If a jet is travelling horizontally at 100m/s at a height of 500m above the ground and a jet drops a bomb to the ground. Where does the bomb strike the ground relative to the point atwhich it is released ?A.100m B.1km C.500m D.5km

The final velocity of the projectile when it strikes the ground below is 198.51 m/s.

The time taken for the projectile to fall to the ground is calculated as follows;

h = vt + ¹/₂gt²

where;

265 = (185 x sin45)t + (0.5)(9.8)t²

265 = 130.8t + 4.9t²

4.9t² + 130.8t - 265 = 0

solve the quadratic equation using formula method,

t = 1.89 s

vyf = vyi + gt

where;

vyf = (185 x sin45) + (9.8 x 1.89)

vyf = 149.322 m/s

vxf = vxi

where;

vxf = 185 x cos(45)

vxf = 130.8 m/s

vf = √(vyf² + vxf²)

where;

vf = √(149.322²  +  130.8²)

vf = 198.51 m/s

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The Brundtland Commission spotted as a matter of urgency, the interconnection between natural resource use and the economy.

The Brundtland Commission was set up with the intention to achieve sustainable development. The tenure of the commission lasted from 1984 to 1987.

The commission identified the pillars of sustainable growth as economic growth, environmental protection, and social equality. The committee opined that environmental problem emanates from poverty in the southern region and reckless consumption of resources in the northern region.

The Brundtland Commission spotted as a matter of urgency, the interconnection between natural resource use and the economy.

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Answer:

D. 20m

Explanation:

this answer is 20 m.

wavelength = velocity / frequency

The magnitude of the average force exerted on the ball by the wall is 225.469 N.

The change in momentum is equal to the product of impact force applied while colliding and time for that impact.

Impulse F. t = m (Vf -Vi)

where, Vf is the final velocity and Vi is the initial velocity.

Given is a 3.42 kg steel ball strikes a massive wall at 14.3 m/s at an angle of α = 61.5° with respect to the plane of the wall. It bounces off with the same speed and angle. The ball is in contact with the wall for 0.207 s

Substitute the values into the expression, we get

Impulse = 2 x mvcosθ

Impulse= 2 x 3.42 x 14.3 x cos 61.5°

Impulse = 46.672 kg.m/s

The impact force can be written as

F.t = I

Put the given values, we have

F = 46.672 /  0.207

F = 225.469 N

Thus, the magnitude of force exerted by the wall is 225.469 N

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Answer:

See the answer Explain why aircraft are carefully designed so that parts do not resonate. Expert Answer This virtually takes place, however maximum usually in small piston-engined airplanes, in particular dual-engined airplanes. The resonant frequency of the fuselage of a small plane goes to have numerous nodes, withinside the low loads of hertz.

Answer:

Explanation:

Force:

⇒ Mass × Acceleration

Work Done:

⇒ Force × Distance

Calculations:

⇒ Force = 0.0083 × 7000 N

⇒ Work done = 0.0083 × 7000 × 5.8 × 10^4 Joules

Answer:

71m/s

Explanation:

when you convert it you get 158.822 miles per hour

The two factors that affect current carrying capacity of an accumulator will be accumulator size and ambient temperature

Utilizing the compressible and decompressible properties of nitrogen gas, an accumulation vessel is used to store hydraulic pressure.

These key deciding elements are:

1. Accumulator Size: The current capacity increases as the circumference of the conductor increases.

2. Ambient Temperature: The greater the ambient temperature, the less heat is needed to raise the insulation's maximum operating temperature.

3. Accumulator identification:

4. Conditions for Installation:

Hence, the two factors that affect current carrying capacity of an accumulator will be accumulator size and ambient temperature

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The difference between polar covalent bonds and nonpolar covalent bonds using these two molecules - H2 and HCl are HCl is a polar covalent compound because the chloride ion is extra electronegative than the hydrogen ion.

HCl is a polar molecule. This is due to the fact the Chlorine (Cl) atom withinside the HCl molecule is extra electronegative and does now no longer proportion the bonding electrons similarly with Hydrogen (H). But H2 And Cl2 are nonpolar because of comparable electronegativity of each the atoms withinside the molecule H2 And Cl2 .

Hydrogen chloride is a diatomic molecule, such as a hydrogen atom H and a chlorine atom Cl related with the aid of using a polar covalent bond. The chlorine atom is an awful lot extra electronegative than the hydrogen atom, which makes this bond polar.

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Answer: A) Ball A

Explanation:

Density = m/v where m is mass and v is volume

This means the greater the mass, the greater the density.

