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2 years ago

Which of the following are dwarf planets? Check all that apply. Ceres Namaka Eris Charon Haumea Makemake Pluto

2 answers:

6 0

Ceres: Yes!
Namaka: No!
Eris: Yes!
Charon: No. (it's a satellite, and dwarf planet's can't be satellites!)
Haumea: Yes!
Makemake: Yes!
Pluto: Yes!

Glad To Help;)

lorasvet [3.4K]2 years ago

3 0

Answer:

Dude above is correct

Explanation:

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A force of 150 N accelerates a 25 kg wooden chair across a wood floor at 4.3 m/s2 . How big is the frictional force on the block

solniwko [45]

We can first calculate the net force using the given information.

By Newton's second law, F(net) = ma:

F(net) = 25 * 4.3 = 107.5

We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):

f = F(net) - F(app) = 150 - 107.5 = 42.5 N

Now we can calculate the coefficient of friction, u, using the normal force, F(N):

f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17

4 0

2 years ago

A person weighing 0.70 kn rides in an elevator that has an upward acceleration of 1.5 m/s2. what is the magnitude of the normal

creativ13 [48]

First of all, we can find the mass of the person, since we know his weight W:
$W=mg=0.70 kN=700 N$
And so
$m= \frac{W}{g}= \frac{700 N}{9.81 m/s^2}=71.4 kg$

We know for Newton's second law that the resultant of the forces acting on the person must be equal to the product between the mass and the acceleration a of the person itself:
$\sum F = ma$
There are only two forces acting on the person: his weight W (downward) and the vincular reaction Rv of the floor against the body (upward). So we can rewrite the previous equation as
$R_v -W = ma$
We know the acceleration of the system, $a=1.5 m/s^2$ (upward, so with same sign of Rv), so we can solve to find the value of Rv, the normal force exerted by the elevator's floor on the person:
$R_v = ma+W=(71.4 kg)(1.5 m/s^2)+700 N =807N$

8 0

2 years ago

At room temperature, a typical person loses energy to the surroundings at the rate of 62 W. If this energy loss has to be made u

Alex_Xolod [135]

To solve this problem it is necessary to use the given proportions of power and energy, as well as the energy conversion factor in Jules to Calories.

The power is defined as the amount of energy lost per second and whose unit is Watt. Therefore the energy loss rate given in seconds was

$P = \frac{E}{t} \rightarrow E= Energy, t = time$

$P = 62W = 62 \frac{J}{s}$

The rate of energy loss per day would then be,

$P = 62\frac{J}{s} (\frac{86400s}{1day})$

$P = 5356800 \frac{J}{day}$

That is to say that Energy in Jules per lost day is 5356800J

By definition we know that $1KCal = 4.184*10^{6}J$

In this way the energy in Cal is,

$E = 5356800J \frac{1KCal}{4.184*10^{6}J}$

$E = 1279.694 KCal$

The number of kilocalories (food calories) must be 1279.694 KCal

4 0

2 years ago

Your town is installing a fountain in the main square. If the water is to rise 26.0 m (85.3 feet) above the fountain, how much p

Brums [2.3K]

Answer:

$P = 3.55 \times 10^5 Pa$

Explanation:

As we know that water from the fountain will raise to maximum height

$H = 26.0 m$

now by energy conservation we can say that initial speed of the water just after it moves out will be

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

$v = \sqrt{2gH}$

$v = \sqrt{2(9.81)(26)}$

$v = 22.6 m/s$

Now we can use Bernuolli's theorem to find the initial pressure inside the pipe

$P = P_0 + \frac{1}{2}\rho v^2$

$P = 10^5 + \frac{1}{2}(1000)(22.6^2)$

$P = 3.55 \times 10^5 Pa$

6 0

2 years ago

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a

Lapatulllka [165]

Answer:

Amplitude, A = 0.049 meters

Explanation:

Given that,

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a function of time according to the equation :

$y = 0.049 \cos(7t)$ .......(1)

The general equation of a wave is given by :

$y=A\cos(\omega t)$ .......(2)

A is amplitude of wave

On comparing equation (1) and (2) we get :

A = 0.049 meters

So, the amplitude of the wave is 0.049 meters.

3 0

2 years ago