Answer:
0.375
Explanation:
For incompressible flow, we know that;
ρ1•v1•A1 = ρ2•v2•A2
Where;
ρ1 = density of fluid at position A
v1 = speed of fluid at position A
A1 = area of tube
ρ2 = density of fluid at position B
v2 = speed of fluid at position B
A2 = area of tube
We want to find ratio of the density of the fluid at position A to the density of the fluid at position B.
Thus;
ρ1/ρ2 = (v2•A2)/(v1•A1)
Now, the tube will have the same height.
But we are given;
diameter of A = 12.00 cm = 0.12 m
diameter of B = 6 cm = 0.06 m
Thus;
A1 = π(d²/4)h = πh(0.12²/4)
A2 = πh(0.06²/4)
We are also given;
v1 = 12 m/s
v2 = 18 m/s
Thus;
ρ1/ρ2 = (18 × πh(0.06²/4))/(12 × πh(0.12²/4))
πh/4 will cancel out to give;
ρ1/ρ2 = (18 × 0.06²)/(12 × 0.12²)
ρ1/ρ2 = 0.375
1
Select the correct answer.
Which type of energy is thermal energy a form of?
A
chemical energy
B.
kinetic energy
C. magnetic energy
D. potential energy
Reset
Next
Answer:
B. kinetic energy
Explanation:
Thermal energy (It’s a low form of energy ) is a form of kinetic energy as it is produced as a result of motion of particles either if they vibrate at their position or they move along longer paths.
Carl works hard to get a grades on his report card because his mother pays him 25 dollars for each semester he earns straight as Carl’s behavior is being influenced by
. Assume that the batter does hit the ball. If the bat's instantaneous angular velocity is 30 rad/s at the instant of contact, and the distance from the sweet spot on the bat to the axis of rotation is 1.25 m, what is the instantaneous linear velocity of the sweet spot at the instant of ball contact
Answer:
37.5 m/s
Explanation:
Using,
Formula
v = ωr....................... Equation 1
Where ω = instantaneous angular velocity, v = instantaneous linear velocity, r = radius or distance from the sweet spot of the bat to the axis of rotation.
From the question,
Given: ω = 30 rad/s, r = 1.25 m
Substitute these values into equation 1
v = 30(1.25)
v = 37.5 m/s.
Hence the instantaneous linear velocity of the sweet spot at the instant of ball contact is 37.5 m/s
Highest density of electrostatic charges in a metal is found where
I don't know the answer but I just want points sorry
prove that d1=R(d1-d2) in relative density
During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.145 kg baseball crashing through the pane of a second-floor window in a nearby building. The ball strikes the glass at 14.5 m/s , shatters the glass as it passes through, and leaves the window at 10.9 m/s with no change of direction. What is the direction of the impulse that the glass imparts to the baseball
Answer:
J = -0.522 m/s
Explanation:
Given that,
The mass of the baseball, m = 0.145 kg
Initial velocity, u = 14.5 m/s
Final velocity, v = 10.9 m/s
aWe need to find the direction of the impulse that the glass imparts to the baseball. Impulse is equal to the change in momentum such that,
[tex]J=m(v-u)[/tex]
Substitute all the values,
[tex]J=0.145\times (10.9-14.5)\\\\=-0.522\ kg-m/s[/tex]
The direction of impulse is opposite to the direction of velocity.
Mechanical energy is the most concentrated form of energy.
a. true
b. false
a pendulum clock having Copper keeps time at 20 degree Celsius it gains 15 second per day if cooled to 0°C celsius calculate the coefficient of linear expansion of copper.
?.............................
Fig 1 shows a pendulum of length L = 1.0 m. Its ball has speed of vo=2.0
m/s when the cord makes an angle of 30 degrees with the vertical. What
is the speed (V) of the ball when it passes the lowest position?