Because of Ball A's greater mass it would have the greater density.

The net force on q2 will be 1.35 N

A force in physics is an effect that has the power to alter an object's motion. A mass-containing object's velocity can vary, or accelerate, as a result of a force. Intuitively, a push or a pull can also be used to describe force. Being a vector quantity, a force has both magnitude and direction.

Given Particles q1, q2, and q3 are in a straight line. Particles q1 = -5.00 x 10-6 C,q2 = +2.50 x 10-6 C, and q3 = -2.50 x 10-6 C. Particles q₁ and q2 are separated by 0.500 m. Particles q2 and q3 are separated by 0.250 m.

We have to find the net force on q2

At first we will find Force due to q1

F = 9 × 10⁹ × 5 × 10⁻⁶ × 2.5 × 10⁻⁶ / 0.5²

F = 450 × 10⁻³

F₁ = 0.45 N (+)

Now we will find Force due to q2

F = 9 × 10⁹ × 5 × 10⁻⁶ × 2.5 × 10⁻⁶ / 0.25²

F = 1800 × 10⁻³

F₂ = 1.8 N (-)

So net force (F) will be

F = F₂ - F₁

F = 1.8 - 0.45

F = 1.35 N

Hence the net force on q2 will be 1.35 N

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Answer:

kg

Explanation:

the highr and jebad to kilogram mizans

Answer:

w=m×g

Explanation:

w is given and take g as 9.8m/s^2

(a) The equilibrant C for force of vector A and B is 3.43 N.

(b) The equilibrant C for fx of vector A and B is 2.1 N.

(c) The equilibrant  C, for fy of vector A and B is 2.12 N.

An equilibrant force is a single force that will bring other bodies into equilibrium.

Vector A: mass = 0.2 kg, θ = 20⁰

Vector B: mass = 0.15 kg, θ = 80⁰

Fx = mg cosθ

Fy = mg sinθ

where;

Force of A due to its weight

F(A) = mg

F(A) = 0.2 x 9.8 = 1.96 N

Fx = (0.2 x 9.8) cos(20) = 1.84 N

Fy = (0.2 x 9.8) sin(20) = 0.67 N

R = √(0.67² + 1.84²)

R = 1.96 N

Force of B due to its weight

F(B) = mg

F(B) = 0.15 x 9.8

F(B) = 1.47 N

Fx = (0.15 x 9.8) cos(80) = 0.26 N

Fy = (0.15 x 9.8) sin(80) = 1.45 N

R = √(0.26² + 1.45²)

R= 1.47 N

Equilibrant force:

Force, C = 1.96 N + 1.47 N

Force, C = 3.43 N

Equilibrant FX:

Fx, C = Fx(A) + Fx(B)

Fx, C = 1.84 N + 0.26 N = 2.1 N

Equilibrant FY:

Fy, C = Fy(A) + Fy(B)

Fy, C =0.67 N + 1.45 N = 2.12 N

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We need Net resistance

solve from right words

[tex]\\ \rm\Rrightarrow R_1=20+10=30\Omega[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{1}{R_2}=\dfrac{1}{30}+\dfrac{1}{50}=\dfrac{5+3}{150}[/tex]

[tex]\\ \rm\Rrightarrow R_2=\dfrac{150}{8}=18.75\Omega[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{1}{R_3}=\dfrac{1}{30}+\dfrac{1}{40}=\dfrac{4+3}{120}[/tex]

[tex]\\ \rm\Rrightarrow R_3=\dfrac{120}{7}=17.14\Omega[/tex]

[tex]\\ \rm\Rrightarrow R_{net}=17.14+18.75=35.89\Omega[/tex]

Use ohm's law

[tex]\\ \rm\Rrightarrow I=\dfrac{V}{R}[/tex]

[tex]\\ \rm\Rrightarrow I=\dfrac{9}{35.89}[/tex]

[tex]\\ \rm\Rrightarrow I\approx 0.25A[/tex]

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[tex]\large \bold {ANSWER}[/tex]

Considering that the pulley is fixed, the force applied should be equal to the weight of the object - of 400N.

[tex]\large \bold {EXPLANATION}[/tex]

Pulleys or pulleys are mechanical tools used to assist in the movement of objects and bodies. There are two types of pulleys: fixed and movable. While the fixed pulley changes the direction of force, the moving pulley helps to decrease the force needed to move the object or body in question.

As the statement only tells us a pulley, we must consider that it is fixed, because generally when it is mobile, this information is highlighted in the question.

In this way, a fixed pulley only changes the direction of the applied force. Thus, the force must have the same magnitude as the weight of the object to be moved. If the bucket weighs 400N, the force applied to the pulley must be 400N.

