What is the name of the ionic compound Cs2S

A man on the moon with a mass of 90 kilograms weighs 146 newtons. The radius of the moon is 1.74 x 106 meters, find the mass of

melisa1 [442]

Answer:

7.36 × 10^22 kg

Explanation:

Mass of the man = 90kg

Weight on the moon = 146N

radius of the moon =1.74×10^6

Weight =mg

g= weight/mass

g= 146/90 = 1.62m/s^2

From the law of gravitational force

g = GM/r^2

Where G = 6.67 ×10^-11

M = gr^2/G

M= 1.62 × (1.74×10^6)^2/6.67×10^-11

= 4.904×10^12/6.67×10^-11

=0.735×10^23

M= 7.35×10^22kg. (approximately) with option c

3 0

3 years ago

Read 2 more answers

Calcula la resistencia atravesada por una corriente con una intensidad de

Reil [10]

Hhhgcfghhhhhhhhhhhhhgh

5 0

3 years ago

One of the signs a chemical reaction is taking place

ladessa [460]

Answer: One of the signs a chemical reaction is taking place is the presence of bubbles. :)

6 0

3 years ago

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve

Gemiola [76]

Answer:

(a) v = 5.42m/s

(b) vo = 4.64m/s

(c) a = 2874.28m/s^2

(d) Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

$v=\sqrt{2gh}$ (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

$v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}$

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

$h_{max}=\frac{v_o^2}{2g}$ (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

$v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}$

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}$

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}$

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

$\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m$

THe compression of the ball when it strikes the floor is 5.11*10^-3m

4 0

3 years ago

Could someone please explain to me Archimede's principle without using the term displace (or at least explaining what it means)?

disa [49]

Does this help?

When an object is immersed in a fluid (in this case water, but may include both liquids and gases) the fluid exerts an upward force on the object which is called buoyancy force or up-thrust. Archimedes’ Principle states that the buoyant force (upward push or force) applied to an object is equal to the weight of the fluid that the object takes the space of by that object. Thus when an object is placed in water the rise in the water level is dictated by the mass of that object.

So for example if you fill a bucket with water and you drop a stone in that bucket, if you measure the weight of the water that overflows from the bucket due to the stone being dropped into the bucket is equivalent to the pushing force that the water has on the stone (as the stone drops to the bottom of the bucket the water is pushing it to stay afloat but the rock is more dense than water and as such its downthrust exceeds water's upthrust).

4 0

3 years ago