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3 years ago

Discuss the role of global ization in the development of SI unit

Physics

1 answer:

Delvig [45]3 years ago

8 0

Answer:

Sharing of information

Explanation:

The development of SI unit has helped in the sharing of scientific as well as techical information internationally.

HOPE THIS HELPED

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An object with mass M is attached to the end of a string and is raised vertically at a constant acceleration of g 10 . If it has

RoseWind [281]

Answer:

Work done against gravity will be

W = Mgℓ

Explanation:

Work done to raise the mass from ground to given height is against gravity

So here work done is given by the formula

$W = F.d$

here we know that

$F = Mg$

it is the force due to gravity which is also known as weight

so here distance moved by the object is given as

d = ℓ

now work done is given as

W = Mg ℓ

8 0

3 years ago

Read 2 more answers

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.8 1010 m (inside the orbit

creativ13 [48]

Answer:

Explanation:

From the given information:

Distance $d_i = 4.8 \times 10^{10} \ m$

Speed of the comet $V_i = 9.1 \times 10^{4} \ m/s$

At distance $d_2 = 6 \times 10^{12} \ m$

where;

mass of the sun = $1.98 \times 10^{30}$

$G = 6.67 \times 10^{-11}$

To find the speed $V_f$ :

Using the formula:

$E_f = E_i + W \\ \\ where; \ \ W = 0 \ \ \text{since work done by surrounding is zero (0)}$

$E_f = E_i + 0 \\ \\ K_f + U_f = K_i + U_i \\ \\ = \dfrac{1}{2}mV_f^2 + \dfrac{-GMm}{d^2} = \dfrac{1}{2}mV_i^2+ \dfrac{-GMm}{d_i} \\ \\ V_f = \sqrt{V_i^2 + 2 GM \Big [ \dfrac{1}{d_2}- \dfrac{1}{d_i}\Big ]}$

$V_f = \sqrt{(9.1 \times 10^{4})^2 + 2 (6.67\times 10^{-11}) *(1.98 * 10^{30} ) \Big [ \dfrac{1}{6*10^{12}}- \dfrac{1}{4.8*10^{10}}\Big ]}$

$\mathbf{V_f =53.125 \times 10^4 \ m/s}$

3 0

2 years ago

Please help me

Grace [21]

Answer:

a. 60 N*s

b. 60 (kg*m)/s

c. 3 m/s

Explanation:

Givens:

m = 20 kg

v_i = 0 m/s

t = 10 s

F = 6 N

a) Impulse:

I = F*t

I = 6 N*10 s

I = 60 N*s

b) Momentum:

p = v*m

F = m(a)

a = F/m

a = 6 N/20 kg

a = 0.3m/s^2

a = (v_f -v_i)/t

v_f = (0.3 m/s^2)*10 s

v_f = 3.0 m/s

p = 3 m/s*20 kg

p = 60 (kg*m)/s

c. Final velocity

a = (v_f -v_i)/t

v_f = (0.3 m/s^2)*10 s

v_f = 3.0 m/s

6 0

3 years ago

A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each sec

Dennis_Churaev [7]

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 8=50.24rad/sec$

Velocity of the string wave is equal to $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec$

Power of wave propagation is equal to $P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt$

So power of the wave will be equal to 5.464 watt

6 0

3 years ago

Two +1 C charges are separated by 30000 m, what is the magnitude of the force?

Kipish [7]

Answer:

The magnitude of the force is 10 N

Explanation:

Coulomb's Law

The electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge

d= The distance between the objects

We have two identical charges of q1=q2=1 c separated by d=30000 m, thus the magnitude of the force is:

$\displaystyle F=9\cdot 10^9\frac{1*1}{30000^2}$

F = 10 N

The magnitude of the force is 10 N

7 0

3 years ago

Discuss the role of global ization in the development of SI unit​

Discuss the role of global ization in the development of SI unit