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3 years ago

What is the mass of a 50 kg person on earth?

Physics

1 answer:

olganol [36]3 years ago

3 0

Answer:

a. 50 Kg hope it helps :)

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Two remote control cars with masses of 1.16 kilograms and 1.98 kilograms travel toward each other at speeds of 8.64 meters per s

Black_prince [1.1K]

The initial momentum of the system can be expressed as,

$p_i=m_1u_1+m_{2_{}}u_2$

The final momentum of the system can be given as,

$p_f=m_1v_1+m_{2_{}}v_2$

According to conservation of momentum,

$p_i=p_f$

Plug in the known expressions,

$\begin{gathered} m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ m_2v_2=m_1u_1+m_2u_2-m_1v_1 \\ v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2} \end{gathered}$

Initially, the second mass move towards the first mass therefore the initial speed of second mass will be taken as negative and the recoil velocity of first mass is also taken as negative.

Plug in the known values,

$\begin{gathered} v_2=\frac{(1.16\text{ kg)(8.64 m/s)+(1.98 kg)(-3.34 m/s)-(1.16 kg)(-2.16 m/s)}}{1.98\text{ kg}} \\ =\frac{10.02\text{ kgm/s-}6.61\text{ kgm/s+}2.51\text{ kgm/s}}{1.98\text{ kg}} \\ =\frac{5.92\text{ kgm/s}}{1.98\text{ kg}} \\ \approx2.99\text{ m/s} \end{gathered}$

Thus, the final velocity of second mass is 2.99 m/s.

3 0

1 year ago

The vertical height of an ocean wave from a crest to a trough is 10 meters. What is the amplitude of the ocean wave?

natima [27]

The amplitude of the wave would be 5 meters.

6 0

3 years ago

Read 2 more answers

I NEED HELP PLEASE, THANKS! :)

mrs_skeptik [129]

Answer:

1. Largest force: C; smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

$F=K\dfrac{ q_{1}q_{2}}{r^{2}}$

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write

$F=\dfrac{k}{r^{2}}$

1. Net force on each particle

Let's

Call the distance between adjacent charges d.
Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

$\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(b) Force on B

$\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(C) Force on C

$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$

2. Ratio of largest force to smallest

$\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}$

7 0

3 years ago

How much heat is lost by 2.0 grams of water if the temperature drops from 31 °C to 29 °C? The specific heat of water is 4.184 J/

Elanso [62]

Given :

Mass of water, m = 2 grams.

The temperature of water drops from 31 °C to 29 °C .

The specific heat of water is 4.184 J/(g • °C).

To Find :

Amount of heat lost in this process.

Solution :

We know, heat lost is given by :

$Heat\ lost,H = ms( T_f - T_i)\\\\H = 2\times 4.184 \times ( 31 - 29 )\ J\\\\H = 16.736\ J$

Therefore, amount of heat lost in this process is 16.736 J.

4 0

3 years ago

One event occurs at the origin at t equal to zero, and a second events occurs at the point x=5m along the x-axis at time with ct

Anastasy [175]

Answer:

The separation will be spacelike.

The first event can't cause the second event, as there exist an frame of reference in which both happens at the same time, in different positions, so, if there were causally connected, it will imply an instant connection, this is faster than light.

Explanation:

We can define the separation between two events (using the + - - - signature) as :

$(\Delta s )^2 = (ct_2 - c t_1 )^2 - (x_2 - x_1)^2$

where the separation will be lightlike if is equal to zero, timelike if is positive and spacelike if is negative.

For our problem

$c t_1 = 0$

$x_1 = 0$

$ct_2 = 4 \ m$

$x_2 = 5 \ m$

$(\Delta s )^2 = (4 \ m - 0 )^2 - ( 5 \ m - 0)^2$

$(\Delta s )^2 = (4 \ m )^2 - ( 5 \ m 0)^2$

$(\Delta s )^2 = 16 \ m^ 2 - 25 \ m^2$

$(\Delta s )^2 = - 9\ m^2$

So the separation will be spacelike, and the first event can't cause the second event, as there exist an frame of reference in which both happens at the same time, in different positions, so, if there were causally connected, it will imply an instant connection, this is faster than light.

8 0

3 years ago