A tank is full of water. Find the work (in J) required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1,000 kg/m3 as the density of water. Round your answer to the nearest whole number.)

Answer:

Explanation:

English alphabets numbered fro 1 to 26

and the numbers 1 to10 so they are written in roman numbers as

1 - I

2 - II

3 - III

4 - IV

5 -V

6 - VI

7 -VII

8 - VIII

9 - IX

10 -X

11 - XI

12 - XII

13 - XIII

14 - XIV

15 - XV

16 - XVI

17 - XVII

18 - XVIII

19 - XIX

20- XX

21 - XXI

22 - XXII

23 - XXIII

24 - XXIV

25 - XXV

26 - XXVI  

Answer:

2) when acceleration triples force triples,  3) a diagram with dynamic friction force in the opposite direction of movement of the car

4)  a = 2.44 ft / s², 5)  fr = 894.3 N

Explanation:

In this exercise you are asked to answer some short questions

2)  Newton's second law is

         F = m a

when acceleration triples force triples

3) Unfortunately, the diagrams are not shown, but the correct one is one where the axis of movement has a friction force in the opposite direction of movement, as well as indicating that the car slips, the friction coefficient of dynamic.

The correct answer is: a diagram with dynamic friction force in the opposite direction of movement of the car

4) let's use the scientific expressions

          v = v₀ - a t

as the car stops v = 0

          a = v₀ / t

let's reduce the magnitudes

          v₀ = 40 mile / h ([tex]\frac{5280 ft}{1 mile}[/tex]) ([tex]\frac{1 h}{3600 s}[/tex]) = 58.667 ft / s

          a = 58.667 / 24

          a = 2.44 ft / s²

5) let's use Newton's second law

           fr = m a

We must be careful not to mix the units, we will reduce the acceleration to the system Yes

           a = 2.44 ft / s² (1 m / 3.28 ft) = 0.745 m / s²

           fr = 1200  0.745

           fr = 894.3 N

Answer:

[tex]M_b=6kg[/tex]

Explanation:

From the question we are told that:

Coefficient of restitution [tex]\mu=1.00[/tex]

Mass A [tex]M_a=6kg[/tex]

Initial Velocity of A [tex]U_a=6m/s[/tex]

Initial Velocity of B [tex]U_b=0m/s[/tex]

Generally the equation for Coefficient of restitution is mathematically given by

 [tex]\mu=\frac{V_b-V_a}{U_a-U_b}[/tex]

 [tex]1=\frac{v_B}{6}[/tex]

 [tex]V_b=6*1[/tex]

 [tex]V_b=6m/s[/tex]

Generally the equation for conservation of linear momentum  is mathematically given by

 [tex]M_aU_a+M_bU_b=M_aV_a+M_bV_b[/tex]

 [tex]6*6+=M_b*6[/tex]

 [tex]M_b=6kg[/tex]

Answer:

Option (a), (d) are correct.

Explanation:

Frequency, f = 220 Hz

Resonant frequency = 660 Hz

The next frequency of stopped organ pipe is

2f, 3 f, 4 f ....

= 2 x 220 , 3 x 220 , 4 x 220 ....

= 440 Hz, 660 Hz, 880 Hz

So, the option (a) is correct.

The next resonant frequency of open organ pipe is

3 f, 5 f,...

= 3 x 220, 5 x 220 , ..

= 660 Hz, 1100 Hz,...

So, option (d) is correct.

Answer:

Radio waves are used to carry satellite signals. These waves travel at 300,000 km/s (the speed of light). This means that a signal sent to a satellite 38,000 km away takes 0.13 s to reach the satellite and another 0.13 s for the return signal to be received back on Earth.

Explanation:

hope it help

Answer:

The answer to this question is plasma

Answer:

Plasma

Explanation:

Answer:

3.1m

Explanation:

Since we only care about the y direction we only need to find vy. Once u draw your vector you will realize that vy= 4.5sin60=3.897m/s.

use vf²=v²+2a(y)

At the maximum height the velocity is 0 and since the object is in freefall, a=-g

Plug in all values

0=15.1875-2*9.8(y)

solve for y

-15.1875*2/-9.8=y

y=3.1m

Answer:

0.774m

Explanation:

The formula for maximum height is given by:

hmax = ∨₀² ₓ Sin (α)² / 2 × g

where;

∨₀ = initial velocity

Sin (α) = angle of launch

g = gravitational acceleration which is equal to 9.8m/s²

Plugging in our values, we will have:

hmax = (4.50m/s)² × (Sin 60.0)² / 2 × 9.8m/s²

hmax= 20.25m/s × 0.75 / 19.8m/s²

hmax = 15.1875 / 19.8

hmax = 0.774m

Explanation:

Conservation of energy, principle of physics according to which the energy of interacting bodies or particles in a closed system remains constant

hope it is helpful to you

Answer:

No, it will not affect the results.

