All categories

4 years ago

What is the weight, in newtons, of a 50.-kg person on the Moon?

Physics

2 answers:

solniwko [45]4 years ago

5 0

A 50kg object on earth weighs 81.67 on the moon

pav-90 [236]4 years ago

3 0

1 kg (kilogram) is equal to approx. 9.80665 n (newtons) so a 50 kg object would be 490.33250 newtons

You might be interested in

A small charge q is placed near a large spherical charge Q. The force experienced by both charges is F. The electric eld created

anastassius [24]

The electric field created by Q at the position of q is $\frac{F}{Q}$ .

The given parameters:

<em>Magnitude of charge, = q</em>
<em>Spherical charge, = Q</em>
<em>Force experienced by both charges, = F</em>

The electric field created by Q at the position of q is calculated as follows;

$E = \frac{F}{Q} \\\\$

where;

<em>E is the magnitude of electric field strength </em>
<em>F is the force experienced by both charges</em>
<em>Q is the charge</em>

Thus, the electric field created by Q at the position of q is $\frac{F}{Q}$ .

Learn more about electric field here: brainly.com/question/14372078

7 0

3 years ago

In a 5000 m race, the athletes run 12 1/2 laps; each lap is 400 m.Kara runs the race at a constant pace and finishes in 17.9 min

Ksju [112]

Answer:

No. of laps of Hannah are 7 (approx).

Solution:

According to the question:

The total distance to be covered, D = 5000 m

The distance for each lap, x = 400 m

Time taken by Kara, $t_{K} = 17.9 min = 17.9\times 60 = 1074 s$

Time taken by Hannah, $t_{H} = 15.3 min = 15.3\times 60 = 918 s$

Now, the speed of Kara and Hannah can be calculated respectively as:

$v_{K} = \frac{D}{t_{K}} = \frac{5000}{1074} = 4.65 m/s$

$v_{H} = \frac{D}{t_{H}} = \frac{5000}{918} = 5.45 m/s$

Time taken in each lap is given by:

$(v_{H} - v_{K})t = x$

$(5.45 - 4.65)\times t = 400$

$t = \frac{400}{0.8}$

t = 500 s

So, Distance covered by Hannah in 't' sec is given by:

$d_{H} = v_{H}\times t$

$d_{H} = 5.45\times 500 = 2725 m$

No. of laps taken by Hannah when she passes Kara:

$n_{H} = \frac{d_{H}}{x}$

$n_{H} = \frac{2725}{400} = 6.8$ ≈ 7 laps

3 0

3 years ago

The initial concentration of N2O4 is 0.45M The system is heated and allowed to reach equilibrium. At equilibrium concentration o

Damm [24]

Answer : The value of equilibrium constant $K_c$ for the following reaction will be, 2.4

Explanation : Given,

Initial concentration of $N_2O_4$ = 0.45 M

Equilibrium concentration of $NO_2$ = 0.30 M

$K_c$ is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The given balanced equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

As we know that the concentrations of pure solids are constant that means they do not change. Thus, they are not included in the equilibrium expression.

Actual concentration of $NO_2$ at equilibrium = $2\times 0.30M=0.6M$

Concentration of $N_2O_4$ at equilibrium = $[N_2O_4]-[NO_2]=0.45-0.30=0.15M$

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

Now put all the given values in this expression, we get :

$K_c=\frac{(0.6)^2}{(0.15)}$

$K_c=2.4$

Therefore, the value of equilibrium constant $K_c$ for the following reaction will be, 2.4

3 0

3 years ago

Two forces act on wooden box with mass of (30kg) (if F1,=85.7N east and F270.7N west):

tresset_1 [31]

The acceleration of the wooden box is 6.17 m/s² towards the west.

<h3 /><h3>What is acceleration?</h3>

Acceleration can be defined as the rate of change of velocity.

To calculate the acceleration of the wooden box can be calculated using the formula below.

Formula:

F₂-F₁ = ma.................. Equation 1

Where:

F₂ = Force on the wooden box acting towards west
F₁ = Force on the wooden box acting towards the east
m = Mass of the wooden box.
a = Acceleration of the wooden box

Make a the subject of the equation

a = (F₂-F₁)/m........................ Equation 2

From the question,

Given:

F₂ = 270.7 N
F₁ = 85.7 N
m = 30 kg

Substitute these values into equation 2

a = (270.7-85.7)/30
a = 185/30
a = 6.17 m/s² towards the west.

Hence, The acceleration of the wooden box is 6.17 m/s² towards the west.

Learn more about acceleration here: brainly.com/question/605631

7 0

2 years ago

The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were obser

Dafna1 [17]

A) Orbital speed: $v=\sqrt{\frac{GM}{R}}$

B) Kinetic energy: $K= \frac{GmM}{2R}$

D) The orbital period is $T=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

F) The angular momentum is $L=m\sqrt{GMR}$

G) Exponent of radial dependence:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Explanation:

We know that for a satellite in circular orbit around a planet of mass M, the gravitational force between the satellite and the planet is

$F=G\frac{mM}{R^2}$

where m is the mass of the satellite.

This force provides the centripetal force needed for the circular motion, which is

$F=m\frac{v^2}{R}$

where v is the orbital speed.

Since the gravitational force provides the centripetal force, we can equate the two expressions:

$G\frac{mM}{R^2}=m\frac{v^2}{r}$

And solving for v, we find

$v=\sqrt{\frac{GM}{R}}$

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

In this problem,

m is the mass of the satellite

$v=\sqrt{\frac{GM}{R}}$ is the speed of the satellite (found in part A)

Substituting, we find an expression for the kinetic energy of the satellite:

$K=\frac{1}{2}m(\sqrt{\frac{GM}{R}})^2 = \frac{GmM}{2R}$

The orbital speed of the satellite can be rewritten as the ratio between the distance covered during one orbit (the circumference of the orbit) divided by the period of revolution:

$v=\frac{2\pi R}{T}$