Answer:
1.991 × 10^(8) N/m²
Explanation:
We are told that its volume increases by 9.05%.
Thus; (ΔV/V_o) = 9.05% = 0.0905
To find the force per unit area which is also pressure, we will use bulk modulus formula;
B = Δp(V_o/ΔV)
Making Δp the subject gives;
Δp = B(ΔV/V_o)
Now, B is bulk modulus of water with a value of 2.2 × 10^(9) N/m²
Thus;
Δp = 2.2 × 10^(9)[0.0905]
Δp = 1.991 × 10^(8) N/m²
Answer:
163.8 ft
Explanation:
In triangle ABD
= 155 ft


Using Pythagorean theorem in triangle ADC

= distance between the anchor points
distance between the anchor points is given as

Answer:
Explanation:
From newton's equation of motion of uniform acceleration
v = u + at
where v is final velocity , u is initial velocity , a is acceleration and time is t .
putting the values
v = 0 + .5 x 3 x 60 ( time in second = 3 x 60 s )
= 90 m /s
So , final velocity is 90 m /s .
When you are on a huge water slide, the force present as you slide is the gravitational force. It is because the gravity enables you to slide down the water slide. The net force is the overall forces of the object, so as you slide the water slide, you may experience the net force once you slide down with the gravity and water sliding you down.
Answer:
Distance, d = 778.05 m
Explanation:
Given that,
Force acting on the car, F = 981 N
Mass of the car, m = 1550 kg
Initial speed of the car, v = 25 mi/h = 11.17 m/s
We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

Let d is the distance covered by car. Using second equation of motion as :

So, the car will cover a distance of 778.05 meters.