Explanation:

For elastic collisions in an isolated system, when a collision occurs, it means that the systems objects total momentum will be conserved under the condition that there will be no net external forces that act upon the objects.

What that means is that if the pucks start spinning after the collision, we are not told that there was any net external force acting on the puck and thus momentum will be conserved because momentum before collision will be equal to the momentum after the collision.

Answer:

W = 4.75 KW

Explanation:

First, we will calculate the heat to be removed:

Q = (No. of students)(Metabolic Power of Each Student)

Q = (190)(125 W)

Q = 23750 W = 23.75 KW

Now the formula of COP is:

[tex]COP = \frac{Q}{W}\\\\W = \frac{Q}{COP}\\\\W = \frac{23.75\ KW}{5}\\\\[/tex]

W = 4.75 KW

Answer:

The tension in the second string is 226.7 N.

Explanation:

Length is L, mass per unit length = m

T = 510 N

Let the tension in the second string is T'.

second harmonic of the first string = third harmonic of the second string

[tex]2 f = 3 f'\\\\2\sqrt{\frac{T}{m}} = 3 \sqrt {\frac{T'}{m}}\\\\4 T = 9 T'\\\\4\times 510 = 9 T'\\\\T' = 226.7 N[/tex]

Answer:

[tex]{ \bf{F = { \tt{ \frac{4}{3} \pi {r}^{3}v gM}}}}[/tex]

Explanation:

Given

Ball is projected horizontally from a building of height [tex]h=19.6\ m[/tex]

time taken to reach ground is given by

[tex]\text{Cosidering vertical motion}\\\Rightarrow h=ut+0.5at^2\\\Rightarrow 19.6=0+0.5\times 9.8t^2\\\Rightarrow t^2=4\\\Rightarrow t=2\ s[/tex]

(b) Line joining the point of projection and the point where it hits the ground makes an angle of [tex]45^{\circ}[/tex]

From the figure, it can be written

[tex]\Rightarrow \tan 45^{\circ}=\dfrac{h}{x}\\\\\Rightarrow x=h\cdot 1\\\Rightarrow x=19.6[/tex]

Considering horizontal motion

[tex]\Rightarrow x=u_xt\\\Rightarrow 19.6=u_x\times 4\\\Rightarrow u_x=4.9\ m/s[/tex]

(c) The vertical velocity with which it strikes the ground is given by

[tex]\Rightarrow v^2-u_y^2=2as\\\Rightarrow v^2-0=2\times 9.8\times 19.6\\\Rightarrow v=\sqrt{384.16}\\\Rightarrow v=19.6\ m/s[/tex]

Thus, the ball strikes with a vertical velocity of [tex]19.6\ m/s[/tex]

Explanation:

Given

Ball is projected horizontally from a building of height  

time taken to reach ground is given by

(b) Line joining the point of projection and the point where it hits the ground makes an angle of  

From the figure, it can be written

Considering horizontal motion

(c) The vertical velocity with which it strikes the ground is given by

Thus, the ball strikes with a vertical velocity of

Answer:

x ’= 368.61 m,  y ’= 258.11 m

Explanation:

To solve this problem we must find the projections of the point on the new vectors of the rotated system  θ = 35º

            x’= R cos 35

            y’= R sin 35

The modulus vector can be found using the Pythagorean theorem

            R² = x² + y²

            R = 450 m

we calculate

            x ’= 450 cos 35

            x ’= 368.61 m

            y ’= 450 sin 35

            y ’= 258.11 m

Answer:

hope it helps

much as you can

Answer:

I could not find the answer or do it myself if I did find it I would defenetly share

Answer:

Option A. –5 m/s²

Explanation:

From the question given above, the following data were obtained:

Initial velocity (v₁) = 9 m/s

Initial time (t₁) = 0 s

Final velocity (v₂) = –6 m/s

Final time (t₂) = 3 s

Acceleration (a) =?

Next, we shall determine the change in the velocity and time. This can be obtained as follow:

For velocity:

Initial velocity (v₁) = 9 m/s

Final velocity (v₂) = –6 m/s

Change in velocity (Δv) =?

ΔV = v₂ – v₁

ΔV = –6 – 9

ΔV = –15 m/s

For time:

Initial time (t₁) = 0 s

Final time (t₂) = 3 s

Change in time (Δt) =?

