All categories

3 years ago

A hard material is easily scratched. a. True b. False

1 answer:

8 0

<span>The statement, a hard material is easily scratched, is false, the answer is letter F. Harder materials have strong bonding between their neighboring atoms and it takes a greater amount of energy for them to break apart, even a scratch. The relative hardness of a mineral can be tested by the Moh’s scale or the mineral hardness. The Moh’s scale can be characterized by the ability of the mineral to resist scratch resistance by scratching a harder or softer mineral. It has a scale of 1 to 10 where 1 being the softest and 10 being the hardest.</span>

You might be interested in

If a net horizontal force of 7.5 N is applied to a baby carriage whose mass is 12.5 kg what acceleration is

11111nata11111 [884]

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{7.5}{12.5} \\$

We have the final answer as

Hope this helps you

8 0

3 years ago

A 4.4-µF capacitor is initially connected to a 5.1-V battery. Once the capacitor is fully charged the battery is removed and a 2

Grace [21]

Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S

2) t = 1 mu S

3) t = 50 mu S

1) $22.4 \mu C$

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

$C=\frac{Q_0}{V_0}$

where

C is the capacitance

Q0 is the initial charge stored

V0 is the initial potential difference across the capacitor

When the capacitor is connected to the battery, we have:

$C=4.4\mu F = 4.4\cdot 10^{-6}F$

$V_0 = 5.1 V$

Solving for $Q_0$ ,

$Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C$

So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.

2) $20.0\mu C$

To find the charge on the capacitor at any other time t, we use the equation:

$Q(t) = Q_0 e^{-\frac{t}{RC}}$

where

$Q_0 = 22.4 \mu C$

t is the time

$R=2.0 \Omega$ is the resistance

$C=4.4\mu F$ is the capacitance

Therefore, at time $t=1 \mu s$ , we have:

$Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C$

3) $0.08 \mu C$

As before, we use again the equation:

$Q(t) = Q_0 e^{-\frac{t}{RC}}$

However, here the time to consider is

$t=50 \mu C$

Substituting into the formula,

$Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C$

4 0

3 years ago

In a race, a runner traveled 12 meters in

Nastasia [14]

Since it is a race, the runner start from rest, so we can use the Velocity Hourly Equation to describle yours moviment.

$S=S_{o}+V_{o}t+ \frac{at^2}{2} \\ S=\frac{at^2}{2}$

Entering the unknowns, we have:

$S=\frac{at^2}{2} \\ 12*2=a*4^2 \\ \boxed {a=1.5m/s^2}$

Number 2

If you notice any mistake in my english, please let me know, because i am not native.

3 0

3 years ago

A force of 100 N acts on a body and moves at a distance of 2 m in the direction of the force. How much work has been done?

Flauer [41]

Answer:

200 joules

Explanation:

work=force×distance

5 0

3 years ago

How much work was done on a charge of 3.0 to move it through 12v electric potential difference

erik [133]

Answer:

Hence, the work done in moving the charge is 45 J.

<h2>( Too lazy to translate it to Spanish T_T )</h2>

8 0

2 years ago

Read 2 more answers