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Diano4ka-milaya [45]
3 years ago
13

The New Horizons probe that passed by Pluto during July 2015 is one of the fastest spacecraft ever assembled. It was moving at a

bout 14 km/s when it went by Pluto. If it maintained this speed, how long would it take New Horizons to reach the nearest star, Proxima Centauri, which is about 4.3 light-years away? (Note: It isn’t headed in that direction, but you can pretend that it is.)
Physics
1 answer:
LenKa [72]3 years ago
7 0

Answer:

2.91 x 10¹² sec

Explanation:

d = distance of nearest star, Proxima Centauri  = 4.3 ly = 4.3 x 9.46 x 10¹⁵ m

v = speed of new horizon probe = 14 km/hr = 14000 m/s

t = time taken for the new horizon probe to reach nearest star, Proxima Centauri = ?

Using the equation

d = v t

Inserting the values given

4.3 x 9.46 x 10¹⁵ = (14000) t

t = 2.91 x 10¹² sec

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4. A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exer
Rudik [331]

Answer:

0.61°

Explanation:

Since the box move at constant velocity, it means there is no acceleration then we can say it has a balanced force system.

Pulling force= resistance force

From the formula for pulling force,

F(x)= Fcos(θ)

= 425×cos(35.2)

=347N

The force exerted downward at an angle of 35.2° below the horizontal= Fsin(θ)= 425sin(35.2)

=425×0.567=245N

Resistance force= (325N+ 245N) (α)= 570N(α)

We can now equates the pulling force to resistance force

570 (α)= 347N

(α)= 347/570

= 0.61

3 0
3 years ago
Recall from Chapter 1 that a watt is a unit of en- ergy per unit time, and one watt (W) is equal to one joule per second ( J·s–1
harkovskaia [24]

Answer:

Explanation:

The energy of a photon is given by the equation E_p=h f, where h is the <em>Planck constant</em> and f the frequency of the photon. Thus, N photons of frequency f will give an energy of E_N=N h f.

We also know that frequency and wavelength are related by f=\frac{c}{\lambda}, so we have E_N=\frac{N h c}{\lambda}, where c is the <em>speed of light</em>.

We will want the number of photons, so we can write

N=\frac{\lambda E_N}{h c}

We need to know then how much energy do we have to calculate N. The equation of power is P=E/t, so for the power we have and considering 1 second we can calculate the total energy, and then only consider the 4% of it which will produce light, or better said, the N photons, which means it will be E_N.

Putting this paragraph in equations:

E_N=(\frac{4}{100})E=0.04Pt=(0.04)(100W)(1s)=4J.

And then we can substitute everything in our equation for number of photons, in S.I. and getting the values of constants from tables:

N=\frac{\lambda E_N}{h c}=\frac{(520 \times10^{-9}m) (4J)}{(6.626\times10^{-34}Js) (299792458m/s)}=1.047 \times10^{19}

3 0
3 years ago
Helppp pleaseeeeeeee. NO LINKS. HELP HELP HELP
Lera25 [3.4K]

Answer:

Choice A

Explanation:

The lower the point the higher the kinetic energy because Mechanical energy is conserved and the Gravitational Potential Energy gets lower when the height is lower

3 0
2 years ago
Which is a characteristic of a physical change? (1 point)
krek1111 [17]
The answer would be a.

a chemical change is a change to the chemical makeup of a substance so if the bonds are unchanged it would be a characteristic of a physical change
6 0
3 years ago
An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and stic
Vika [28.1K]

Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

R =u\cos\theta t

Put the value into the formula

\dfrac{230}{6} = u\cos\theta

u\cos\theta=38.33.....(I)

We need to calculate the height

Using vertical component

H=u\sin\theta t-\dfrac{1}{2}gt^2

Put the value in the equation

16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2

u\sin\theta=\dfrac{16+9.8\times18}{6}

u\sin\theta=32.06.....(II)

Dividing equation (II) and (I)

\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}

\tan\theta=0.8364

\theta=\tan^{-1}0.8364

\theta=39.90^{\circ}

(a). We need to calculate the initial speed

Using equation (I)

u\cos\theta\times t=38.33

Put the value into the formula

u =\dfrac{230}{6\times\cos39.90}

u=49.96\ m/s

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

5 0
3 years ago
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