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Whitepunk [10]
2 years ago
9

A student says that a speed of 50 m/s is faster than a speed of 140 km/h because the number is bigger. What would you say to the

m?
Pls can someone help me pls!
It’s so important!!!!
Physics
1 answer:
ladessa [460]2 years ago
4 0

The units are not consistent - 1 m/s is not the same as 1 km/h.

First thing to do would be to convert from one unit of speed to the other, say km/h to m/s. There are 1000 meters (m) for every kilometer (km) and 3600 seconds (s) for every hour (h), so

1\,\dfrac{\mathrm{km}}{\mathrm h}\cdot\dfrac{10^3\,\mathrm m}{\mathrm{km}}\cdot\dfrac{\mathrm h}{3600\,\mathrm s}\approx0.278\,\dfrac{\mathrm m}{\mathrm s}

So in fact 1 km/h is about 4 times slower than 1 m/s.

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A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial
Nataly_w [17]

Answer:

0.092 m

Explanation:

A charged moving particle immersed in a region with magnetic field follows a circular trajectory at constant speed (uniform circular motion), since the magnetic forces acts perpendicular to the direction of motion of the particle.

Since the magnetic force acts as centripetal force, we can write:

qvB=m\frac{v^2}{r}

where

q is the charge of the particle

v is its velocity

B is the strength of the magnetic field

m is the mass of the particle

r is the radius of the orbit

Solving the equation for r,

r=\frac{mv}{qB}

For the ion of oxygen-16, we have:

m_A=2.66\cdot 10^{-26}kg

q_A = 1.6\cdot 10^{-19}C (it is singly charged)

v_A=2.90\cdot 10^6 m/s

B_A=1.30 T

So the radius of its orbit is

r_A=\frac{m_A v_A}{q_A B_A}=\frac{(2.66\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.371 m

For the ion of oxygen-18, we have:

m_B = \frac{18}{16}m_A = 2.99\cdot 10^{-26}kg

q_B = 1.6\cdot 10^{-19}C (it is singly charged)

v_B=2.90\cdot 10^6 m/s

B_B=1.30 T

So the radius of its orbit is

r_B=\frac{m_B v_B}{q_B B_B}=\frac{(2.99\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.417 m

After each ion has travelled a semicircle, the separation between the two ions will be twice the difference in their radius, so:

d=2(r_B-r_A)=2(0.417-0.371)=0.092 m

3 0
2 years ago
charge, q1 =5.00μC, is at the origin, a second charge, q2= -3μC, is on the x-axis 0.800m from the origin. find the electric fiel
IRISSAK [1]

Answer:

Explanation:

Electric field due to charge at origin

= k Q / r²

k is a constant , Q is charge and r is distance

= 9 x 10⁹ x 5 x 10⁻⁶ / .5²

= 180 x 10³ N /C

In vector form

E₁ = 180 x 10³ j

Electric field due to q₂ charge

= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²

= 30.33 x 10³ N / C

It will have negative slope θ with x axis

Tan θ = .5 / √.5² + .8²

= .5 / .94

θ = 28°

E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j

= 26.78 x 10³ i - 14.24 x 10³ j

Total electric field

E = E₁  + E₂

= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j

= 26.78 x 10³ i + 165.76 X 10³ j

magnitude

= √(26.78² + 165.76² ) x 10³ N /C

= 167.8 x 10³  N / C .

3 0
3 years ago
A certain frictionless simple pendulum having a length l and mass m swings with period t. If both l and m are doubled, what is t
iVinArrow [24]

If l and m both are doubled then the period becomes √2*T

what is a simple pendulum?

It is the one which can be considered to be a point mass suspended from a string or rod of negligible mass.

A pendulum is a weight suspended from a pivot so that it can swing freely.

Here,

A certain frictionless simple pendulum having a length l and mass m

mass of pendulum = m

length of the pendulum = l

The period of simple pendulum is:

T = 2\pi \sqrt{\frac{l}{g} }

Where k is the constant.

Now the length and mass are doubled,

m' = 2m

l' = 2l

T' = 2\pi \sqrt{\frac{2l}{g} }

T' = \sqrt{2}* 2\pi \sqrt{\frac{l}{g} }

T' = \sqrt{2} * T

Hence,

If l and m both are doubled then the period becomes √2*T

Learn more about Simple Harmonic Motion here:

<u>brainly.com/question/17315536</u>

#SPJ4

8 0
1 year ago
The dimensions of the disk of our milky way galaxy are:.
Katen [24]

Answer:

According to studies, the milky way is approximately, "170,000–200,000 light-years (52–61 kpc) in diameter and, on average, approximately 1,000 ly (0.3 kpc) thick."

With that being said, it is safe to say that the dimensions are somewhere around 100,000 by 1,000

6 0
2 years ago
PLEASE HELP!!! WILL GIVE 30 POINTS!! HAS TO BE CORRECT!
belka [17]

Answer:

Disruption to electricity power grid

Explanation:

We're looking a a solar flare. This will whip solar particles at high velocity into space and, If they are near earth, will interact with the earth's magnetic field. These magnetic changes will be measurable in the electric grid. Whether they are strong enough to cause "disruption" depends on a huge number of factors such as strength of and angles of the interacting magnetic fields and location of grid infrastructure,

4 0
2 years ago
Read 2 more answers
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