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Darina [25.2K]
3 years ago
10

How large a net force is required to accelerate a 1600-kg SUV from rest to a speed of 25 m/s in a distance of 200 m

Physics
1 answer:
Mars2501 [29]3 years ago
6 0

Answer:

F=2496 N

Explanation:

Given that,

Mass of SUV, m = 1600 kg

Initial speed, u = 0

Final speed, v = 25 m/s

Distance, d = 200 m

We need to find the net force. Firstly, let's find acceleration using equation of motion.

v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(25)^2-(0)^2}{2\times 200}\\\\a=1.56\ m/s^2

Net force, F = ma

F=1600\times 1.56\\\\F=2496\ N

So, the net force is 2496 N.

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The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming a
Vera_Pavlovna [14]

Answer:

The radius of the gold nucleus is 7.1x10⁻¹⁵m

Explanation:

The nearest distance is:

r=\frac{kze^{2} }{mpV^{2} } (eq. 1)

Where

z = atomic number of gold = 79

e = electron charge = 1.6x10⁻¹⁹C

k = electrostatic constant = 9x10⁹Nm²C²

energy of the particle = 32 MeV = 5.12x10⁻¹²J

At the potential energy is zero, all the energy will be kinetic energy:

E_{k} =\frac{1}{2} m V^{2}

Where

m = 4 mp = mass of proton

5.12x10^{-12} =\frac{1}{2} *4*m_{p}* V^{2} \\m_{p}* V^{2} = 2.56x10^{-12}

Replacing in equation 1

r=\frac{9x10^{9}*79*(1.6x10^{-19})^{2}   }{2.56x10^{-12} } =7.1x10^{-15} m

8 0
3 years ago
A simple model for a person running the 100 m dash is to assume the sprinter runs with constant acceleration until reaching top
Trava [24]

Answer:

He will complete the race in total time of T = 10 s

Explanation:

Total distance moved by the sprinter in 2.14 s is given as

s = \frac{(v_{in} + v_{f})}{2} time

s = \frac{(0 + 11.2)}{2} (2.14)

s = 11.98 m

now the distance remaining to move

d = 100 - 11.98 = 88 m

now he will move with uniform maximum speed for the remaining distance

so we will have

time = \frac{d}{v}

time = \frac{88}{11.2} = 7.86 s

so the total time to complete the race is given as

T = 7.86 + 2.14 = 10 s

6 0
3 years ago
25 POINTS FOR ANSWER How are Newton’s Laws used to describe the motion of planets? Justify your response in two or more complete
Alexus [3.1K]

Pour la seule et simple raison qu'elle s'exerce entre tous les corps de l'univers ( objet, astres etc..

Si on tient compte des frottements liés aux chocs successifs des billes les une sur les autre, au bout d'un certain temps, le mouvement va cesser.

Si on dit que toute l'énergie potentielle de pesanteur est transformée en énergie cinétique, et réciproquement, donc que l'énergie mécanique est conservée au fil des chocs et des rebonds, alors, le mouvement est perpétuel. Le nombre de billes qui remontent est toujours égal au nombre de billes qu'on a lâchées.

La première loi concerne des systèmes immobiles, ou plutôt on considère des systèmes dit "isolé", c'est à dire qu'ils ne sont pas soumis à d'autre force que celle que l'on connait.

Ce qu'il faut retenir de celui ci c'est ça :

Si j'ai un système en mouvement rectiligne uniforme OU immobile, alors :

Avec F1 F2 F3, trois forces s'exercant sur mon système

Attention ! Ici je n'ai pas mit les flèches sur les différentes forces mais elles sont obligatoires ! On parle de vecteur force !

Pour la deuxième loi :

C'est le même principe, la différence c'est que l'on est en mouvement.

 

Avec a le vecteur accélération. Il y a beaucoup de ressource sur le net, n'hésite pas à regarder, la j'ai simplement pu te donner les expressions les plus connus. Mais il faudra les manipuler, et sans exercice sur lequel se baser, c'est plus difficile ! 

La troisième loi est bien moins importante que les deux autres, mais n'hésite pas à regarder sur le net, tu trouveras l'énoncé. C'est la même logique.

4 0
3 years ago
What is the anomalous expansivity of water
salantis [7]

Answer:

The anomalous expansion of water is an abnormal property of water whereby it expands instead of contracting when the temperature goes from 4°C to 0°C, and it becomes less dense. The density becomes less and less as it freezes because molecules of water normally form open crystal structures when in solid form.

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3 years ago
What is the half-life for the radioactive decay of particle that
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Answer:

ytyug

Explanation:

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