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3 years ago

What is the mass of an atom that has 2 protons, 3 neutrons, and 2 electrons

1 answer:

3 0

You can find the mass of an atom by adding the number of protons and neutrons. In this case the atom has 2 protons and 3 neutrons so the mass is 5.

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Zak, helping his mother rearrange the furniture in their living room, moves a 60 kg sofa 5.9 m with a constant force of 40 N. Wh

Len [333]

Answer:236 J

Explanation:

Given

Mass of sofa $m=60 kg$

Displacement $s=5.9 m$

Force $F=40 N$

neglecting Friction

Work done is given by

$W=F\cdot s$

$W=F\cdot s\cos 0$

$W=40\cdot 5.9$

$W=236 J$

4 0

3 years ago

Read 2 more answers

1. The string shown is 1.5 meters long and is vibrating as the first harmonic. The string vibrates up and down with 33 cycles in

svet-max [94.6K]

Answer:

Explanation:

6 0

3 years ago

In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra

Arturiano [62]

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

$m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}$

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

$v=\sqrt{\frac{GM}{R+h}}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$

$M=1.40\cdot 10^{16}kg$

$R=8.20 km=8200 m$

$h=6.00 km = 6000 m$

Substituting into the formula,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=8.11 m/s$

B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$

$M=1.40\cdot 10^{16}kg$

$R=8.20 km=8200 m$

$h=6.00 km = 6000 m$

Substituting into the formula, we find:

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=11.47 m/s$

5 0

3 years ago

What is the speed of an object that travels 20 meters in 4 seconds?

Iteru [2.4K]

Given:-

Distance covered by the particle (s) = 20 m
Time taken (t) = 4 seconds

To Find: Speed of it.

We know,

v = s/t

where,

v = Speed,
s = Distance &
t = Time taken.

Putting the values,

v = (20 m)/(4 s)

→ v = 5 m/s (Ans.)

5 0

3 years ago

0.180-kg stone rets on a frictionless, horizontal surface. A bullet of mass 7.50 g, traveling horizontally at 390 m/s, strikes t

Musya8 [376]

Answer:

The magnitud of the velocity is

$8.46m/s$

and the direccion:

$-28.3$ degrees from the horizontal.

Explanation:

Fist we define our variables:

$m_{s}=0.18kg\\m_{b}=7.5g = 0.0075kg\\v_{1b}= 390m/s -i\\v_{1s}=0m/s\\v_{2b}=210m/s -j\\v_{2s}=?$

The letters i and j represent the direction of the movement, i in this case is the horizontal direction, and j is perpendicular to i.

velocities with sub-index 1 are the speeds before the crash, and with sub-index 2 are the velocities after the crash.

Using conservation of momentum:

$m_{s}v_{1s}+m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}\\v_{1s}=0, so\\m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}$

Clearing for the velocity of the stone after the crash:

$v_{2s}=\frac{m_{b}v_{1b}-m_{b}v_{2b}}{m_{s}}$

Substituting known values:

$v_{2s}=\frac{0.0075kg(((390m/s-i)-210m/s-j)}{0.18kg}\\v_{2s}=(16.25m/s-i) - (8.75m/s-j)$

The magnitud of the velocity is :

$|v_{2s}|=\sqrt{(16.25m/s-i)^2 + (8.75m/s-j)^2}\\|v_{2s}|=18.46m/s$

and the direction:

$tan^{-1}(-8.75/16.25)=-28.3$

this is -28.3 degrees from the +i direction or the horizontal direcction.

Note: i and j can also be seen as x and y axis.

3 0

4 years ago