Answer:
v = 2.57 m / s
Explanation:
For this exercise let's use conservation of energy
starting point. When it is at an angle of 30º
Em₀ = K + U = ½ m v₁² + m g y₁
final point. Lowest position
Em_f = K = ½ m v²
as there is no friction, the energy is conserved
Em₀ = Em_f
½ m v₁² + m g y₁ = ½ m v²
Let's find the height(y₁), which is the length of the thread minus the projection (L ') of the 30º angle
cos 30 = L ’/ L
L ’= L cos 30
y₁ = L -L '
y₁ = L- L cos 30
we substitute
½ m v₁² + m g L (1- cos 30) = ½ m v²
v = [tex]\sqrt{ v_1^2 +2gL(1-cos30 )}[/tex]
let's calculate
v = [tex]\sqrt{ 2^2 + 2 \ 9.8 \ 1.0 (1- cos 30)}[/tex]
v = 2.57 m / s
At the base of a hill, a 90 kg cart drives at 13 m/s toward it then lifts off the accelerator pedal). If the cart just barely makes it to the top of this hill and stops, how high must the hill be?
Answer:
8.45 m
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 90 Kg
Initial velocity (u) = 13 m/s
Final velocity (v) = 0 m/s
Height (h) =?
NOTE: Acceleration due to gravity (g) = 10 m/s²
The height of the hill can be obtained as follow:
v² = u² – 2gh (since the cart is going against gravity)
0² = 13² – (2 × 10 × h)
0 = 169 – 20h
Rearrange
20h = 169
Divide both side by 20
h = 169/20
h = 8.45 m
Therefore, the height of the hill is 8.45 m
it takes 560s for a runner to complete one circular lap, moving at a speed of 6.00 m/s. what is the radius of a track?
Answer:
534.8 meters
Explanation:
Use T=(2*pi*r)/v
560=(2*pi*r)/6
3360=2*pi*r
1680=pi*r
534.8 meters=radius
It takes 560s for a runner to complete one circular lap, moving at a speed of 6.00 m/s. The radius of a track is 534.7 m.
What is Distance?The distance covered by a body is equal to the sum of total path covered. It is equal to the total path traveled by an object during its entire journey.This quantity is always positive. It can't be 0 or a negative number.It is defined as a scalar quantity.
Mathematically, it can be calculated as follows :
distance = speed × time
The formula relating distance (d), speed (s), and time (t) is
d = st
First, Calculating the distance,
d = 560 s × 6 m·s⁻¹
= 3360 m
When, Calculating the track radius,
The distance travelled is the circumference of a circle,
C = 2пr
r = 3360/2п
= 534.7 m
The radius of the track is 534.7 m.
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(1) Which appliance is designed to transfer electrical energy to kinetic energy?
D)
A food mbuer
BB kettle
Clamp
D radio
Answer:
bb kettle
Explanation:
it transfres electricsl to kinetic
What is the only difference between the reactant and product side of a chemical reaction?
Answer:
Products is the result. Reactants produce the result
Explanation:
,,
What do you think will happen to the Lunar phases if the moon was hit by an asteroid?
1. A perspex box has a 10 cm square base and contains water to a height of 10 cm. A piece of rock of mass 600g is lowered into the water and the level rises to 12 cm.
(a) What is the volume of water displaced by the rock?
(b) What is the volume of the rock?
