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2 years ago

6. Contaminated rags are recommended for disposal in landfills. A) True B) False

Physics

2 answers:

Inga [223]2 years ago

6 0

<h2>Answer:</h2>

<u>The statement is </u><u>False</u>

<h2>Explanation:</h2>

Contaminated rags may contain plastic and other materials that cannot be defused in soil and they remain there for a long period of time. Another reason to avoid disposal of contaminated material in earth is that it may also contain toxic materials which may result into land pollution. So it is not recommended to dispose contaminated rags in landfills.

RideAnS [48]2 years ago

6 0

False because it would depend on what the contaminant is

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The magnitude of the magnetic force on a current-carrying wire held in a magnetic

Kamila [148]

Answer:

All of the above

Explanation:

The magnitude of the magnetic force on a current-carrying wire held in a magnetic is given by the equation $F = BIlsin \theta$

Where B = Strength of the magnetic field

I = The current carried by the wire

l = length of the wire in the magnetic field

θ = Angle between the wire and the magnetic field

Based on the relationship written above, the magnitude of the magnetic force on the current - carrying wire in the magnetic field depends on the strength of the magnetic field (B), length of the wire(l), current in the wire (I).

All the options are correct.

3 0

3 years ago

There are four springs stretched by the same mass.

brilliants [131]

Spring C stretches 100 cm.

Explanation:

The spring constant is simply the stiffness of the spring. The higher the spring constant the more stiff the spring is.

Spring constant shows the force needed to stretch a spring from it's equilibrium position. If a material requires more force to cause it to stretch, it will have a high spring constant.

According to hooke's law "the force needed to extended an elastic material is directly proportional to its extension"

F = ke

k is the spring constant

e is the extension

We see that the spring that stretches by 100 is the less stiff compared to other springs. It has the smallest spring constant.

Learn more;

Force brainly.com/question/8882476

#learnwithBrainly

8 0

2 years ago

A small block with a mass of 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. 6

Rina8888 [55]

Answer:

a) 0.147 N

b) 9.408 N

c) 9.261 N

Explanation:

The tension on the cord is the only force keeping the block in circular motion, thus representing the entirety of its centripetal force $\frac{mv^{2} }{r}$ . Plugging in values for initial and final states and we get answers for a and b. The work done by the person causes the centripetal force to increase, and thus is the difference between the final tension and the initial tension.

7 0

2 years ago

A team of ecologists are studying four different ecosystems with varying levels of biodiversity. The ecologists categorize the d

Ann [662]

Answer:

B. Ecosystem B, because its high species diversity could have resulted from increased competition among its members.

Explanation:

This is because, in the ecosystem with varying level of biodiversity, Ecosystem B has medium level of species diversity found in them with High medium level of habitat diversity which causes increasing competitions among them.

4 0

3 years ago

I need homework help

KIM [24]

1) the weight of an object at Earth's surface is given by $F=mg$ , where m is the mass of the object and $g=9.81 m/s^2$ is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is
$F=mg=(2.2 kg)(9.81 m/s^2)=21.6 N$

2) On Mars, the value of the gravitational acceleration is different: $g=3.7 m/s^2$ . The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: $F=mg=(2.2 kg)(3.7 m/s^2)=8.1 N$

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:
$g= \frac{F}{m}= \frac{ 19.6 N}{2.2 kg}=8.9 m/s^2$

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:
$g= \frac{F}{m} = \frac{11.55 N}{0.5 kg} =23.1 m/s^2$

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as
$g= \frac{F}{m} = \frac{0.3 N}{0.5 kg} =0.6 m/s^2$

<span>6) On Earth, the gravity acceleration is </span> $g=9.81 m/s^2$ <span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is
</span> $F=mg=(0.5 kg)(9.81 m/s^2)=4.9 N$ <span>
</span>

5 0

2 years ago