Therefore, having a fixed pulley, the force applied must be equal to the weight of the object, and will be 400N.

The total mass of the galaxy is 443.4 Solar mass

Orbital velocity ([tex]v[/tex]) = [tex]\sqrt{\frac{MG}{R} }[/tex]

where M= weight of galaxy

G= gravitational constatnt = [tex]6.674*10^-^1^1[/tex] (given)

R = distance from centre = [tex]41200[/tex] Light years (given)= [tex]4.12*9.5*10^1^6[/tex]  km (1 ly= [tex]9.5*10^3[/tex] billion km)

v= orbital velocity = [tex]275[/tex]  [tex]km/s[/tex] (given)

∴ According to the formula

[tex](2.75*10^2)^2[/tex] = [tex]\frac{M*6.674*10^-^1^1}{4.12*9.5*10^1^6}[/tex]

⇒ [tex]7.56*10^4*4.12*9.5*10^1^6=M*6.674*10^-^1^1[/tex] (cross multiplying and expanding)

⇒ [tex]29.59*10^2^1=M*6.674*10^-^1^1[/tex]

⇒ [tex]\frac{29.59*10^2^1*10^1^1}{6.674}=M[/tex]

⇒ [tex]4.434*10^3^2=M[/tex]

1 solar mass = [tex]1.989*10^3^0 kg[/tex]

⇒ Mass in solar mass =443.4 Solar mass

⇒ M = 443.4 Solar mass

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• The speed change between T and S is greater between S and R.

• The speed of m at X is  greater that at Q.

• The size of the total force on m at P is less at U.

• The mechanical energy of m at P is equal to that at V.

• The size of the total force on m at S is greater at P.

• The velocity of m at X is equal to that at V.

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

Given is a small mass m slides with negligible friction down an incline at an angle of 25.76° with respect to the horizontal. It then drops down to a horizontal surface and bounces elastically back up as shown.

The picture shows the position of the mass at equal time intervals starting from rest at T. The height of the mass at X is the same as at V.

Between T to S and S to R, the mass is under constant acceleration. Time taken to move from T to S is greater than S to R. Thus, the speed change between T and S is greater than between S and R.

At Q, there is only a horizontal velocity component, but at X. the speed will be greater and has both vertical and horizontal component. Thus, the speed of m at X is greater than that at Q.

Force is given as the rate of change of momentum with time. At U, change in momentum is large compared to P. Thus, the size of the total force on m at P is less at U.

There is no friction acting on the system. So the energy remains conserved. Mechanical energy at P = V.

The force on mass m at S is only the gravity force. The remaining forces are cancelled by the normal force. Thus, size of the total force on m at S is greater at P.

The energy is conserved at each point of motion of mass. If X and V are at same height, they have same potential energy and so their kinetic energy. Thus, velocity of m at X is equal to that at V.

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The diagram shows the speed and distance of a free-falling item dropped from rest. In the attachment, there is a graph.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The velocity and the time at the different points are;

A(0 sec ,0 m/sec)

B(1,9.8)

C(2,19.6)

D(3,29.4)

E(4,39.2)

F(5,49.0)

The point is plotted on the graph and get the graph.

The graph is attached in the attachment.

The complete question is;

Velocity and distance of free-falling object dropped from rest are given in the digrame. Draw the velocity-time graph for the given conditions.

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(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².

(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.

(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.

The work done in stretching the spring is calculated as follows;

W = ¹/₂kx²

W(1 to 2) = ¹/₂K₂Δx²

where;

W(1 to 2)  =  ¹/₂(250)(0.65 - 0.35)²

W(1 to 2)  = 11.25 J

W(0  to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃

where;

W(0  to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)

W(0  to 3) = 64.28 J

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The distance x will the ball land after flies off with a horizontal initial velocity  is 3.0635 m.

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

A small metal ball with a mass of m = 62.0 g is attached to a string of length l = 1.85 m. It is held at an angle of θ = 48.5° with respect to the vertical.

The ball is then released. When the rope is vertical, the ball collides head-on and perfectly elastically with an identical ball originally at rest. This second ball flies off with a horizontal initial velocity from a height of h = 3.76 m, and then later it hits the ground.

The conservation of energy principle states that total mechanical energy remains conserved in all situations where there is no external force acting on the system.