(c) Calculate the density of the rock
Answer:
(a) The volume of water is 100 cm³
(b) The volume of the rock is 20 cm³
(c) The density of the rock is 30 g/cm³
Explanation:
The given parameters of the perspex box are;
The area of the base of the box, A = 10 cm²
The initial level of water in the box, h₁ = 10 cm
The mass of the rock placed in the box, m = 600 g
The final level of water in the box, h₂ = 12 cm
(a) The volume of water in the box, 'V', is given as follows;
V = A × h₁
∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³
The volume of water in the box, V = 100 cm³
(b) When the rock is placed in the box the total volume, [tex]V_T[/tex], is given by the sum of the rock, [tex]V_r[/tex], and the water, V, is given as follows;
[tex]V_T[/tex] = [tex]V_r[/tex] + V
[tex]V_T[/tex] = A × h₂
∴ [tex]V_T[/tex] = 10 cm² × 12 cm = 120 cm³
The total volume, [tex]V_T[/tex] = 120 cm³
The volume of the rock, [tex]V_r[/tex] = [tex]V_T[/tex] - V
∴ [tex]V_r[/tex] = 120 cm³ - 100 cm³ = 20 cm³
The volume of the rock, [tex]V_r[/tex] = 20 cm³
(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)
∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³
A 38.0 kg box initially at rest is pushed 4.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N. If the coefficient of friction between box and floor is 0.300, find the following. (a) the work done by the applied force J (b) the increase in internal energy in the box-floor system due to friction J (c) the work done by the normal force J (d) the work done by the gravitational force J (e) the change in kinetic energy of the box J (f) the final speed of the box m/s
Answer:
a) Wapp = 520 N
b) ΔUf = 447 N
c) Wn = 0
d) Wg = 0
e) ΔK = 73 J
f) v = 1.96 m/s
Explanation:
a)
Applying the definition of work, as the dot product between the applied force and the displacement, since both are parallel each other, the work done on the box by the applied force can be written as follows:[tex]W_{app} = F_{app} * \Delta X = 130 N * 4.0 m = 520 N (1)[/tex]
b)
The change in the internal energy due to the friction, is numerically equal to the work done by the force of friction.This work is just the product of the kinetic force of friction, times the displacement, times the cosine of the angle between them.As the friction force opposes to the displacement, we can find the work done by this force as follows:[tex]W_{ffr} = F_{fr} * \Delta X * cos (180) (2)[/tex]
The kinetic force of friction, can be expressed as the product of the kinetic coefficient of friction times the normal force.If the surface is level, this normal force is equal to the weight of the object, so we can write (2), as follows:[tex]W_{ffr} = F_{fr} * \Delta X * cos (180) = -\mu_{k} * m* g* \Delta X = \\ -0.300*38.0kg9.8 m/s2*4.0m = -447 J (3)[/tex]
So, the increase in the internal energy in the box-floor system due to the friction, is -Wffr = 447 Jc)
Since the normal force (by definition) is normal to the surface, and the displacement is parallel to the surface, no work is done by the normal force.d)
Since the surface is level, the displacement is parallel to it, and the gravitational force is always downward, we conclude that no work is done by the gravitational force either.e)
The work-energy theorem says that the net work done on the object, must be equal to the change in kinetic energy.We have two forces causing net work, the applied force, and the friction force.So the change in kinetic energy must be equal to the sum of the work done by both forces, that we found in a) and b).So, we can write the following expression:[tex]\Delta K = W_{app} + W_{ffr} = 520 J - 447 J = 73 J (4)[/tex]
f)
Since the object starts at rest, the change in kinetic energy that we got in e) is just the value of the final kinetic energy.So, replacing in (4), we finally get:[tex]\Delta K = 73 J = \frac{1}{2}*m*v^{2} (5)[/tex]
Solving for v:[tex]v_{f} = \sqrt{\frac{2*\Delta K}{m} } = \sqrt{\frac{2*73J}{38.0kg}} = 1.96 m/s (6)[/tex]a) The work done by the applied force [tex]W_{AP}=520\ J[/tex]
b) The change in the internal energy [tex]\Delta U=447\ J[/tex]
c) Work done by normal force [tex]W_n=0[/tex]
d) Work done by gravitation [tex]W_g=0[/tex]
e) The change in KE will be [tex]\Delta KE=73\ J[/tex]
f) The final speed v = 1.96 m/s
What will be the work done?The work done on any object can be defined as the force applied on the object and its displacement due the effect of the force.
If the object achieve movement due to the work then the energy in the object will be kinetic energy.
If the object attains some height against the gravity then the energy in the object will be potential energy.
