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GenaCL600 [577]
4 years ago
14

How would the number 56,780,000,000 be written in scientific notation?

Physics
1 answer:
kipiarov [429]4 years ago
3 0

5.678 * 10^10 is the answer

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A 300 cm rope under a tension of 120 N is set into oscillation. The mass density of the rope is 120 g/cm. What is the frequency
Vikki [24]

Answer:

Explanation:

f = \sqrt{T/(m/L)} / 2L

T = 120 N

L = 3.00 m

(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m

                                                  (wow that's massive for a "rope")

f = \sqrt{120/12} /(2(3)))

f = \sqrt{10\\}/6 = 0.527 Hz

This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.

A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N

5 0
3 years ago
Can you help me with a science test
Semenov [28]

Answer:

yes just give me questions and put me brainliest

Explanation:

8 0
3 years ago
Read 2 more answers
EARTH SCIENCE --the line of longitude used as the origin in a system of coordinates
hichkok12 [17]
The answer is prime meridan
4 0
3 years ago
Archimedes supposedly was asked to determine whether a crown made for the king consisted of puregold. According to legend, he so
DENIUS [597]

Answer:

the crown was not made of pure gold

Explanation:

Mass of gold = weight in air/ g = 7.84N/10ms-2= 0.784 Kg or 0.8Kg

From Archimedes principle:

Upthrust= weight in air- weight in a fluid

Upthrust= volume × density × g

Note density of water = 1000kgm-3

7.84-6.84= V × 1000kgm-3×10ms-2

V= 1/10000= 1×10-4 m^3

Density = mass/ volume= 0.8/1×10-4

= 8×10^3 Kgm-3

But we know the density of gold to be 19.3 ×10^3 kgm-3

Hence the crown was not made of pure gold

4 0
3 years ago
7. A car is travelling along a road at 30 ms when a pedestrian steps into the road 55 m ahead. The
trasher [3.6K]

Answer: Car collide with man

Explanation:

Given

Speed of car is u=30\ m/s

Distance of the man from the car is s=55\ m

Reaction time t_r=0.5\ s

Rate of deceleration a_d=-10\ m/s^2

Distance traveled in the reaction time d_o=30\times 0.5=15\ m

Net effective distance to cover d=55-15=40\ m

Distance required to stop the car

\Rightarrow v^2-30^2=2(-10)(s)\\\Rightarrow 0-900=-20s\\\Rightarrow s=45\ m

Require distance is more than that of net effective distance. Hence, car collides with the man.

6 0
3 years ago
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