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saw5 [17]
3 years ago
8

A snail can move approximately.3 meters per minute. How many meters can a snail travel in 45min?

Physics
1 answer:
Aloiza [94]3 years ago
6 0

Answer:

13.5

Explanation:

.3×45=13.5 See that is correct u can check with a calculator, I used a calculator

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Find the final velocity if the initial velocity of 8 m/s with an acceleration of 7 m/s2 over a 3 second interval?
LenKa [72]

I don't know about it your answer will give another people

5 0
2 years ago
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A mercury thermometer is used to measure the temperature of boiling water.<br>Why?​
Rama09 [41]

Answer:

It has very high density, so a small bulb of a thermometer can contain much mercury. Mercury remains liquid state over a quite wide range of temperature because it freezes at 39°C and boils at 357°C.

Explanation:

6 0
3 years ago
A single point charge is placed at the center of an imaginary cube that has 10 cm long edges. The electric flux out of one of th
12345 [234]

Answer:

Explanation:

<u>Given:</u>

Length of each side of the cube, L=10\ cm

The Elecric flux through one of the side of the cube is, \phi =-1 kNm^2/C.

The net flux through a closed surface is defined as the total charge that lie inside the closed surface divided by \epsilon_0

Since Flux is a scalar quantity. It can added to get total flux through the surface.

\phi_{total}=\dfrac{Q_{in}}{\epsilon_0}\\6\times {-1}\times 10^3=\dfrac{Q_{in}}{\epsilon_0}\\\\Q_{in}=-6}\times 10^3\epsilon_0\\Q_{in}=-6}\times 10^3\times8.85\times 10^{-12}\\Q_{in}=-53.1\times10^{-9}\ C

So the the charge at the centre is calculated.

5 0
3 years ago
A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00°C, the resistance of the carbon r
dusya [7]

Answer:

28 degree C

Explanation:

We are given that

T_1=4.00^{\circ}

R_1=217.7 \Ohm

R_2=215.1\Ohm

\alpha=-5.00\times 10^{-4}C^{-1}

We have to find the temperature on a spring day when resistance is 215.1 ohm.

We know that

\alpha(T_2-T_1)=\frac{R_2}{R_2}-1

Using the formula

-5.00\times 10^{-4}(T_2-4)=\frac{215.1}{217.7}-1

-5\times 10^{-4}(T_2-4)=0.988-1=-0.012

T_2-4=\frac{0.012}{5\times 10^{-4}}=24

T_2=24+4=28^{\circ}C

Hence, the temperature  on a spring day 28 degree C.

7 0
3 years ago
An electron is brought from rest infinitely far away to rest at point P located at a distance of 0.041 m from a fixed charge q.
torisob [31]

Answer:

The electric potential in volts is 1.618 x 10⁻¹⁷ V

Explanation:

The electric potential, in volts, at point P, can be calculated as follow;

Electric potential is the work done in moving a unit positive charge from infinity to a particular point in the electric field.

Thus, the work done in this process in moving the charge to point p is 101eV.

Convert this Volts = 101 × 1.602 x 10⁻¹⁹ V

                              = 1.618 x 10⁻¹⁷ V

Therefore, the electric potential in volts is 1.618 x 10⁻¹⁷ V

3 0
3 years ago
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