A space craft is moving relative to the earth , an observer on the earth finds that, between 1pm and 2pm according to her clock, 3601 seconds elapse on the space craft clock . What is the space craft speed relative to the earth?c=2.998×10^8ms

The volume of the helium balloon in order to lift the weight is 17,760m³.

To find the answer, we need to know about the buoyant force.

     => V= 3179/0.179 = 17,760m³

Thus, we can conclude that the volume of the helium balloon in order to lift the weight is 17,760m³.

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If you speed through a construction zone while workers are present, your fines could be as much as one thousand dollars.

This is referred to as the amount which is instructed by a court or authority to be paid as a result of it being a penalty for various types of offences. each crime has its specific fine which helps to serve as deterrent for unlawful behavior in the community.

it is always best not to speed when within a construction zone which has workers present in the area. This is ideal as it helps to prevent incidences of accidents or death.

it is therefore the reason why a fine of 1000 dollars is to be paid so that people can think of such high amount before performing certain types of activities when driving and makes it the most appropriate choice.

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The angle of projection given that the range is 256√3 m and the maximum height reached is 64 m is 30°

R = u²Sine(2θ) / g

256√3 = u²Sine(2θ) / 9.8

Cross multiply

256√3 × 9.8 = u²Sine(2θ)

Divide both sides by Sine(2θ)

u² = 256√3 × 9.8 / Sine(2θ)

H = u²Sine²θ / 2g

64 = [256√3 × 9.8 / Sine(2θ)] × [Sine²θ / 2 × 9.8]

64 = [256√3 / Sine(2θ)] × [Sine²θ / 2]

Recall

Sine²θ = SineθSineθ

Sine2θ = 2SineθCosθ

Thus,

64 = [256√3 / 2SineθCosθ] × [SineθSineθ / 2]

64 = 256√3 × Sineθ / 4Cosθ

Recall

Sineθ / Cosθ = Tanθ

Thus,

64 = 256√3 / 4 × Tanθ

Divide both side by 256√3 / 4

Tanθ = 64 ÷ 256√3 / 4

Tanθ = 0.5774

Take the inverse of Tan

θ = Tan⁻¹ 0.5774

θ = 30°

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The tension in the cable is equal to 323.5 N.

The tension, T in the cable is determined by taking moments about the pivot  marked X.

The angles of the boom and the cable with the horizontal are first calculated.

Angle of the boom with horizontal, θ = tan⁻¹(5/10) = 26.56°

The angle of cable with horizontal, B = tan⁻¹(4/10) = 21.80

Taking moments about the pivot:

175.5 * cos 26.56 + 94.7 * cos 26.56 * 0.5 = T (sin(26.56 + 21.80) * 1

Tension = 241.68/0.747

Tension = 323.5 N

In conclusion, the tension in the cable helps to suspend the crate.

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The amount of heat needed to raise the temperature of 6.00 g of copper by 15.0°C is 34.65J (option A).

The amount of heat absorbed or released by a substance can be calculated using the following formula:

Q = mc∆T

Where;

Q = 6 × 0.385 × 15

Q = 90 × 0.385

Q = 34.65J

Therefore, the amount of heat needed to raise the temperature of 6.00 g of copper by 15.0°C is 34.65J.

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(a) The magnitude of the magnetic field at point P1 , midway between the wires is  1.005 x 10⁻⁴ T and the direction will be out of the page.

(b) The magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 is 2.67 x 10⁻⁶ T and the direction is into the page.

(c) The magnitude of the magnetic field at point P3 , 20.0 cm directly above P1  is 3.88 x 10⁻⁶ T and the direction is downwards.

B = μ/2π[I₁/0.5r + I₂/0.5r]

B = (μ/2π) x (I/0.5r + I/0.5r)

B = (μ/2π) x (2I/0.5r)

B = μI/0.5r

B = 2μI/r

where;

B = (2 x 4π x 10⁻⁷ x 4)/(0.1)

B = 1.005 x 10⁻⁴ T

The direction of the magnetic field is out of the page.

B = μI/2πd

d = 5 cm + 25 cm = 30 cm

B =  (4π x 10⁻⁷ x 4)/(2π x 0.3)

B = 2.67 x 10⁻⁶ T

The direction of the magnetic field is into the page towards P1.

B = μI/2πd

d = √(20² + 5²)

d = 20.62 cm

B =  (4π x 10⁻⁷ x 4)/(2π x 0.2062)

B = 3.88 x 10⁻⁶ T

The direction of the magnetic field is downwards towards P1.

Thus, the magnitude of the magnetic field at point P1 , midway between the wires is  1.005 x 10⁻⁴ T and the direction will be out of the page.

The magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 is 2.67 x 10⁻⁶ T and the direction is into the page.

The magnitude of the magnetic field at point P3 , 20.0 cm directly above P1  is 3.88 x 10⁻⁶ T and the direction is downwards.

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Answer:

36 cm

Explanation:

Mass of stick; m1 = 0.20kg at midpoint.

Total length; L=1.0 m

Pivot at 0.40m

Atached mass m2 = 0.50kg

Applying rotational equilibrium we have;

Ʈnet = 0

(m1g) • r1 = (m2g) • r2

(0.2) (0.1m) = (0.5)(x)

x = 0.04m =4cm

measured away from 40cm mark gives a position on the stick of; 40cm - 4cm = 36 cm

The original kinetic energy of the player and the average power is required to stop him are 1200 J and 1200 W respectively.

The energy possessed by a body in motion is known as Kinetic Energy. The S. I unit is Joule.

Given that an 87-kg football player traveling 5.2 m/s is stopped in 1.0 s by a tackler. The given parameters are;

The original kinetic energy of the player can be calculated by using the formula K.E = 1/2m[tex]v^{2}[/tex]

Substitute all the parameters into the formula

K.E = 1/2 x 87 x [tex]5.2^{2}[/tex]

K.E = 1176.24

K.E = 1200 J

Power is the rate at which work is done.

Work done = energy

The average power is required to stop him can be calculated by using the formula P = E/t

Substitute all the parameters into the formula

P = 1200/1

P = 1200 W

Therefore, the original kinetic energy of the player and the average power is required to stop him are 1200 J and 1200 W respectively.

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The kinetic energy of the projectile at the highest point of its motion will be E/4.

When a projectile will be thrown obliquely near the surface of the earth, it travels a curved path under uniform acceleration directed toward the center of the Earth. The path of a particle is called a projectile while the motion of a projectile is projectile motion.

Given, the angle of projection  with horizontal, [tex]\theta = 60 ^o[/tex]

Consider that 'E' is the initial value of the kinetic energy of the projectile.

The equation for the initial kinetic energy is : [tex]E =\frac{1}{2}mu^2[/tex]

where m is the mass of the given projectile.

The component of the velocity of the projectile in the horizontal direction:

uₓ = u cosθ

uₓ = u cos 60°

uₓ = u/2

From the equation of motion: v = u +at

v = (u/2) + (0) t

v = u/2

The final kinetic energy of the projectile:

[tex]E_f = \frac{1}{2}mv^2[/tex]

[tex]E_f = \frac{1}{2}m(\frac{u}{2} )^2[/tex]

[tex]E_f = \frac{1}{4} (\frac{1}{2}mu^2 )[/tex]

[tex]E_f = \frac{E}{4}[/tex]

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The vertical component of force exerted by the hi.nge on the beam will be,142.10N.

To find the answer, we need to know more about the tension.

                      [tex]F_V+T sin\alpha -mg=0\\F_V=mg-Tsin\alpha \\[/tex]

                     [tex]Tlsin\alpha - mg\frac{l}{2}sin\beta =0\\ \\Tlsin\alpha = mg\frac{l}{2}sin\beta\\\\Tsin57=\frac{mg}{2}sin90\\\\T=\frac{mg}{2sin57} =169.43N[/tex]

                [tex]F_V=(29*9.8)-(169.43*sin57)=142.10N[/tex]

Thus, we can conclude that, the vertical component of force exerted by the hi.nge on the beam will be,142.10N.

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The hi.nge will apply a force of 142.10N on the beam in the vertical direction.

We must learn more about the tension in order to find the solution.

                           [tex]F_V=mg-Tsin\alpha[/tex]

                            [tex]Tlsin\alpha =mg\frac{l}{2}sin\beta \\T=\frac{mgsin90}{2sin57} =169.43N[/tex]

                     [tex]F_v=(29*9.8)-(169.43*sin57)=142.10N[/tex]

This leads us to the conclusion that the vertical component of the force the hi.nge exerts on the beam will be 142.10N.

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(a) The initial vertical velocity of the stone is 16.85 m/s and the initial horizontal velocity is 14.14 m/s.

(b) The final vertical velocity is 0 and the final horizontal velocity is 14.14 m/s.

The initial vertical velocity of the stone is calculated as follows;

Vi = Vsinθ

Vi = 22 x sin(50)

Vi = 16.85 m/s

Vxi = V cosθ

Vxi = 22 x cos(50)

Vxi = 14.14 m/s

Vf = Vi - gt

where;

Vfx = Vxi = 14.14 m/s

Thus, the initial vertical velocity of the stone is 16.85 m/s and the initial horizontal velocity is 14.14 m/s.

The final vertical velocity is 0 and the final horizontal velocity is 14.14 m/s.