Now it is given in the question that
The horizontal force [tex]F_h=130\N[/tex]
mass of the object m= 38 kg
Coefficient of friction [tex]\mu=0.3[/tex]
Displacement of the object [tex]\delta x=4\ m[/tex]
(a) The work done will be
[tex]W=F_h\tines \Delta x[/tex]
[tex]W=130\times 4=520\ J[/tex]
(b) The increase in the internal energy
The increase in the internal energy of the box is due to the energy generated by the force of friction so
[tex]W_f=F_f\times \Delta x\times Cos(180)[/tex]
here [tex]F_f[/tex] is the frictional force and is given as
[tex]\mu=\dfrac{F_f}{R}[/tex]
Here R is the normal reaction and its value will be weight of the box in opposite direction.
[tex]\mu=\dfrac{F_f}{-mg}[/tex]
[tex]F_f=-mg\times \mu[/tex]
[tex]W_f=F_f\times \Delta x\times Cos180=-mg\times\mu \times cos180[/tex]
[tex]W_f=-38\times 9.81\times 0.3\times4=-447\J\ J[/tex]
(c) The work done by the normal force will be zero because the displacement is horizontal against the normal work so the work done will be zero.
(d) The work done by the gravitational force will also be zero. Because the displacement is horizontal and the gravitational force acts downward.
(e) The change in the KE of the box.
The change in the KE of the box will be the net energy of the box so from the work energy theorem the net energy will be
[tex]\Delta KE =W_{AP}-W_f=520-447=73\ J[/tex]
(f) The speed of the box
The KE of the box will be
[tex]KE=\dfrac{1}{2} mv^2[/tex]
[tex]v=\sqrt{ \dfrac{2\times KE}{m}[/tex]
[tex]v=\sqrt{\dfrac{2\times73}{38} }=1.96\ \dfrac{m}{s}[/tex]
Thus
a) The work done by the applied force [tex]W_{AP}=520\ J[/tex]
b) The change in the internal energy [tex]\Delta U=447\ J[/tex]
c) Work done by normal force [tex]W_n=0[/tex]
d) Work done by gravitation [tex]W_g=0[/tex]
e) The change in KE will be [tex]\Delta KE=73\ J[/tex]
f) The final speed v = 1.96 m/s
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A planet of mass M has a moon of mass m in a circular orbit of radius R. An object is placed between the planet and the moon on the line joining the center of the planet to the center of the moon so that the net gravitational force on the object is zero. How far is the object placed from the center of the planet
Answer:
r =[tex]\frac{ 1 \pm \sqrt{ \frac{m}{M} } }{1 - \frac{m}{M} }[/tex]
Explanation:
Let's apply the universal gravitation law to the body (c), we use the indications 1 for the planet and 2 for the moon
∑ F = 0
-F_{1c} + F_{2c} = 0
F_{1c} = F_{2c}
let's write the force equations
[tex]G \frac{m_c M}{r^2} = G \frac{m_c m}{(d-r)^2}[/tex]
where d is the distance between the planet and the moon.
[tex]\frac{M}{r^2} = \frac{m}{(d-r)^2}[/tex]
(d-r)² = [tex]\frac{m}{M} \ \ r^2[/tex]
d² - 2rd + r² = \frac{m}{M} \ \ r^2
d² - 2rd + r² (1 - [tex]\frac{m}{M}[/tex]) = 0
(1 - [tex]\frac{m}{M}[/tex]) r² - 2d r + d² = 0
we solve the second degree equation
r = [2d ± [tex]\sqrt{ 4d^2 - 4 ( 1 - \frac{m}{M} ) }[/tex] ] / 2 (1- [tex]\frac{m}{M}[/tex])
r = [2d ± 2d [tex]\sqrt{ \frac{m}{M} }[/tex]] / 2d (1- [tex]\frac{m}{M}[/tex])
r =[tex]\frac{ 1 \pm \sqrt{ \frac{m}{M} } }{1 - \frac{m}{M} }[/tex]
there are two points for which the gravitational force is zero
The distance between object from planet will be "[tex]\frac{R}{[1+\sqrt{\frac{m}{M} } ]}[/tex]".
