Which of the following statements is true about the variation of pressure in function of the depth? O Pressure decreases exponentially with the depth O Pressure increases exponentially with the depth O Pressure decreases linearly with the depth o Pressure increases linearly with the depth O None of the above
Answers
Answer:
The answer is: Pressure increases linearly with the depth
Explanation:
In this case, the definition of pressure is:
where F = mg is the weight of the fluid over the body, and A is the area of the surface to which the force is exerted. If we consider , then
.
Volume can be expressed as V = A*h, where A is the cross section of the column of the fluid over the body and h is the height of the column, in other words, the depth.
So
,
which means that pressure increases linearly with the depth in a factor of .
Related Questions
A rocket exhausts fuel with a velocity of 1500m/s, relative to the rocket. It starts from rest in outer space with fuel comprising 80 per cent of the total mass. When all the fuel has been exhausted its speed is:________
The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge q? ( k = 1/4πε 0 = 8.99 × 10 9 N ∙ m 2/C 2)
An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 × 10-6 C-1.
An object of mass m moves to the right with a speed v. It collides head-on with an object of mass 3m moving with speed v/3 in the opposite direction. If the two objects stick together, what is the speed of the combined object, of mass 4m, after the collision
Answers
Answer:
a) Revolutions per minute = 2.33
b) Centripetal acceleration = 11649.44 m/s²
Explanation:
a) Angular velocity is the ratio of linear velocity and radius.
Here linear velocity = 72 m/s
Radius, r = 0.89 x 0. 5 = 0.445 m
Angular velocity
Frequency
Revolutions per minute = 2.33
b) Centripetal acceleration
Here linear velocity = 72 m/s
Radius, r = 0.445 m
Substituting
Centripetal acceleration = 11649.44m/s²
Answers
Answer:
A. The wires exert equal magnitude attractive forces on each other.
Explanation:
Magnetic field due to current i on current 2i
B₁ = 10⁻⁷ x 2 i / r where r is distance between the two wires
Force on wire II due to wire I per unit length
= magnetic field x current in wire II
= B₁ x 2 i
= [ 10⁻⁷ x 2 i / r ] x 2i
= 4 x 10⁻⁷ i² / r
Magnetic field due to current 2i on current i
B₂ = 10⁻⁷ x 4 i / r where r is distance between the two wires
Force on wire I due to wire II per unit length
= magnetic field x current in wire I
= B₂ x i
= [ 10⁻⁷ x 4 i / r ] x i
= 4 x 10⁻⁷ i² / r
So final forces on each wire are same .
This force will be attractive in nature . The direction of force can be known from fleming's right hand rule .
Answers
Answer:
elative magnitude of the two forces is the same and they are applied in a constant direction.
Explanation:
Newton's second law states that the sum of the forces is equal to the mass times the acceleration
∑ F = m a
in this case there are two forces on the x axis
F_applied - fr = 0
since they indicate that the velocity is constant, consequently
F_applied = fr
the relative magnitude of the two forces is the same and they are applied in a constant direction.
A high powered projectile is fired horizontally from the top of a cliff at a speed of 638.6 m/s. Determine the magnitude of the velocity (in m/s) after 5 seconds.
Take gravitational acceleration to be 9.81 m/s2.
2-
A man throws a ball with a velocity of 20.9 m/s upwards at 33.2° to the horizontal. At what vertical distance above the release height (in metres) will the ball strike a wall 13.0 m away ?
Take gravitational acceleration to be 9.81 m/s2.
3-
A particle is moving along a straight path and its position is defined by the equation s = (1t3 + -5t2 + 3) m, where t is measured in seconds. Determine the average velocity (in m/s) of the particle when t = 5 seconds.
4-
A particle has an initial speed of 26 m/s. The particle undergoes a deceleration of a = (-9t) m/s2, where t is measured in seconds. Determine the distance (in metres) the particle travels before it stops. When t = 0, s = 0.
Answers
Answer:
1.V= 640.48 m/s :total velocity in t= 5s
2. Y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away
3. v =25m/s
4. s= (-1.5t³+26t ) m
Explanation:
1. Parabolic movement in the x-y plane , t=5s
V₀=638.6 m/s=Vx :Constant velocity in x
Vy=V₀y +gt= 0+9.8*5 = 49 m/s : variable velocity in y
V= 640.48 m/s : total velocity in t= 5s
2.
x=v₀x*t
13=v₀x*t
13=17.49*t
t=13/17.49=0.743s : time for 13.0 m away
th=v₀y/g=11.44/9.8= 1,17s :time for maximum height
at t=0.743 sthe ball is going up ,then g is negative
y=v₀y*t - 1/2 *g¨*t²
y=11.44*0.743 -1/2*9.8*0.743²
y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away
3. s = (1t3 + -5t2 + 3) m
v=3t²-10t=3*25-50=75-50=25m/s
at t=0, s=3 m
at t=5s s=5³-5*5²+3
4. a = (-9t) m/s2
a=dv/dt=-9t
dv=-9tdt
v=∫ -9tdt
v=-9t²/2 + C1 equation (1)
in t=0 , v₀=26m/s ,in the equation (1) C1= 26
v=-9t²/2 + 26=ds/dt
ds=( -9t²/2 + 26)dt
s= ∫( -9t²/2 + 26)dt
s= -9t³/6+26t+C2 Equation 2
t = 0, s = 0 , C2=0
s= (-9t³/6+26t ) m
s= (-1.5t³+26t ) m
Answers
Explanation:
It is given that,
Mass of the passenger, m = 75 kg
Acceleration of the rocket,
(a) The horizontal component of the force the seat exerts against his body is given by using Newton's second law of motion as :
F = m a
F = 3675 N
Ratio,
So, the ratio between the horizontal force and the weight is 5 : 1.
(b) The magnitude of total force the seat exerts against his body is F' i.e.
F' = 3747.7 N
The direction of force is calculated as :
Hence, this is the required solution.
Final answer:
The horizontal component of the force the seat exerts against the passenger's body is 3675 N. The ratio of this force to the passenger's weight is 5. The total force the seat exerts has a magnitude of 3793 N.
Explanation:
(a) To calculate the horizontal component of the force the seat exerts against the passenger's body, we can use Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the passenger is 75.0 kg and the acceleration of the rocket sled is 49.0 m/s2. So the force exerted by the seat is:
Force = mass * acceleration
Force = 75.0 kg * 49.0 m/s2
Force = 3675 N
Now let's compare this force to the passenger's weight. The weight of an object is given by the formula:
Weight = mass * gravitational acceleration
Weight = 75.0 kg * 9.8 m/s2
Weight = 735 N
To find the ratio, we divide the force exerted by the seat by the weight of the passenger:
Ratio = Force / Weight
Ratio = 3675 N / 735 N
Ratio = 5
(b) The total force the seat exerts against the passenger's body has both a horizontal and vertical component. The direction of the total force is the same as the direction of the acceleration of the rocket sled. The magnitude of the total force can be found using the Pythagorean theorem:
Total Force = √(horizontal component2 + vertical component2)
Total Force = √(36752 + 7352)
Total Force = 3793 N
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b. energy
c. amplitude
d. velocity
Answers
Answer:wrong. Jts not velocity. Its period.
Explanation:
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