Answer:
ΔP = (640 N/cm^2)
Explanation:
Given:-
- The volume increase, ΔV/V0 = 4 ✕ 10^-3
- The Bulk Modulus, B = 1.6*10^9 N/m^2
Find:-
Calculate the force exerted by the moonshine per square centimeter
Solution:-
- The bulk modulus B of a material is dependent on change in pressure or Force per unit area and change in volume by the following relationship.
B = ΔP / [(ΔV/V)]
- Now rearrange the above relation and solve for ΔP or force per unit area.
ΔP = B* [(ΔV/V)]
- Plug in the values:
ΔP = (1.6*10^9)*(4 ✕ 10^-3)
ΔP = 6400000 N/m^2
- For unit conversion from N/m^2 to N/cm^2 we have:
ΔP = (6400000 N/m^2) cm^2 / (100)^2 m^2
ΔP = (640 N/cm^2)
Newton's motion laws state that if an object is at rest or in movement, it will tend to maintain its basal state.
<h3>What are Newton's motion laws?</h3>
Newton's motion laws are a set of scientific statements aimed at explaining the physical property of movement.
These laws explain why objects in movement tend to maintain the same velocity for a short period of time.
In conclusion, Newton's motion laws state that if an object is at rest or in movement, it will tend to maintain its basal state.
Learn more about Newton's motion laws here:
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Answer:
El avión recorrió 45 km en los 180 s.
Explanation:
La relación entre velocidad, distancia y tiempo se da de la siguiente manera;
Por lo cual los parámetros dados son los siguientes;
Velocidad = 900 km/h = 250 m / s
Tiempo = 180 s
Estamos obligados a calcular la distancia recorrida
De la ecuación para la velocidad dada arriba, tenemos;
Distancia recorrida = Velocidad pf viaje × Tiempo de viaje
Distancia recorrida = 900 km/h × 180 s = 900
Distancia recorrida = 900 km/h × 1 h/60 min × 1 min/60 s × 180 s = 45 km
Por lo tanto, el avión viajó 45 km en 180 s.
Answer:
Time period of the osculation will be 2.1371 sec
Explanation:
We have given mass m = (B+25)
And the spring is stretched by (8.5 A )
Here A = 13 and B = 427
So mass m = 427+25 = 452 gram = 0.452 kg
Spring stretched x= 8.5×13 = 110.5 cm
As there is additional streching of spring by 3 cm
So new x = 110.5+3 = 113.5 = 1.135 m
Now we know that force is given by F = mg
And we also know that F = Kx
So 
Now we know that 
So 
Answer:
In general, the annual sea surface temperatures(SSTs) in the Bay of Bengal(BOB) are higher than the Arabian sea(AS). because, there are two main reasons for higher SST in the Bay of Bengal than the Arabian Sea. they are 1. stratification, 2.strong mixing
stratification is nothing but a phenomenon which stratifies(layers) the sea water when different density water(fresh water, rain water) add into the sea water. So the stratification in the bay of Bengal is comparatively high than the Arabian sea due to the high river discharge and precipitation in the BOB than the AS. the mixing process over the Arabian sea is higher than the Bay of Bengal due to the prevailing of strong winds over the AS (because of the presence of the mountains of east Africa) than Bay of Bengal (because of the winds over the BOB are sluggish in nature then the AS). But generally winds over the sea mixes easily the normal sea water than stratified/stabilized sea water column. That's why less mixing will takes place over the surface of BOB than the AS. So due to the presence of less mixing over the surface of the Bay of Bengal than the Arabian sea, the SST values over the Arabian sea are always lower than the Bay of Bengal. that's why the Arabian sea is colder than the Bay of Bengal.
Explanation:
