(10p+15,15-10q)=(25,5)

Which will ensure laboratory safety during the experiment? Check all that apply. using beaker tongs to handle the hot beaker rea

Aleks [24]

Answer: • using beaker tongs to handle the hot beaker.

• checking the beaker for chips prior to heating on the hot plate.

• Turning off the hot plate after use

Explanation:

The options that will ensure laboratory safety during the experiment will be:

• using beaker tongs to handle the hot beaker.

• checking the beaker for chips prior to heating on the hot plate.

• Turning off the hot plate after use.

We should note that the beaker tongs are simply used in the holding of the beakers that have hot liquids in them. Also, it s vital for the hot plate to be turned off after its use so as to prevent accident.

7 0

3 years ago

Read 2 more answers

A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

SIZIF [17.4K]

The solution to the questions are given as

$t=40.39 \mathrm{sec}$
$\varepsilon &=(0.12v)e^{0.057t}$
the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}$

$\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$$

$\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}$

$\varepsilon &=(0.12v)e^{0.057t}$

(b) $Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$$

$\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}$

c)

In conclusion, the direction of the induced current will be Counterclockwise.

Read more about current

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4 0

1 year ago

Consider and mass of a material

Leni [432]

Answer:Density is the mass of an object divided by its volume. Density often has units of grams per cubic centimeter (g/cm3). ... You probably have an intuitive feeling for density in the materials you use often. For example, sponges are low in density; they have a low mass per unit volume.

Explanation:

3 0

2 years ago

How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to an altitude of 5r miles above t

Cloud [144]

Let the data is as following

mass of payload = "m"

mass of Moon = "M"

now we know that we place the payload from the position on the surface of moon to the position of 5r from the surface

So in this case we can say that change in the gravitational potential energy is equal to the work done to move the mass from one position to other

so it is given by

$W = U_f - U_i$

we know that

$U_f = -\frac{GMm}{6r}$

$U_i = -\frac{GMm}{r}$

now from above formula

$W = -\frac{GMm}{6r} + \frac{GMm}{r}$

$W = \frac{5GMm}{6r}$

so above is the work done to move the mass from surface to given altitude

7 0

2 years ago

In a third class lever, the distance from the effort to the fulcrum is ____________ the distance from the load/resistance to the

padilas [110]

The complete sentence is:
In a third class lever, the distance from the effort to the fulcrum is SMALLER the distance from the load/resistance to the fulcrum.
In fact, in a third class lever, the fulcrum is on one side of the effort and the load/resistance is on the other side, so the effort is located somewhere between the two of them. This means that the distance effort-fulcrum is smaller than the distance load-fulcrum.

4 0

2 years ago

(10p+15,15-10q)=(25,5)​

(10p+15,15-10q)=(25,5)