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3 years ago

Give an order-of-magnitude estimate for the time in seconds of a year

1 answer:

4 0

We all know that there are 365 days in a year, 24 hours in a day, 60 minutes in an hour and lastly 60 seconds in a minute. Round that off to 100, 10, 100 and 100.That will be 10^2 x 10^2 x 10^1 x 10^2 = 10^7 seconds.

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A neutron star and a black hole are 3.34 x 1012 m from each other at a certain point in their orbit. The neutron star has a mass

m_a_m_a [10]

Answer:

F=1.65 x 10²⁶ N

Explanation:

Given that

Distance ,R= 3.34 x 10¹² m

Mass m₁= 2.78 x 10³⁰ kg

Mass ,m₂= 9.94 x 10³⁰ kg

we know that gravitational force F given as

$F=G\dfrac{m_1m_2}{R^2}$

G=Constant

G=6.67 x 10⁻¹¹ Nm²/kg²

Now by putting the values

$F=6.67\times 10^{-11}\times \dfrac{2.78\times 10^{30}\times 9.94\times 10^{30}}{(3.34\times 10^{12})^2}\ N$

F=1.65 x 10²⁶ N

Therefore the force between these two mass will be 1.65 x 10²⁶ N.

5 0

3 years ago

Can someone explain with steps please

bonufazy [111]

Answer:

The speed of other projectile is $3.1m/s$

Explanation:

Range of projectile is given by the equation

$\mathrm{R}=\frac{\mathrm{u}^{2} \cdot \sin 2 \theta}{\mathrm{g}}$

Here we have same range

Hence

$\frac{\mathrm{2.5}^{2} \cdot \sin (2 \times 65)}{\mathrm{g}}=\frac{\mathrm{u}^{2} \cdot \sin (2 \times 15)}{\mathrm{g}}\\\\u^2=\frac{2.5^2\sin130}{\sin30} \\\\u=3.10m/s$

5 0

3 years ago

Read 2 more answers

Help science 70 points

nirvana33 [79]

I didn't want to comment but don't trust those links it's a scam or a virus

3 0

3 years ago

Read 2 more answers

Describe what happens to the magnitude of the net electrostatic force on the electron as the electron

solniwko [45]

Answer:

hm my mom said the net force changes

7 0

3 years ago

Two physics students are standing on skateboards and facing each other. Their masses, including the skateboards, are 95 kg and 5

andrey2020 [161]

Answer:

1) $\sum \vec p=0\ kg.m.s^{-1}$

2) $v_2=55.078\ m.s^{-1}$

Explanation:

Given:

mass of heavier student, $m_1=95\ kg$

mass of lighter student, $m_2=58\ kg$

velocity of the heavier student after the mutual push, $v_1=3.1\ m.s^{-1}$

Since they push each other and we are neglecting any friction that means we can consider this as a perfectly elastic collision. So the momentum of the two bodies will be equal and in opposite direction.

Since momentum is a vector quantity so the total momentum of the system as a whole will be zero.

<u>From the condition of elastic collision:</u>

$p_1=p_2$

$m_1.v_1=m_2.v_2$

$95\times 3.1=58\times v_2$

$v_2=55.078\ m.s^{-1}$

6 0

4 years ago