According to the question,
Let,
Object is "x" m from planet center = R - xGravitational force = 0Mass of object = m₁As we know,
→ [tex]Prerequisites-Gravitational \ force = \frac{GMm}{r^2}[/tex]
Now,
→ [tex]\frac{GMm_1}{x^2} = \frac{Gmm_1}{(R-x)^2}[/tex]
→ [tex]\frac{(R-x)^2}{x^2} = \frac{m}{M}[/tex]
→ [tex]\frac{R-x}{x} =\sqrt{\frac{m}{M} }[/tex]
→ [tex]x = \frac{R}{[1+ \sqrt{\frac{m}{M} } ]}[/tex]
Thus the answer above is appropriate.
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A ball on a string in uniform circular motion has a velocity of 8 meters per second, a mass of 2 kilograms, and the radius of the circle is 0.5 meters. What is the centripetal force keeping the ball in the circle?
Answer:
256 N
Explanation:
formula of centripetal force = mv²/r
m= 2kg
v= 8m/s
r= 0.5m
mv²/r = 2×8²/0.5 = 256N
३.रात में घूमने वाला write one word substitute
Explanation:
रात में घूमने वाला arthaarat निशाचर
A cylindrical body has 6 m height and its radius is 2 metre calculate its volume. Ans :75.428m3
Answer:
75.4
Explanation:
r= 2
h= 6
v= 22/7 *r*r*h
v= 75.42
why doesn't a radio operating with two batteries function when one of the batteries is reversed?
Answer:
If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. However, batteries aren't like that. The slightest difference in voltages mean that current will flow.
Explanation:
Which of the following is true for the entropy of the universe?
A It is always decreasing.
B It is always increasing.
C It is always negative.
D It is always a constant.
Answer:
B It is always increasing.
Explanation:
In Physics, entropy can be defined as the tendency or ability of a substance to reach maximum disorder i.e to be randomly distributed.
This ultimately implies that, entropy is a thermodynamic quantity that measures the degree of maximum disorder or randomness of a system.
The S.I unit used for the measurement of the degree of maximum order or randomness of a system is Joules per Kelvin (JK¯¹). An example of entropy is the mixing of ideal gases.
Generally, the entropy in an irreversible process always increases and as such the change in entropy has a positive value.
Hence, the entropy of the universe is always increasing because its energy flow is considered to be in a downward direction rather than upward i.e from a hot region to a cold region; making the energy to be evenly distributed.
How fast were both runners traveling after 4 seconds?
40
Distance (in yards)
30
20
10
1
2.
3
0
Time in seconds
Answer:
they were fast ⛷⛷
An astronaut has a mass of 75 kg and is floating in space 500 m from his 125,000 kg spacecraft. What will be the force of gravitational attraction between the two? Since there is no force opposing him, he will accelerate toward the ship. Find his acceleration.
Answer:
1. 2.5×10¯⁹ N
2. 3.33×10¯¹¹ m/s²
Explanation:
1. Determination of the force of attraction.
Mass of astronaut (M₁) = 75 Kg
Mass of spacecraft (M₂) = 125000 Kg
Distance apart (r) = 500 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Force of attraction (F) =?
The force of attraction between the astronaut and his spacecraft can be obtained as follow:
F = GM₁M₂ /r²
F = 6.67×10¯¹¹ × 75 × 125000 / 500²
F = 2.5×10¯⁹ N
Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N
2. Determination of the acceleration of the astronaut.
Mass of astronaut (m) = 75 Kg
Force (F) = 2.5×10¯⁹ N
Acceleration (a) of astronaut =?
The acceleration of the astronaut can be obtained as follow:
F = ma
2.5×10¯⁹ N = 75 × a
Divide both side by 75
a = 2.5×10¯⁹ / 75
a = 3.33×10¯¹¹ m/s²
Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²
According to Newton's first law, an object at rest will _____.
never move
stay at rest forever
start moving
stay at rest unless moved by force
Define Mechanical advantage
fe effort of 2125N is used to lift a Lead of 500N
through a Verticle high of 2.N using a buly System
if the distance Moved by the effort is 45m
Calculate 1. Work done on the load
2. work done by the effort
3. Efficiency of the System
Answer:
1) 1000Nm
2) 95,625Nm
3) 1.05%
Explanation:
Mechanical Advantage is the ratio of the load to the effort applied to an object.
