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BaLLatris [955]
3 years ago
14

What are the SI base units for length and mass?

Physics
2 answers:
notka56 [123]3 years ago
7 0

Answer:

meter and kilogram

Explanation:

As we know that there are seven fundamental SI units which are used to fine derived units

these are as following

1). Mass - Kilogram

2). Length - meter

3). Time - Seconds

4). Electric Current - Ampere

5). Amount of substance - Moles

6). Intensity of light - Candela

7). Temperature - Kelvin

hoa [83]3 years ago
6 0
The answer is:
Meter and Kilogram

I hope that i could be of assistance! ;)
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The images show two different types of Galapagos tortoises that scientists believe descended from the same species. The first ty
vichka [17]

Answer:

C. turtles with genes for long necks had a better chance of surviving to reach reproductive age.

Explanation:

The turtles that had long necks were more fit to the environmnet in which they were lovated and were able to grow larger and have more reproductive time because of their ability to feed on grass and small shrubs, this helped them always haev food available, and made them the dominant gene eventually.

3 0
3 years ago
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Mars has a mass of about 6.4 x 1023 kg, and its moon Phobos has a mass of about 9.6 x 1015 kg. If the magnitude of the gravitati
Sliva [168]

Answer:9.55×10^-4 metre

Explanation:

8 0
3 years ago
PLZ HELP!!!!
noname [10]
I believe the answer would be B) ...... sorry if i'm wrong
4 0
3 years ago
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Question 1 (13 marks) A charge Q is located on the top-left corner of a square. Charges of 2Q, 3Q and 4Q are located on the othe
Colt1911 [192]

Answer:

Explanation:

First of all we shall calculate electric field near charge 2Q .

electric field due to charge Q = K x Q /  (5 x 10⁻² )²

E₁  = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis

E₁  = KQ x 10⁴  i / 25  

Similarly electric field due to charge 3Q near 2Q

=  3KQ x 10⁴  i / 25 . It is acting along y-axis

E₂ = 3KQ x 10⁴  j / 25

Similarly electric field due to charge 4Q near 2Q

=  4KQ x 10⁴  j / (25 x 2 )

= 2 KQ x 10⁴  / 25 . It is acting acting along north east direction

unit vector in north east direction = ( i + j )/ √2

So E₃ can be represented by

E₃ = 2 KQ x 10⁴  ( i + j )  / 25 x √2

Total field =  KQ x 10⁴  i / 25 + 3KQ x 10⁴  j / 25 + 2 KQ x 10⁴  ( i + j )  / ( 25 x √2 )

= KQ x 10⁴  [ i + 3 j + √2 i + √2 j ) / 25

= 400 KQ ( 2.414 i + 4.414 j )  N / C

Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j )  x 2Q  N

= 800 KQ² ( 2.414 i + 4.414 j ) N

= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N

= 108.63 ( i + 2 j ) N .

Magnitude of this force

= 108.63 x √5

= 243 N approx .

4 0
3 years ago
A 17-mm-wide diffraction grating has rulings of 530 lines per millimeter. White light is incident normally on the grating. What
Sedbober [7]

Answer:

377 nm

Explanation:

Number of lines per meter is, N &=530 \times 1000 \\ &=530000 \text { lines } / \mathrm{m} \end{aligned}

Grating element is, d=\frac{1}{N}

=1.8868 \times 10^{-6} \mathrm{~m}[tex]
Order is, n=5
Condition for maximum intensity is, [tex]d \sin \theta=n \lambda

 \lambda &=\frac{1.8868 \times 10^{-6}}{5(\sin 90)} \\
&=0.377 \times 10^{-6} \mathrm{~m} \\
&=377 \mathrm{~nm}

7 0
3 years ago
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