The theory of evolution does not:

A plate drops onto a smooth floor and shatters into three pieces of equal mass.Two of the pieces go off with equal speeds v at r

Firlakuza [10]

Answer:

Speed of the this part is given as

$v_3 = \sqrt2 v$

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

Explanation:

As we know by the momentum conservation of the system

we will have

$P_1 + P_2 + P_3 = P_i$

here we know that

$P_1 = P_2$

the momentum of two parts are equal in magnitude but perpendicular to each other

so we will have

$P_1 + P_2 = \sqrt{P^2 + P^2}$

$P_1 + P_2 = \sqrt2 mv$

now from above equation we have

$P_3 = -(P_1 + P_2)$

$mv_3 = -(\sqrt 2 mv)$

$v_3 = \sqrt2 v$

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

6 0

2 years ago

Un movil de masa 12Kg sobre el cual estan actuando varias fuerzas F_1=48N, F_2=60N y F_3=30N Calcular la aceleracion con la cual

Nikitich [7]

Answer:

Lamentablemente el problema está incompleto, pues no sabemos la dirección en la que se aplican las fuerzas. Por ello, voy a resolver el problema asumiendo dos casos. (abajo se puede ver una imagen donde se describe cada caso)

1) Todas las fuerzas están en la misma dirección.

Entonces la fuerza neta será la suma de las 3 fuerzas, entonces:

F = 48N + 60N + 30N = 138N

Y por la segunda ley de Newton sabemos que:

F = m*a

fuerza igual a masa por aceleración.

Entonces la aceleración está dada por:

a = F/m = 138N/12kg = 11.5 m/s^2

2) Segundo caso, suponemos que F1 es opuesta a F2 y F3

En este caso, la fuerza neta será:

F = F2 + F3 - F1 = 60N + 30N - 48N = 42N

En este caso, la aceleración será:

a = 42N/12kg = 3.5 m/s^2

7 0

2 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

2 years ago

What is the apparent magnitude of the brightest star in the Big Dipper?

alex41 [277]

The magnitude of Alioth ( the brightest star in the big dipper ) is 1.76 and it is about 81 light years distant from Earth.

4 0

2 years ago

Read 2 more answers

The engine of a 2000 kg Mercedes going up Pike’s Peak delivers energy to its drive wheel at the rate 100 kW . Neglecting air res

kykrilka [37]

Answer:

v = 10 m/s

Explanation:

given,

Mass of Mercedes engine = 2000 Kg

Power delivered = 100 kW

angle made with horizontal = 30°

acceleration due to gravity = 10 m/s²

largest speed car can sustain = ?

we know,

Power = Force x velocity

P = F x v

P = mg sinθ x v

P = mg sin 30° x v

P = 0.5 mg x v

$v = \dfrac{100 \times 10^3}{0.5\times 2000 \times 10}$

v = 10 m/s

hence, the maximum velocity is equal to v = 10 m/s

8 0

2 years ago