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mylen [45]
3 years ago
9

Don't take your heart rate with your thumb because the thumb has it's own pulse.

Physics
2 answers:
pogonyaev3 years ago
6 0

Explanation:

yeah that's true.........

swat323 years ago
5 0

Answer:

True

Explanation:

You might be interested in
Can an electron be diffracted? Can it exhibit interference?
Serggg [28]

Answer:

Yeah, it can be diffracted. Though it depends on a diffracting medium.

It must have some magnetic fields .

Forexample; X-ray diffraction where electrons are diffracted to the target filament.

5 0
3 years ago
A box of weight w=2.0N accelerates down a rough plane that is inclined at an angle ϕ=30∘ above the horizontal, as shown (Figure
Nesterboy [21]

Answer:

<h2>The work done is 0.882 Joules.</h2>

Explanation:

To calculate the work, we need to find all forces that are involved in the movement.

As you can analyse, the body is the force that make the box to move, and the friction force is opposite to it, we cannot forget about the friction. So, we have to calculate the resultant force to this context.

F_{R}=W_{x}-F_{k}

So, to find W_{x} we use: W_{x}=W.sen\phi; where W=2N;\phi = 30\°

W_{x}=2N.sin30\°=1\frac{1}{2}N=1N

The friction force would be:

F_{k}=\mu_{k}N=(0.30)(1.7N)=0.51

Then, the resultant force is:

F_{R}=W_{x}-F_{k}

F_{R}=1N-0.51N=0.49N

Now, we calculate the work: W=F_{R} d

W=(0.49N)(1.8m)=0.882J

Therefore, the work done is 0.882 Joules.

7 0
4 years ago
Read 2 more answers
The masses of the earth and moon are 5.98 x 1024 and 7.35 x 1022 kg, respectively. Identical amounts of charge are placed on eac
dybincka [34]

Answer:

The magnitude of charge on each is 5.707\times 10^{13} C

Solution:

As per the question:

Mass of Earth, M_{E} = 5.98\times 10^{24} kg

Mass of Moon, M_{M} = 7.35\times 10^{22} kg

Now,

The gravitational force of attraction between the earth and the moon, if 'd' be the separation distance between them is:

F_{G} = \frac{GM_{E}M_{M}}{d^{2}}        (1)

Now,

If an identical charge 'Q' be placed on each, then the Electro static repulsive force is given by:

F_{E} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}           (2)

Now, when the net gravitational force is zero, the both the gravitational force and electro static force mut be equal:

Equating eqn (1) and (2):

\frac{GM_{E}M_{M}}{d^{2}} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}

(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22}) = (9\times 10^{9}){Q^{2}}

\sqrt{\farc{(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22})}{9\times 10^{9}}} = Q

Q = \pm 5.707\times 10^{13} C

7 0
3 years ago
if an object is being pulled by two forces, one 4n to the left, and the other 2 n to the right, what is the net force acting on
Naddika [18.5K]

Answer:

2 N to the left

Explanation:

6 0
3 years ago
A point charge Q at the center of a sphere of radius R produces an electric flux of Φ coming out of the sphere. If the charge r
Dafna1 [17]

Remains the same

Explanation:

According to Gauss's law, the electric flux through a closed surface is proportional to the charge enclosed by the surface. So no matter how big or small we make the surface that encloses the charge, the electric flux remains the same because it only depends on the enclosed charge, not surface area.

3 0
3 years ago
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