Ask question

Ask question

All categories

4 years ago

11

If you could travel to a star 25 light-years away and return to earth at nearly the speed of light, how much time would elapse o

n earth?

2 answers:

anastassius [24]4 years ago

7 0

Millions and millions of years

lyudmila [28]4 years ago

6 0

50 years, would it not be, because time moves on no matter what, so 25ish plus 25ish would be 50 years. Good question

You might be interested in

What respiratory structure controls breathing?

shutvik [7]

Lungs is what helps u breath

3 0

3 years ago

Read 2 more answers

A glider with mass m = 0.230 kg sits on a frictionless horizontal air track, connected to a spring with force constant k = 4.50

loris [4]

Answer

given,

mass of glider = 0.23 Kg

spring constant = k = 4.50 N/m

spring stretched to 0.130 m

The springs potential energy =

$U = \dfrac{1}{2}kx^2$

$U = \dfrac{1}{2}\times 4.5 \times 0.13^2$

U = 0.038 J

at x = 0,the only energy will be kinetic .

$\dfrac{1}{2}mv^2=0.038$

$\dfrac{1}{2}\times 0.23 \times v^2=0.038$

v² = 0.3304

v = 0.575 m/s

displacement of the glider

using conservation of energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

$x =v\sqrt{\dfrac{m}{k}}$

$x =3\times \sqrt{\dfrac{0.23}{4.5}}$

x = 0.678 m

8 0

3 years ago

A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch

Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

$x=V_{o}cos \theta t$ ----------------------(1)

$y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}$ ---------(2)

$V=V_{o}-gt$ ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

$V_{o}$ is the golf ball's initial velocity

$\theta=0\°$ is the angle (it was a horizontal shot)

t is the time

y is the final height of the ball

$y_{o}$ is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

$t=\sqrt{\frac{2 y_{o}}{g}}$

$t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}$

$t=1.597$ s

Substituting (6) in (1):

$67.1 =V_{o} cos(0\°) 1.597$ -------------------(4)

Step 2: Finding $V_{o}$ :

From equation(4)

$67.1 =V_{o}(1) 1.597$

$V_0 = \frac{6.71}{1.597}$

$V_{o}=42.01$ m/s (8)

Substituting $V_{o}$ in (3):

$V=42.01 -(9.8)(1.597)$

v =42 .01 - 15.3566

V=26.359 m/s

5 0

3 years ago

Power selection feature for resistors to become water modules 10 liters of water at 25°C to đến

Mila [183]

Answer:

P = 2439.5 W = 2.439 KW

Explanation:

First, we will find the mass of the water:

Mass = (Density)(Volume)

Mass = m = (1 kg/L)(10 L)

m = 10 kg

Now, we will find the energy required to heat the water between given temperature limits:

E = mCΔT

where,

E = energy = ?

C = specific heat capacity of water = 4182 J/kg.°C

ΔT = change in temperature = 95°C - 25°C = 70°C

Therefore,

E = (10 kg)(4182 J/kg.°C)(70°C)

E = 2.927 x 10⁶ J

Now, the power required will be:

$Power = P = \frac{E}{t}$

where,

t = time = (20 min)(60 s/1 min) = 1200 s

Therefore,

$P = \frac{2.927\ x\ 10^6\ J}{1200\ s}$

<u>P = 2439.5 W = 2.439 KW</u>

7 0

3 years ago

Which of these pollutants is transferred from soil to water by organic pesticides and fertilizer runoff from farms?

Sedbober [7]

I would say your answer to this question would be D

4 0

3 years ago

Read 2 more answers