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3 years ago

Which is a unit of length in the International System of units?

Physics

1 answer:

REY [17]3 years ago

7 0

Answer:

Source: National Bureau of Standards Wall Chart. 1 inch, = 25 millimetres. 1 foot, = 0.3 metre. 1 yard, = 0.9 metre. 1 mile ...

Explanation:Source: National Bureau of Standards Wall Chart. 1 inch, = 25 millimetres. 1 foot, = 0.3 metre. 1 yard, = 0.9 metre. 1 mile ...

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When are the two days of equinox

galina1969 [7]

March 20 and September 22

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3 years ago

Read 2 more answers

On a 100km track , a train travels the first 30km with a speed of 30km/h . How fast the train travel the next 70 km if the avera

nirvana33 [79]

Solution :-

Given :

Distance 1 = 30 km

Distance 2 = 70 km

We know that speed = distance/time

and, Average speed = total distance/total time taken

When the train acquired a speed of 30 km/hr, the time taken = 30/30 = 1 hour

Average speed = 9distance 1 + distance 2)/(time 1 + time 2)

AS time 2 or t2 is time taken for the second part of the journey of 70 km

⇒ 40 = 100/(1 + t2)

⇒ 40 + 40t2 = 100

⇒ 40t2 = 100 - 40

⇒ 40t2 = 60

⇒ t2 = 60/40

⇒ t2 = 1.5

So, t2 or time taken to travel the second part of the journey is 1.5 hours.

Speed of the second part of the journey = distance 2/time 2

⇒ 70/1.5

⇒ 46.666 km/hr or 46.7 km/hr.

Hence the answer is = 46.666 km/hr or 46.7 km/hr.

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

3 0

3 years ago

A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).

just olya [345]

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles ( $\vec r_{cm}$ ), measured in meters, is defined by this weighted average:

$\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}}$ (1)

Where:

$m_{i}$ - Mass of the i-th particle, measured in kilograms.

$\vec r_{i}$ - Location of the i-th particle with respect to origin, measured in meters.

If we know that $\vec r_{cm} = (-0.500\,m,-0.700\,m)$ , $m_{1} = 1\,kg$ , $\vec r_{1} = (-1.20\,m, 0.500\,m)$ , $m_{2} = 4.50\,kg$ , $\vec r_{2} = (0.600\,m, -0.750\,m)$ and $m_{3} = 4\,kg$ , then the coordinates of the third particle are:

$(-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}$

$(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})$

$(4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)$

$(x_{3},y_{3}) = (-1.562\,m,-0.944\,m)$

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

5 0

3 years ago

Quasars are thought to be the nuclei of active galaxies in the early stages of their formation. A typical quasar radiates energy

uranmaximum [27]

Answer:

1.7531 smu/yn

Explanation:

check the attached file below for answer and explanation.

6 0

3 years ago

Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a

malfutka [58]

Answer:

a) $p_i=1.568\hat{i}+0.752 \hat{j}$

b) $v_{fx}=1.668\ m.s^{-1}$

c) $v_{fy}=0.7999\ m.s^{-1}$

Explanation:

Given masses:

$m_1=0.49\ kg$

$m_2=0.47\ kg$

Velocity of mass 1, $v_1=3.2 \hat{i}\ m.s^{-1}$

Velocity of mass 2, $v_2=1.6 \hat{j}\ m.s^{-1}$

Initial momentum:

$p_i=m_1.v_1+m_2.v_2$

$p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}$

$p_i=1.568\hat{i}+0.752 \hat{j}$

magnitude of initial momentum:

$p_i=\sqrt{1.568^2+0.752 ^2}$

$p_i=1.739\ kg.m.s^{-1}$

From the conservation of momentum:

$p_f=p_i$

$m_f.v_f=1.739$

$v_f=\frac{1.739}{0.49+0.47}$

$v_f=1.85\ m.s^{-1}$ is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

$tan\theta=\frac{0.752 }{1.568}$

$\theta=25.62^{\circ}$

$\therefore v_{fx}=1.85\ cos25.62^{\circ}$

$v_{fx}=1.668\ m.s^{-1}$

Vertical component of final velocity:

$v_{fy}=1.85\ sin 25.62^{\circ}$

$v_{fy}=0.7999\ m.s^{-1}$

6 0

3 years ago