MA = Load/Effort
1) Workdone on the load = Force(Load) * distance covered by the load
Workdone on the load = 500N * 2m
Workdone on the load = 1000Nm
2) work done by the effort = Effort * distance moves d by effort
work done by the effort = 2125 * 45
work done by the effort = 95,625Nm
3) Efficiency = Workdone on the load/ work done by the effort * 100
Efficiency = 1000/95625 * 100
Efficiency = 1.05%
Hence the efficiency of the system is 1.05%
How much work is done when 100 N of force is applied to a rock to move it 20 m
Answer: 2000 J
Explanation: work W = F s
A girl weighing 45kg is standing on the floor, exerting a downward force of 200N on the floor. The force exerted on her by the floor is ..............
Select one:
a.
No force exerted
b.
Less than 2000N
c.
Equal to 200 N
d.
Greater than 200 N
Answer:
c.
Equal to 200 N..........
On a 10 kg cart (shown below), the cart is brought up to speed with 50N of force for 7m, horizontally. At this point (A), the cart begins to experience an average frictional force of 15N throughout the ride.
Find:
a) The total energy at (A)
b) The velocity at (B)
c) The velocity at (C)
d) Can the cart make it to Point (D)? Why or why not?
A 41.0-kg crate, starting from rest, is pulled across level floor with a constant horizontal force of 135 N. For the first 15.0 m the floor is essentially frictionless, whereas for the next 12.0 m the coefficient of kinetic friction is 0.320. (a) Calculate the work done by all the forces acting on the crate, during the entire 27.0 m path. (b) Calculate the total work done by all the forces. (c) Calculate the final speed of the crate after being pulled these 27.0 m.
Answer:
Explanation:
From the information given;
mass of the crate m = 41 kg
constant horizontal force = 135 N
where;
[tex]s_1 = 15.0 \ m \\ \\ s_2 = 12.0 \ m[/tex]
coefficient of kinetic friction [tex]u_k[/tex] = 0.28
a)
To start with the work done by the applied force [tex](W_f)[/tex]
[tex]W_F = F\times (s_1 +s_2) \times cos(0) \ J[/tex]
[tex]W_F = 135 \times (12 +15) \times cos(0) \ J \\ \\ W_F = (135 \times 37 )J \\ \\ W_F =4995 \ J[/tex]
Work done by friction:
[tex]W_{ff} = -\mu\_k\times m \times g \times s_2 \\ \\ W_{ff} = -0.320 \times 41 \times 9.81 \times 12 \ J \\ \\ W_{ff} = -1544.49 \ J[/tex]
Work done by gravity:
[tex]W_g = mg \times (s_1+s_2) \times cos (90)} \ J \\ \\ W_g = 0 \ j[/tex]
Work done by normal force;
[tex]W_n = N \times (s_1 + s_2) \times cos (90) \ J[/tex]
[tex]W_n = 0 \ J[/tex]
b)
total work by all forces:
[tex]W = F \times (s_1 + s_2) + \mu_k \times m \times g \times s_2 \times 180 \\ \\ W = 135 \times (15+12) \ J - 0.320 \times 41 \times 9.81 \times 12[/tex]
W = 2100.5 J
c) By applying the work-energy theorem;
total work done = ΔK.E
[tex]W = \dfrac{1}{2}\times m \times (v^2 - u^2)[/tex]
[tex]2100.5 = 0.5 \times 41 \times v^2[/tex]
[tex]v^2 = \dfrac{2100.5}{ 0.5 \times 41 }[/tex]
[tex]v^2 = 102.46 \\ \\ v = \sqrt{102.46} \\ \\ \mathbf{v = 10.1 \ m/s}[/tex]