Δt = t₂ – t₁

Δt = 3 – 0

Δt = 3 s

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Change in velocity (Δv) = –15 m/s

Change in time (Δt) = 3 s

Acceleration (a) =?

a = Δv / Δt

a = –15 / 3

a = –5 m/s²

Thus, the acceleration of the object is

–5 m/s².

Answer:

Its the Federal government

Answer:

1. 588 N

2. 738 N

3. 588 N

Explanation:

time, t = 4 s

initial velocity, u = 0

final velocity, v = 10 m/s

mass, m= 60 kg

1.

Weight of passenger before starts

W =m g = 60 x 9.8 = 588 N

2.

When the elevator is speeding up

v = u + a t

10 = 0 + a x 4

a = 2.5 m/s2

Now the weight is

W' = m (a + g) = 60 (9.8 + 2.5) = 738 N

3.

When he reaches the cruising speed, the weight is

W = 588 N

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

A:

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

[tex]A_a(t) = 5m/s^2[/tex]

To get the velocity, we integrate over time:

[tex]V_a(t) = (5m/s^2)*t + V_0[/tex]

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

[tex]V_a(t) = (5m/s^2)*t[/tex]

To get the position equation we integrate again over time:

[tex]P_a(t) = 0.5*(5m/s^2)*t^2 + P_0[/tex]

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

[tex]P_a(t) = 0.5*(5m/s^2)*t^2[/tex]

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

[tex]V_b(t) =20m/s[/tex]

To get the position equation, we can integrate:

[tex]P_b(t) = (20m/s)*t + P_0[/tex]

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

[tex]P_b(t) = (20m/s)*t + 100m[/tex]

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

[tex]P_a(t) = P_b(t)[/tex]

We can solve this for t.

[tex]0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0[/tex]

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

[tex]t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)} }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}[/tex]

We only care for the positive solution, which is:

[tex]t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s[/tex]

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

[tex]P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m[/tex]

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4)  What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

[tex]V_a(t) = (5m/s^2)*11.48s = 57.4 m/s[/tex]

Answer:

The mass and atomic numbers don't change

Explanation:

An excited atom relaxes to the ground state emitting a photon...called a gamma ray.

The answer is that the mass and atomic numbers don't change.

In gamma decay, the product refers to the nucleus resulting from the emission of a gamma ray. Gamma decay occurs when an excited atomic nucleus releases excess energy in the form of a high-energy photon called a gamma ray.

To find the product of gamma decay, you need to identify the nucleus before and after the decay process. The product nucleus is determined by the parent nucleus that undergoes gamma decay.

During gamma decay, the number of protons and neutrons in the nucleus remains unchanged. Therefore, the identity of the element remains the same, but the energy state of the nucleus is altered.

The product nucleus is typically represented by the same chemical symbol as the parent nucleus, followed by a superscript indicating the mass number (total number of protons and neutrons) and a subscript indicating the atomic number (number of protons).

For example, if a parent nucleus with an atomic number of Z and a mass number of A undergoes gamma decay, the product nucleus will have the same atomic number Z and mass number A.

It's important to note that gamma decay does not involve the emission or absorption of any particles, only the release of electromagnetic radiation (gamma ray).

Thus, the product nucleus remains unchanged in terms of atomic number and mass number.

Know more about gamma decay:

https://brainly.com/question/16039775

#SPJ4

Answer:

45 × 8 units = A + B as formular

If both forces are measured in Newtons, then the net force is

F = (18, 7, -12) N + (16, -10, 16) N = (34, -3, 4) N

The toolbox undergoes a displacement (i.e. change in position) in the direction of the vector

d = (17, 14, -1) m - (15, 14, -8) m = (2, 0, -9) m

The total work done by the astronauts on the toolbox is then

F • d = (34, -3, 4) N • (2, 0, -9) m = (68 + 0 - 36) N•m = 32 J

The work done by the two astronauts is equal to 96 J.

work done?Work done is defined as the product of force applied and the distance moved by the force.

The forces applied = 18+16 N, 7+ -10 N, and -12 + 16N

Forces = 34 N, -3 N, and 4N

Distances = (17 - 15, 14 - 14, -1 - - 8) m

Distances = 2, 0, 7

Work done = 34 × 2 + -3 × 0 + 4 × 7

Work done = 96 J

Therefore, the work done by the two astronauts is equal to 96 J.

Learn more about work done at: https://brainly.com/question/25573309

#SPJ6

Answer:

-1.0778×10⁻¹⁰ N/C

Explanation:

Applying,

E = kq/r²................ equation 1

Where E = elctric field, q = charge, r = distance, k = coulomb's law

From the question,

Given: q = -3.0×10 C, r = 5.0 m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values in equation 1

E = (-3.0×10)(8.98×10⁹)/5²

E = -1.0778×10⁻¹⁰ N/C

Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C

Answer:

The expected radius of the Earth is 3.883 meters.

