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2 years ago

5

In astronomy, distances are often expressed in light-years. One light-year is the distance traveled by light in one year. If the

distance to a star is 3.5 light-years, what is this distance in meters

1 answer:

amid [387]2 years ago

5 0

Answer:

dd=4.3 ly=4.3 •9.46•10°15 m=4.1 10°16m.

Explanation:

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A 2kg mass moving at a speed of 3 m/s is stopped by a constant force of 15N. How many seconds must the force act on the mass to

chubhunter [2.5K]

Answer:

t = 0.4 seconds

Explanation:

Given that,

Mass of the object, m = 2 kg

Initial speed of the object, u = 3 m/s

Finally, it stops, v = 0

Constant force applied to the object, F = 15 N

We know that,

Force, F = ma

$F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{2\times (0-3)}{15}\\\\=0.4\ s$

So, it will take 0.4 seconds to stop the mass.

6 0

3 years ago

A car's engine generates 100,000W of power as it exerts a force of 10,000N . How long does it take the car to travel 100m ?

kompoz [17]

Answer: 10 s

Explanation:

Given

Engine generates $100,000\ W$ power

Force exerts $F=10,000\ N$

Distance traveled $d=100\ m$

work done is given by

$W=F\cdot d\\W=10,000\times 100\\W=10^5\ J$

Also, Energy is given as

$\Rightarrow E=P\cdot t$

Insert the values

$\Rightarrow 10^6=100,000\times t\\\\\Rightarrow t=\dfrac{10^6}{10^5}\\\\\Rightarrow t=10\ s$

5 0

2 years ago

Can Hockey players hold the puck with their open hand?

devlian [24]

Answer:

Yes they can stop or bat the puck with their open hand

Explanation:

8 0

3 years ago

Each of the two coils has 320 turns. The average radius of the coil is 6 cm. The distance from the center of one coil to the ele

Temka [501]

Answer:

Magnetic field, $B=2.39\times 10^{-3}\ T$

Explanation:

It is given that,

Number of turns, N = 320

Radius of the coil, r = 6 cm = 0.06 m

The distance from the center of one coil to the electron beam is 3 cm, x = 3 cm = 0.03 m

Current flowing through the coils, I = 0.5 A

We need to find the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils. The magnetic field midway between the coils is given by :

$B=\dfrac{\mu_oINr^2}{(x^2+r^2)^{3/2}}$

$B=\dfrac{4\pi \times 10^{-7}\times 0.5\times 320\times (0.06)^2}{(0.03^2+0.06^2)^{3/2}}$

B = 0.00239 T

or

$B=2.39\times 10^{-3}\ T$

So, the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils is $2.39\times 10^{-3}\ T$ . Hence, this is the required solution.

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3 years ago

Please help me! As quickly as possible

k0ka [10]

Answer:

1. matter

2. kilograms

3. same

4. gravitational

5. gravity

6. space

7. weightlessness

8. Newton

9. weight

10. more

I HOPE THESE ARE CORRECT AND IT HELPS

7 0

2 years ago