

1 answer:

photoshop1234 [79]3 years ago

5 0

H3 in reaction: NH3 + H2O→NH4 + OH-
<span>The NH3 has gained an H - it is a base
true</span>

You might be interested in

Is cold fusion science or pseudoscience

kondor19780726 [428]

The experiments that claimed to demonstrate cold fusion were found
to have been faulty by others who reviewed them. Also, nobody else
was able to reproduce the finding in other laboratories. In the world
of Science, this pretty much says that the initial claims were unfounded.

5 0

3 years ago

If the resistance of a circuit increases, that means that the current has to

Gnoma [55]

Answer:

If resistance increases current decreases.

Explanation:

Current is <em>inversely proportional</em> to the resistance.
from the relation given below, we can clearly see the relation between current and resistance;

V=IR

I ∝ 1/R

This relation shows that when resistance increases,current decreases.

4 0

3 years ago

Help me on question 5

Sveta_85 [38]

I believe it is acceleration!

6 0

3 years ago

Read 2 more answers

During a braking test, a car is brought to rest beginning from an initial speed of 60 mi/hr in a distance of 120 ft. With the sa

maw [93]

Answer:

Explanation:

Given

Initial speed $u=60\ mi/hr\approx 88\ ft/s$

distance traveled before coming to rest $d_1=120\ ft$

using equation of motion

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$0-(88)^2=2\times a\times 120---1$

for $u_2=80\ mi/hr\approx 117.33\ ft/s$

using same relation we get

$0-(117.33)^2=2\times a\times (d_2)----2$

divide 1 and 2 we get

$(\frac{88}{117.33})^2=\frac{120}{d_2}$

$d_2=213.32\ ft$

So a distance if 213.32 ft is required to stop the vehicle with 80 mph speed

8 0

3 years ago

Determine the velocity that a car should have while traveling around a frictionless curve of radius 100m and that is banked 20 d

alex41 [277]

Answer:

$v=18.89\frac{m}{s}$

Explanation:

From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

$\sum F_x:Nsin(20^\circ)=ma_c(1)\\\sum F_y:Ncos(20^\circ)=mg(2)$

Solving N from (2) and replacing in (1):

$N=\frac{mg}{cos(20^\circ)}\\(\frac{mg}{cos(20^\circ)})sin(20^\circ)=ma_c\\gtan(20^\circ)=a_c$

The centripetal acceleration is given by:

$a_c=\frac{v^2}{r}$

Replacing and solving for v:

$gtan(20^\circ)=\frac{v^2}{r}\\v=\sqrt{grtan(20^\circ)}\\v=\sqrt{9.8\frac{m}{s^2}(100m)tan(20^\circ)}\\v=18.89\frac{m}{s}$

4 0

3 years ago