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4 years ago

Which of the following is not a common property of basesp

Physics

1 answer:

Hatshy [7]4 years ago

6 0

Explanation:

Bases taste bitter, feel slippery, and conduct electricity when dissolved in water. Indicator compounds such as litmus can be used to detect bases. Bases turn red litmus paper blue. The strength of bases is measured on the pH scale.

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A fault in which the hanging wall moves down relative to the footwall is a _____.

LuckyWell [14K]

<span>Normal fault. Normal (extensional ) fault is a displacement of a rock as a result of rock-mass movement and occurs when the crust is stretching. Because of the stretching the thickness of the crust is reduced and the crust or horizontally extended. </span>

4 0

3 years ago

Read 2 more answers

The diver is on a board3.00m above the water. She jumps straight up at 3.25m/s. How many seconds later does she hit the water.

tigry1 [53]

Speed =distance/time
3.25=3.00/time
3.25xt=3.00
t=3/3.25
s=0.9s

5 0

4 years ago

a steel ball bearing is released from a height hhh and rebounds after hitting a steel plate to a height hhh. what is true about

lbvjy [14]

In collision of the steel ball and the steel plate, the collision is an inelastic collision and there is loss in the kinetic energy.

<h3>What are collisions?</h3>

Collisions occur when two objects that are moving in the same directions or in different direction meet each other and collide.

There are two types of collisions:

elastic collision - the kinetic energy is conserved
inelastic collision - there is a loss in kinetic energy

In the collision of the steel ball and the steel plate, there is loss in the kinetic energy of the steel ball which is converted to sound energy.

In conclusion, the collision of the steel and steel plate is an inelastic collision.

Learn more about collisions at: brainly.com/question/7694106

#SPJ1

4 0

2 years ago

A car traveling at 60km/h undergoes uniform acceleration at a rate of 2/ms^2 until is velocity reached 120km/h determine the dis

jenyasd209 [6]

Explanation:

Given that,

Initial speed of a car, u = 60 km/h = 16.67 m/s

Acceleration, a = 2m/s²

Final speed, v = 120 km/h = 33.33 m/s

We need to find the distance traveled and the time taken to make the distance.

acceleration = rate of change of velocity

$a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{33.33 -16.67 }{2}\\\\t=8.33\ s$

let the distance be d.

$d=\dfrac{v^2-u^2}{2a}\\\\d=\frac{33.33^{2}-16.67^{2}}{2(2)}\\\\d=208.25\ m$

Hence, the distance traveled and the time taken to make the distance is 208.25 m and 8.33 seconds respectively.

3 0

3 years ago

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tatuchka [14]

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
$W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N$
the force of the wind F, acting horizontally, with intensity
$F=12 N$
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
$T \cos \alpha -F=0$
$T \sin \alpha -W=$
By dividing the second equation by the first one, we get
$\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04$
From which we find
$\alpha = 63.8 ^{\circ}$
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
$T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N$

7 0

4 years ago