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3 years ago

A speeding car collides with an unlucky bug flying across the road

2 answers:

7 0

<span>The car has more momentum. (A)*
Also, the car is made out of more resilient material than the bug is.

* Unless the bug is much bigger than the car.</span>

guapka [62]3 years ago

6 0

Answer:

A: The car has more momentum

Explanation:

When we discuss collisions, we are usually referring to a change in momentum. If a car crashes into a wall, then all the momentum that the car had acts on the wall and either decreases (this is where a car would drive through a wall), or change to zero (where the wall would withstand the force).

In this case, the bug and the car have different values of momentum. The car obviously has more because it is both heavier and can be assumed to be travelling a lot faster, this means that there is a negligible amount of force in the opposite, from the bug.

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A motor has an internal resistance of 12.1 Ω. The motor is in a circuit with a current of

kotykmax [81]

Answer:

Explanation:

V = I * R

V = 4 * 12.1 = 48.4 v

7 0

3 years ago

An asteroid orbits the Sun every 176 years. What is the asteroids average distance from the Sun? P ^ 2 = a ^ 3 where p = period

KatRina [158]

Answer:

The value is $x = 45.99 \ Au$

Explanation:

From the question we are told that

The period of the asteroid is $T = 176 \ years = 176 * 365 * 24 * 60* 60 = 5.55*10^{9}\ s$

Generally the average distance of the asteroid from the sun is mathematically represented as

$R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }$

Here M is the mass of the sun with a value

$M = 1.99*10^{30} \ kg$

G is the gravitational constant with value $G = 6.67 *10^{-11} \ m^3 \cdot kg^{-1} \cdot s^{-2}$

$R = \sqrt[3]{ \frac{6.67 *10^{-11} * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }$

=> $R = 6.88 *10^{12} \ m$

Generally

$1.496* 10^{11} \ m \to 1 Au (Astronomical \ unit )$

$R = 6.88 *10^{12} \ m \ \ \ \ \to \ \ x \ Au$

=> $x = \frac{6.88 *10^{12}}{1.496 *10^{11}}$

=> $x = 45.99 \ Au$

7 0

3 years ago

9. Una jeringa contiene cloro gaseoso, que ocupa un volumen de 95 mL a una presión de 0,96 atm. ¿Qué presión debemos ejercer en

masha68 [24]

Answer:

2.61 atm

Ley de Boyle

Explanation:

$P_1$ = Presión inicial = 0.96 atm

$P_2$ = Presión final

$V_1$ = Volumen inicial = 95 mL

$V_2$ = Volumen final = 35 mL

En este problema usaremos la ley de Boyle.

$\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=\dfrac{0.96\times 95}{35}\\\Rightarrow P_1=2.61\ \text{atm}$

La presión ejercida sobre el émbolo para reducir su volumen es de 2.61 atm.

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3 years ago

Please answer quickly :)

Elena L [17]

Answer:

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Explanation:

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3 years ago

Analogy: atom is to element as ____ is to compound

Masja [62]

Molecule is the answer to what you are looking for

5 0

4 years ago