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3 years ago

What would be Kelley's weight be in newtons if her mass was 70 kilograms

Physics

2 answers:

ad-work [718]3 years ago

8 0

Answer:

Kelly's weight would be 688.47 Newtons.

Explanation:

1 Kilogram would be 9.81 Newtons.

xxMikexx [17]3 years ago

5 0

It all depends on where Kelley is at the moment.

-- If she's on Mars, she weighs 229 N.

-- If she's on the Moon, she weighs 113N.

-- If she's on Earth, she weighs 686 N.

-- If she's in a space capsule coasting from one body to another, then there's no instrument that can measure any weight of her.

So it completely depends on where she happens to be at the moment.

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A glider of mass 0.240 kg is on a frictionless, horizontal track, attached to a horizontal spring of force constant 6.00 N/m. In

Sveta_85 [38]

Answer:

$E_M=0.0492J$ .

Explanation:

The mechanical energy of the system will be the kinetic energy plus the elastic potential energy: $E_M=K+U_e$ .

We know that the equation for the kinetic energy is $K=\frac{mv^2}{2}$ , where m is the mass of the object and v its velocity.

We know that the equation for the elastic potential energy is $U_e=\frac{k\Delta x^2}{2}$ , where k is the spring constant and $\Delta x$ the compression (or elongation) respect to equilibrium.

So for our values we have:

$E_M=K+U_e=\frac{mv^2}{2}+\frac{k \Delta x^2}{2}=\frac{(0.24kg)(0.4m/s)^2}{2}+\frac{(6N/m)(0.1m)^2}{2}=0.0492J$ .

6 0

3 years ago

Consider the following equilibrium at 979 K for the dissociation of molecular iodine into atoms of iodine. I2(g) equilibrium rea

GenaCL600 [577]

Answer: $I_2$ = 0.0050 M

$I$ = 0.0155 M

Explanation:

Initial moles of $I_2$ = 0.072 mole

Volume of container = 3.9 L

Initial concentration of $I_2=\frac{moles}{volume}=\frac{0.072moles}{3.9L}=0.018M$

The given balanced equilibrium reaction is,

$I_2(g)\rightleftharpoons 2I(g)$

Initial conc. 0.018 M 0

At eqm. conc. (0.018-x) M (2x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[I]^2}{[I_2]}$

$K_c=\frac{(2x)^2}{0.2-x}$

we are given : $K_c=1.60\times 10^{-3}$

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

$1.60\times 10^{-3}=\frac{(2x)^2}{(0.018-x)}$

$x=0.0025$

So, the concentrations for the components at equilibrium are:

$[I]=2\times x=2\times 0.0025=0.0050$

$[I_2]=0.018-x=0.018-0.0025=0.0155$

Hence, concentrations of $I_2$ and $I$ are 0.0050 M ad 0.0155 M respectively.

4 0

3 years ago

3) A driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefticient of friction between

Natali5045456 [20]

Answer:

A) 37 m

Explanation:

The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:

$v^2 -u^2 = 2ad$ (1)

where

v = 0 m/s is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d is the length of the skid

We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

$F=ma=-\mu mg$

where

m = 1000 kg is the mass of the car

$\mu = 0.80$ is the coefficient of friction

a is the deceleration of the car

g = 9.8 m/s^2 is the acceleration due to gravity

The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

$a=-\mu g$

And we can substitute it into eq.(1) to find d:

$v^2 -u^2 = 2(\-mu g) d\\d= \frac{v^2-u^2}{-2 \mu g}=\frac{0-(24 m/s)^2}{-2(0.80)(9.8 m/s^2)}=36.7 m \sim 37 m$

7 0

3 years ago

Do longitude lines run horizontally (east-west) or vertically (north-south)?

nlexa [21]

Longitude- Horizontal (East West)
Latitude- Vertical (North South)

3 0

3 years ago

Look at the graph

viva [34]

Here, Carefully look at the graph.
When it is on x=10, it is approximately 10, (slightly less than 10)
Closest value would be 90, so y/x = 90/10 = 9

So, the density of the graph would be 9 g/cm³

In short, Your Answer would be Option D

Hope this helps!

4 0

3 years ago

Read 2 more answers