Kinetic energy  = Potential energy

1/2 mv² =mgh₁

The velocity at the bottom, when the height h = 5m, is

v= √2gh₁...................(1)

The vertical height h₁ = l- lcosθ

h₁ = l- lcosθ

h₁ = 1.85 - 1.85cos48.5°

h₁ =0.6241 m

Putting the values in equation (1), we get

v = √2x 9.81 x0.6241

v = 3.499 m/s

The horizontal distance traveled is

x = vt

x = v x √2h/g

Plug the values, we get

x =  3.499 x √2x3.76 / 9.81

x = 3.0635 m

Thus, the horizontal distance ball travels is  3.0635 m.

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The weight of an object on pluto will be 15.56. Mass multiplied by the gravitational acceleration gives weight.

Gravitation is a natural law by which all things with all matter are attracted towards one another. Gravity is responsible for large-scale structures present in the Universe.

By dividing the object's weight on Earth by 9.8 m/s², as illustrated below, one may calculate the object's mass.

m = 250 N / 9.8 m/s²

m = 25.51 kg

Multiply the acquired mass by Pluto's gravitational acceleration (g);

W = (25.51 kg) x (0.61 m/s²)  

W = 15.56 N

Thus, the item will only be 15.56 N in weight.

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Answer: B.

15.6 newtons

Explanation: edmentum

The stress on the wire is determined as  6.37 x 10⁶ N/m².

The stress on the wire is calculated as follows;

σ = F/A

where;

Let the diameter of the wire = 2 mm

Area = πr²

Area = π(1 x 10⁻³)²

Area = 3.142 x 10⁻⁶ m²

Stress = 20/(3.142 x 10⁻⁶)

Stress = 6.37 x 10⁶ N/m²

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Answer:

full explanation and answer is on the picture

Answer:

V = ?

C = 3 × 10^8

wavelength = 2.10 × 10^-7

wavelength = C/V

V = C/wavelength

V = 3 × 10^8/2.10 × 10^-7

V = 1.43 × 10^ 15

The forces a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

The application of a force may be used to describe the steel as an elastic element within a certain range of applied force.

Given data;

Young modulus, E=2.10 × 10¹¹ N/m²

Cross-sectional area,A

Final length,[tex]\rm L_f = 1.45 m = 1450 \ mm[/tex]

Initial length,[tex]\rm L_i = 7.5 mm[/tex]

[tex]\rm F = \frac{(L_f-L_i)(E)(A)}{L_1} \\\\ \rm F = \frac{(1450 - 7.5)(2.0 \times 0^{11})(0.746)}{1450} \\\\ F = 1.4 \times 10^{11} \ N[/tex]

Hence, the force a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

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Answer:

0.6192

Explanation:

1.44 x 0.430=0.6192

Answer:

4.2m/s

Explanation:

0.50x8.4=4.2m/s

Taking into account the definition of calorimetry,  the mass of the coffee is 253.33 g.

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

In this case, you know:

Replacing in the equation that allows to calculate heat exchanges is:

190 cal = 1 [tex]\frac{cal}{gC}[/tex]× m× 0.75 C

Solving:

m= 190 cal÷ (1 [tex]\frac{cal}{gC}[/tex]× 0.75 C)

m=253.33 g

Finally, the mass of the coffee is 253.33 g.

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Answer:

given -

initial velocity = 4.09 m/s

acceleration = 1.01 m/s²

distance = 23.4 m

time = ?

using second formula of motion,

s = ut + 1/2 at².

where, s = distance

u = initial velocity

t = time

a = acceleration

23.4 = 4.09(t) + 1/2(1.01)(t) ²

23.4 = 4.09t + 2.02t²

2.02t² + 4.09t - 23.4 = 0

solving the equation by using quadratic formula

Use the standard form, ax² + bx + c = 0 , to find the coefficients of our equation, :

a = 2.02

b = 4.09

c = -23.4

we get t=2.539 or t= -4.563

time cannot be negative so

t= 2.539 sec = 2.6 Sec is the answer

The speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.

According to the law of conservation of linear momentum before the collision is equal to the momentum after the collision. These laws state how momentum gets conserved.

Unit conversion;

1 km/sec = 1000 m/sec

Given data;

Spaceprobe speed  = 1.795 km/s = 1795 m /sec

Probe mass = 635.0 kg

Fuel mass = 4092.0 kg

Expelled propellent velocity = 4.161 km/s = 41461 m/sec

From the momentum conservation principle;

[tex]\rm P_i = P_f \\\\ (m_p+m_f)v_i = m_pV - m_fv_p \\\\ V = \frac{(635+4092)1795+4092 \times 41461}{635} \\\\ V = 280540.7 \ m/sec \\\\ V = 28.05 m/sec[/tex]

Hence, the speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.

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