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3 years ago

How far will you travel if you run for 10. minutes at 2.0 m/s?

1 answer:

6 0

We know that 1 minute= 60 seconds (or 1 min= 60 s).

10 min* (60 s/ 1 min)* (2.0 m/ 1 s)= 1,200 m.
(Note that the units cancel out so you get the answer)

The final answer is 1,200 m.

Hope this helps~

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Light shines through a single slit whose width is 5.7 x 10-4 m. A diffraction pattern is formed on a flat screen located 4.0 m a

lilavasa [31]

Answer:

$\lambda = 570\ nm$

Explanation:

Given,

Width of slit, W = 5.7 x 10⁻⁴ m

Distance between central bright fringe, L = 4 m

distance between central bright fringe and first dark fringe, y = 4 mm

Diffraction angle

$tan \theta = \dfrac{y}{L}$

$tan \theta = \dfrac{4}{4\times 10^3}$

$\theta = 0.0572$

Now.

$W sin \theta = m \lambda$

m = 1

$5.7 \times 10^{-4} \times sin (0.0572) = 1 \times \lambda$

$\lambda = 569.99 \times 10^{-9}\ m$

$\lambda = 570\ nm$

4 0

3 years ago

A metal rod A and a metal sphere B, on insulating stands, touch each other. They are originally neutral. A positively charged ro

ryzh [129]

Answer:

The sphere is positively charged

Explanation:

This is because when the positively charged rod is brought near the metal rod A, the electrons in metal rod A and sphere B are attracted towards it into metal rod A while the positive charges in the are repelled into sphere B. So, when the charged rod is withdrawn, and metal rod A and sphere B are separated, metal rod A is now negatively charged, but sphere B is positively charged.

So, sphere B is positively charged.

3 0

3 years ago

A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height

AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

$E_i=\frac{1}{2}m_1v_1^2$

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

$E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh$

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

$\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s$

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

$p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3$

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

$v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s$

c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

$E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J$

While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:

$E_f = (m_1 +m_2)gh_{max}$

where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find

$E_i = (m_1+m_2)gh_{max}\\h_{max}=\frac{E_i}{(m_1+m_2)g}=\frac{2070 J}{(75 kg)(9.8 m/s^2)}=2.8 m$

4 0

3 years ago

What conventions are used in SI to indicate units

klemol [59]

Answer:

<u>Conventions used in SI to indicate units are as follows:</u>

Only singular form of units are used. for example: use kg and not kgs.
Do not use full stop after the abbreviations of any unit. for example: do not use kg. or cm.
Use one space between last numeric digit and SI unit. for example: 10 cm, 9 km.
Symbols and words should not be mixed. for example: use Kilogram per cubic and not kilogram/m3.
While writing numerals, only the symbols of the units should be written. for example: use 10 cm and not Ten cm.
Units named after a scientist should be written in small letters. for example: newton, henry.
Degree sign should not be used when the kelvin unit is used. for exmaple: use 37° and not 37°k