Explanation:

The formula for the escape speed is derived from Principle of Energy Conservation and knowing that rocket is initially at rest on the surface of the Earth and final energy is entirely translational kinetic, that is:

[tex]U = K[/tex] (1)

Where:

[tex]U[/tex] - Gravitational potential energy, in joules.

[tex]K[/tex] - Translational kinetic energy, in joules.

Then, we expand the formula by definitions of potential and kinetic energy:

[tex]\frac{G\cdot M\cdot m}{r} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)

Where:

[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.

[tex]M[/tex] - Mass of the Earth. in kilograms.

[tex]m[/tex] - Mass of the rocket, in kilograms.

[tex]r[/tex] - Radius of the Earth, in meters.

[tex]v[/tex] - Escape velocity, in meters per second.

Then, we derive an expression for the escape velocity by clearing it within (2):

[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]

[tex]v = \sqrt{\frac{2\cdot G \cdot M}{r} }[/tex] (3)

If we know that [tex]v = \frac{1}{21}\cdot c[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]M = 5.94\times 10^{24}\,kg[/tex], [tex]G = 6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex] and [tex]M = 5.94\times 10^{24}\,kg[/tex], then the expected radius of the Earth is:

[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]

[tex]r = \frac{2\cdot G \cdot M}{v^{2}}[/tex]

[tex]r = \frac{2\cdot \left(6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.94\times 10^{24}\,kg)}{\left[\frac{1}{21}\cdot \left(3\times 10^{8}\,\frac{m}{s} \right) \right]^{2}}[/tex]

[tex]r = 3.883\,m[/tex]

The expected radius of the Earth is 3.883 meters.

Let the point where A is launched act as the origin, so that the horizontal positions at time t of the respective projectiles are

• A : x = (19.5 m/s) cos(30°) t

• B : x = 62.2 m + (19.5 m/s) cos(60°) t

These positions are the same at the moment one projectile is directly above the other, which happens for time t such that

(19.5 m/s) cos(30°) t = 62.2 m + (19.5 m/s) cos(60°) t

Solve for t :

(19.5 m/s) (cos(30°) - cos(60°)) t = 62.2 m

t = (62.2 m) / ((19.5 m/s) (cos(30°) - cos(60°))

t ≈ 8.71 s

Answer:

(d)

Explanation:

because dU = Q -W so ,that the option d(D) is correct

Answer:

I = 0.25 A

Explanation:

Given that,

Three 15 ohms and two 25 ohms light bulbs and a 24 V battery are connected in a series circuit.

In series combination, the equivalent resistance is given by :

[tex]R=R_1+R_2+R_3+....[/tex]

So,

[tex]R=15+15+15+25+25\\\\=95\ \Omega[/tex]

The current each resistor remains the same in series combination. It can be calculated using Ohm's law i.e.

V = IR

[tex]I=\dfrac{V}{R}\\\\I=\dfrac{24}{95}\\\\I=0.25\ A[/tex]

So, the current of 0.25 A passes through each bulb.

Answer:

a) He found the same value of q/m for different cathode materials.

b)      y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex] ,  c)  v = [tex]\frac{E_o}{B_o}[/tex]

Explanation:

In Thomson's experiments he was able to measure the deflection of the light beam under the effect of the magnetic field and with these results find the e / m relationship, which in all cases is the same, therefore the most important conclusion is that the value e E / m is constant for all materials.

b) In the part of the plates the electrons are accelerated by the electric field,

              F = ma

             - e E = m a

              a = - (e/m)  E₀

the distance traveled is          

X axis

          x = v₀ t

the separation of the plates is x = d

          t = vo / d

Y axis

          y = v_{oy} t + ½ to t²

          y = ½ a t²

          y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex]

c) In this case there is a magnetic field B₀ and the electrons have no deflection

         F = - e E + e v x B

if there is no deviation F = 0

         e E = e v B

         v = [tex]\frac{E_o}{B_o}[/tex]

Answer:

1568J

Explanation:

Since the problem states 80 m from the point of drop, the height relative to the ground will be 100-80=20m.

Use conservation of Energy

ΔUg+ΔKE=0

ΔUg= mgΔh=2*9.8*(20-100)=-1568J

ΔKE-1568J=0

ΔKE=1568J

since KEi= 0 since the object is at rest 100m up, the kinetic energy 20meters above the ground is 1568J