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svlad2 [7]
3 years ago
10

The device used to measure a masses of a body is kilogram . true or false​

Physics
1 answer:
olga nikolaevna [1]3 years ago
6 0

Answer: false

Explanation:

While kilograms are the unit used to measure body mass, the device used is a scale.

Hope it helps :)

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At a certain place, Earth's magnetic field has magnitude B =0.703 gauss and is inclined downward at an angle of 75.4° to the hor
Irina18 [472]

Answer:

The charge flows in coulombs is

dq=1.843x10^{-5}C

Explanation:

The current magnitude of current is given by the resistance and the induced Emf as:

I=N*\frac{dF}{Rdt}

\frac{dq}{dt}=\frac{dF}{Rdt}=dq=N*\frac{dF}{R}

dq=\frac{N*\beta*A*(Cos(\alpha_f)-Cos(\alpha_i}{R}

N=1300, \beta=0.703, A=\pi*r^2=\pi*0.10^2=0.01\pi m^2, R=99.4+202=301.4Ω

\alpha_f=14.6,\alpha_i=165.4

Replacing :

dq=\frac{1300*0.703x10^{-4}*0.01\pi*(0.9667-(-0.9667))}{202+99.4}

dq=1.843x10^{-5}C

5 0
3 years ago
Compared to its weight on Earth, a 5kg object on the moon will weigh
shutvik [7]

Answer:

8.1 N/49 N=0.1653  which means 16.53% of the weight of the object on Earth.

Explanation:

On the Moon, where the gravitational constant is 1.62 \frac{m}{s^2}, the weight of the 5 kg object will be: weight_M=m*g_M = 5 kg * 1.62 \frac{m}{s^2} =8.1 N

Where the answer is in Newtons (N) since all quantities are given in the SI system.

On Earth, on the other hand, the weight of the object is:

weight_E=m*g_E= 5 kg* 9.8 \frac{m}{s^2} = 49N

Therefore the object's weight on the Moon compared to that on Earth will be:

8.1N/49N=0.1653

That is, 16.53% of the weight the object has on Earth.

5 0
4 years ago
Think of a time where you have observed someone's behavior and been encouraged to do the same. What did you do? Was it a good or
ololo11 [35]

Answer:

i'm not sure if you are asking as a  personal question or a book question so i'm taking it personal.

Explanation:

I was doing a simple task that was handed to me to test my responsibility and I agreed (knowing i am responsible :3). my first thought was "man , this is easy!" but then i started seeing the other kids slaking off and quiting their tasks. I thought that was against the rules, but then i saw my bff doing it too and i thought "this should be ok then!" so i did the same. other kids where still doing it. the teacher came, saw the ones still working and smiled... but when the teacher looked at the ones slaking off omg... his face was like * im gonna kill yall* we took one big gulp and whined. the teacher awarded the ones who completed the task... the others , we had to do our original task but doubled... for 3 weeks!!! it was awful!!!

I WOULD NEVER DO THAT AGAIN!!!

5 0
3 years ago
please help! find magnitude and direction (the counterclockwise angle with the +x axis) of a vector that is equal to a + c
-BARSIC- [3]

Answer:

Option (2)

Explanation:

From the figure attached,

Horizontal component, A_x=A\text{Sin}37

A_x=12[\text{Sin}(37)]

     = 7.22 m

Vertical component, A_y=A[\text{Cos}(37)]

    = 9.58 m

Similarly, Horizontal component of vector C,

C_x  = C[Cos(60)]

     = 6[Cos(60)]

     = \frac{6}{2}

     = 3 m

C_y=6[\text{Sin}(60)]

    = 5.20 m

Resultant Horizontal component of the vectors A + C,

R_x=7.22-3=4.22 m

R_y=9.58-5.20 = 4.38 m

Now magnitude of the resultant will be,

From ΔOBC,

R=\sqrt{(R_x)^{2}+(R_y)^2}

   = \sqrt{(4.22)^2+(4.38)^2}

   = \sqrt{17.81+19.18}

   = 6.1 m

Direction of the resultant will be towards vector A.

tan(∠COB) = \frac{\text{CB}}{\text{OB}}

                  = \frac{R_y}{R_x}

                  = \frac{4.38}{4.22}

m∠COB = \text{tan}^{-1}(1.04)

             = 46°

Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.

Option (2) will be the answer.

6 0
3 years ago
HURRY Which change is an example of transforming potential energy to kinetic energy
Jet001 [13]

Answer:

C.  changing nuclear energy to radiant energy

Explanation:

Nuclear energy takes atoms in their potential state, split them (fission) or fuse them (fusion)  creating chain reactions of radiant energy.  Most nuclear electrical power plants use fission, radiant energy heats water making steam to spin turbines.

Or think of the atom bomb.  Definitely potential energy until the fuse starts detonation and chain reactions.  The radiant kinetic energy and shock waves were horrendous.

3 0
3 years ago
Read 2 more answers
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