Nuclear energy is currently used in which three kinds of vehicles?

What is the velocity in meters per second of a runner who runs exactly 110 m toward the beah in 72 seconds?

rjkz [21]

B) 1.53 m/s towards the beach

8 0

4 years ago

Which is an si metric unit of measurement that is used to record the heat transfer of a solution in a classroom investigation?

kumpel [21]

The SI unit for heat energy is joule

4 0

3 years ago

A child in a boat throws a 5.80-kg package out horizontally with a speed of 10.0 m/s. The mass of the child is 24.6kg and the ma

Sergio [31]

Answer:

-0.912 m/s

Explanation:

When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.

$(m_c + m_b)v_b + m_pv_p = 0$

where $m_c = 24.6 kg, m_b = 39 kg, m_p = 5.8 kg$ are the mass of the child, the boat and the package, respectively. $, v_p = 10m/s, v_b$ are the velocity of the package and the boat after throwing.

$(24.6 + 39)v_b + 5.8*10 = 0$

$63.6v_b + 58 = 0$

$v_b = -58/63.6 = -0.912 m/s$

3 0

4 years ago

An 80g meter stick is supported at its 30 cm mark by a string attached to the ceiling. A 20 g mass is hung from the 80 cm mark w

BARSIC [14]

Answer:

$104\; {\rm g}$ , assuming that the meter stick is uniform with the center of mass precisely at the $50\; {\rm cm}$ mark.

Explanation:

Refer to the diagram attached. The meter stick could be considered as a lever. The string at the $30\; {\rm cm}$ mark would then act as the fulcrum of this lever.

The $m_{A} = 20\; {\rm g}$ mass at the $80\; {\rm cm}$ mark is at a distance of $r_{A} = 50\; {\rm cm}$ to the right of the fulcrum at $30\; {\rm cm}$ .

The weight of the $80\; {\rm g}$ meter stick acts like a weight of $m_{B} = 80\; {\rm g}$ attached to the center of mass of this meter stick. Under the assumptions, this center of mass of this meter stick would be at the $50\; {\rm cm}$ mark, which is $r_{B} = 20\; {\rm cm}$ to the right of the fulcrum at $30\; {\rm cm}$ .

Let $m_{C}$ be the mass attached to the meter stick at the $5\; {\rm cm}$ mark. This mass would be at a distance of $r_{C} = 25\; {\rm cm}$ to the left of the fulcrum at $30\; {\rm cm}$ .

At equilibrium:

$\begin{aligned} & m_{C}\, r_{C} && (\text{mass on the left of fulcrum})\\ &= m_{A}\, r_{A} + m_{B} \, r_{B} && (\text{mass on the right of fulcrum})\end{aligned}$ .

Solve for $m_{C}$ , the unknown mass attached to the meter stick at the $5\; {\rm cm}$ mark:

$\begin{aligned}m_{C} &= \frac{m_{A}\, r_{A} + m_{B}\, r_{B}}{r_{C}} \\ &= \frac{20\; {\rm g} \times 50\; {\rm cm} + 80\; {\rm g} \times 20\; {\rm cm}}{25\; {\rm cm}} \\ &= 104\; {\rm g}\end{aligned}$ .