Answer:
3.78 m/s
Explanation:
Recall that the formula for average speed is given by
Speed = Distance ÷ Time taken
Where,
Speed = we are asked to find this
Distance = given as 340m
Time taken = 1.5 min = 1.5 x 60 = 90 seconds
Substituting the values into the equation:
Speed = Distance ÷ Time taken
= 340 meters ÷ 90 seconds
= 3.777777 m/s
= 3.78 m/s (round to nearest hundredth)
Answer:
total number of electron in 1 litter is 3.34 ×
electron
Explanation:
given data
mass per mole = 18 g/mol
no of electron = 10
to find out
how many electron in 1 liter of water
solution
we know molecules per gram mole is 6.02 ×
molecules
no of moles is 1
so
total number of electron in water is = no of electron ×molecules per gram mole × no of moles
total number of electron in water is = 10 × 6.02 ×
× 1
total number of electron in water is = 6.02×
electron
and
we know
mass = density × volume ..........1
here we know density of water is 1000 kg/m
and volume = 1 litter = 1 ×
m³
mass of 1 litter = 1000 × 1 × 
mass = 1000 g
so
total number of electron in 1 litter = mass of 1 litter × 
total number of electron in 1 litter = 1000 × 
total number of electron in 1 litter is 3.34 ×
electron
The type of radiation that can penetrate through paper, but not through wood is called beta rays. Beta rays can penetrate paper and air, but a thin piece of alimony can stop it. Gamma can cut through anything except lead and many inches of concrete. Alpha can be stopped by paper and not penetrated. The correct answer is B.
Hi there!
The period of an orbit can be found by:

T = Period (? s)
r = radius of orbit (6400000 m)
v = speed of the satellite (8000 m/s)
This is the same as the distance = vt equation. The total distance traveled by the satellite is the circumference of its circular orbit.
Let's plug in what we know and solve.

Answer:0.0704 kg
Explanation:
Given
initial Absolute pressure
=210+101.325=311.325



as the volume remains constant therefore



therefore Gauge pressure is 337.44-101.325=236.117 KPa
Initial mass 

Final mass 

Therefore
=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back