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Strike441 [17]
3 years ago
6

What is the acceleration of an 800kg car that has a force of 2400N? A= F/M

Physics
1 answer:
dangina [55]3 years ago
6 0

You habe already the formula, what's wrong then? it's simple

A = F/M

A = 2400/800

A = 24/8

A = 3 m/s²

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a tree truck

is the answer i hope this helps you xD


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All organisms in Kingdom Animalia are multicellular, meaning their bodies are composed of more than one cell. Which other charac
k0ka [10]
Well,

For the first one, the answer would be C, because all organisms in Kingdom Animalia must eat in order to survive.

For the second one, all of the options are in Kingdom Animalia, but worms (A) and clams (C) are invertebrates.  So that leaves options B and D.
4 0
3 years ago
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Pls help ASAP!! Thank you
GaryK [48]

Answer:

Explanation:

The answer you should pick is A. Two magnets will attract each other when their poles are opposite and they are close together.

5 0
3 years ago
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The Hubble Space Telescope has an aperture of 2.4 m and focuses visible light (400-700 nm). The Arecibo radio telescope in Puert
stealth61 [152]

Answer:

y_{hubble} = 77\ \ m

y_{aceribo} = 1.1*10^6 \ \ m

Explanation:

what is the smallest crater that each of these telescopes could resolve on our moon?

For moon ;

s = 3.8 × 10 ⁸ m

y = 1.22 λs/D

where;

λ = 400 nm = 400× 10 ⁻⁹

D = 2.4 m

The smallest crater for the hubble space is calculated as follows:

y_{hubble} = 1.22*400*10^{-9}*3.8*10^8/2.4

y_{hubble} = 77\ \ m

For Aceribo ;

y = 1.22 λs/D

where :

λ = 75 cm = 0.75 m

D = 305 m

y_{acerbo} = 1.22*0.75 *3.8*10^8/305

y_{aceribo} = 1.1*10^6 \ \ m

5 0
2 years ago
A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h
Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

                    \tau = I \times \alpha

and,       I = \sum mr^{2}

So,      I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}

                       = 4 kg m^{2}

      \tau_{1} = 4 kg m^{2} \times \alpha_{1}

     \tau_{2} = I_{2} \alpha_{2}

So,      I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}

                     = 1 kg m^{2}

 as \tau_{2} = I_{2} \alpha_{2}

                   = 1 kg m^{2} \times \alpha_{2}        

Hence,     \tau_{1} = \tau_{2}

                  4 \alpha_{1} = \alpha_{2}

            \alpha_{1} = \frac{1}{4} \alpha_{2}

Thus, we can conclude that the new rotation is \frac{1}{4} times that of the first rotation rate.

8 0
3 years ago
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