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Brilliant_brown [7]
3 years ago
13

What object or class of objects makes up most of the solar system’s mass?

Physics
2 answers:
Vladimir [108]3 years ago
8 0
Solar system is the gravitationally bound system that consists of the sun and the objects that orbit around it directly or indirectly. These objects includes the planets which orbit the sun directly an other small objects such as meteoroids, asteroids, satellites of the planets and numerous comets. The sun makes up most of the solar system' mass.
Serggg [28]3 years ago
7 0
<span>The Sun makes up most of our solar system's mass. It accounts for about 98% of the mass of the whole solar system. Other objects such as planets, moons, asteroids and comets amount to only about 2% of the solar system's mass.</span>
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You are running at a speed of 10km/h and hit a patch of mud. Two seconds later you speed is 8km/h. What is your acceleration in
Vlad1618 [11]

Answer:

0.28 m/s^2

Explanation:

Acceleration is given by

a=\frac{v-u}{t}

where

u is the initial velocity

v is the final velocity

t is the time interval

In this problem:

u = 10 km/h \cdot \frac{1000 m/km}{3600 s/h}=2.78 m/s is the initial velocity

v = 8 km/h \cdot \frac{1000 m/km}{3600 s/h}= 2.22 m/s is the final velocity

t = 2 s is the time

Substituting, we find the acceleration:

a=\frac{2.78-2.22}{2}=0.28 m/s^2

5 0
3 years ago
An electron has a mass of 9.1x10-31 kg. What is
zloy xaker [14]

Answer:

3.185×10^-29 kgm/s

Explanation:

Momentum(p)=mass×velocity

=9.1×10^-31×3.5×10

=3.185×10^-29 kgm/s

4 0
3 years ago
A penny is dropped from the Statue of Liberty
Brrunno [24]

Answer:

a) The velocity is 2.94m/s

b) 0.441

Explanation:

a) Assume gravity is 9.8m/s^2

Use the equation below to solve for the velocity at 0.30 seconds

vf=vi+at,

vf =unknown velocity  vi= initial velocity vi=0m/s  a= 9.8m/s^2 t=0.30seconds

Step 1: Substitute the variables with the knowns

vf=0m/s+(9.8m/s^2)*0.30seconds

Step 2: Solve

vf=2.94m/s

b)

Use the equation below to solve for the displacement at 0.30 seconds

x=vit+\frac{1}{2} at^{2}

Step 1: Substitute the same variables with the knowns

x=\frac{1}{2}*(9.8m/s^2)*(0.30seconds)^{2}

Note that vi*t=0 as vi=0m/s

Step 2: Solve

x=0.441m

5 0
2 years ago
Can someone pleaseEeeeeeee ASAP answer this❗️❗️❗️❗️ I really need help it is my second time posting this❗️❗️❗️❗️
belka [17]

Answer:

Data:-vi=om/s (b/c as in question penny is dropped from building means before coming to ground its initial state or velocity was considered as zero ) now distance or height h=380m and now we have to find the final velocity vf=? and the time t=?

Explanation:

So applying second eq of motion s=vit+1/2×gt² (here we have taken a gravity b/c when ever body is in vertical position then acceleration due to gravity is applied ) s=0×t+1/2×gt² , s=0+1/2×9.8×t² ,380=4.9t² we have to find t so 4.9t²=380 , t²=380÷4.9 , t²=77.55 now sq root on b/s

\sqrt{t }  =  \sqrt{77.55}

so t=8.806s and now apply 1st eq o²f motion to find out vf so vf=vi+gt , vf=0+9.8×8.806 ,vf=86.298 and if you want to verify that either this is answer is correct or not so put the value of t in second eq of motion and if you got distance same as give in the question so your value of t is considered as correct likewise s=vit+1/2gt² , s=0+1/2×9.8(8.806)²,s=4.9×77.55 ,s=380m (proved) I hope it would be helpfull

3 0
3 years ago
An electron and a proton are fixed at a separation distance of 973 nm. Find the magnitude and the direction of the electric fiel
Vilka [71]

Answer:

The magnitude is: |E|=6084.1\: N/C

The direction of E is in the negative x-direction.

Explanation:

The electric field equation is:

E=k\frac{Q}{r^{2}}

Where:

  • Q is the charge (we can choose the electron or the proton)
  • r is the distance (in our case is at the midpoint 973/2 nm)
  • k is the Coulomb constant (9*10^{9}\: Nm^{2}C^{-2})

Using the electron charge (e = -1.6*10^{-19}\: C)

E=-9*10^{9}\frac{1.6*10^{-19}}{(486.5*10^{-9})^{2}}

The magnitude is:

|E|=6084.1\: N/C

The direction of E is in the negative x-direction.

I hope it helps you!

6 0
2 years ago
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