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The horizontal force applied is  160 N while the velocity is  2.03 m/s.

The work done by the car is obtained as the product of the force and the distance;

W = F x

F = ?

x = 30.0 m

W = 4,800 J

F = 4,800 J/30.0 m

F = 160 N

But F = ma

a = F/m

a = 160 N/2.30 ✕ 10^3-kg

a= 0.069 m/s

Now;

v^2 = u^2 + 2as

u = 0/ms because the car started from rest

v = √2as

v = √2 * 0.069 * 30

v = 2.03 m/s

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5.8 moles of nitrogen gas are needed to pressurize the air bag.

= (2.37×60)/(0.0821×298)

= 5.8 moles

Thus, we can conclude that 5.8 moles of nitrogen gas are needed to pressurize the air bag.

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5.48 V is the potential drop across the cable for a copper transmission cable of 50.0 km long and 10.0 cm in diameter carries a current of 105 A

Ohm's Law states that the potential drop is determined by the equation: V = IR, where I is the current and R is the wire resistance.
R=PL/A
Under the assumption that all physical parameters and temperatures remain constant, Ohm's law asserts that the voltage across a conductor is directly proportional to the current flowing through it.

Only when the given temperature and the other physical variables remain constant does Ohm's law apply. Increasing the current causes the temperature to rise in some components. The filament of a light bulb serves as an illustration of this, where the temperature increases as the current increases. Ohm's law cannot be applied in this situation. The filament of the lightbulb defies Ohm's Law.

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The force F, necessary to support an object with a mass of 991 kg placed on piston 2. is 20.61J.

It should be noted that by Pascal Law, the pressure on piston 1 will have the same value as the pressure on piston 2.

This will be:

(991 × 10) /(π × 0.0946/2)²

= 9910/0.022

= 450454.6 Pa

F1 = A1 × 450454.6 = 3.14 × (0.0187/2)² × 450454.6

= 123.64

F = 123.64/6

F = 20.61

Therefore, the force F, necessary to support an object with a mass of 991 kg placed on piston 2. is 20.61J.

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The deceleration of the bullet over 0.2 second, given the data from the question is –2500 m/s²

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

a is the acceleration

v is the final velocity

u is the initial velocity

t is the time

NOTE: Deceleration is the opposite of acceleration

With the above equation for acceleration, we can obtain the deceleration of the bullet. Details below:

a = (v – u) / t

a = (0 – 500) / 0.2

a = –500 / 0.2

a = –2500 m/s²

Thus, the deceleration of the bullet is –2500 m/s²

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An object is considered to be in a condition of equilibrium when it is balanced with regard to all external forces.

Equilibrium:

An object is considered to be in equilibrium if both its angular acceleration and the acceleration of its center of mass are equal to zero. In layman's terms: The item must either be at rest or moving at a constant speed if it is not accelerating because F = ma (force = mass x acceleration). Even in motion, a body can be in equilibrium. This kind of equilibrium is referred to as a dynamic equilibrium.

A weight suspended by a spring or a brick laying on a flat surface is an example. The equilibrium is unstable if the force with the smallest deviation tends to increase the displacement. As an example, imagine a ball bearing on the edge of a razor blade.

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The magnitude of the magnetic field and the direction of the magnetic field at point P is mathematically given as

B=1.9*10^{-5}T

To determine the magnetic field direction, use the right-hand -rule on the page magnetic field is going.

Generally, the equation for Biot savant law is  mathematically given as

[tex]B=(\frac{u}{4\pi}*{\frac{Ilsin\theta}{r^2})[/tex]

Their net field is

Bn=2B

[tex]Bn=2* B=(\frac{u}{4\pi}*{\frac{Ilsin\theta}{r^2})[/tex]

Hence

[tex]B=(\frac{4*\p *10^{-7}}{4\pi}*{\frac{(30)(2*10^{-3}sin45)}{\sqrt{(3*10^{-2})^2+((3*10^{-2})^2)}/2})[/tex]

B=1.9*10^{-5}T

In conclusion, To determine the magnetic field direction, use the right-hand -rule on the page magnetic field is going.

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Answer:

The magnitude of the magnetic field at the point P is [tex]1.57*10^{-5}T[/tex] and the field is pointing into the page.

Explanation:

The general form of a similar question to this is:

[tex]\vec{B} = \frac{\mu _{0} }{4\pi } * \oint \frac{Id\vec{l} \times \hat{r}}{r^{2} }[/tex]

where [tex]\vec{B}[/tex] is the vector of the Magnetic Field, [tex]\mu _{0}[/tex] is the Free Space Permeability Constant (equal to [tex]4\pi * 10^{-7} \frac{N}{A^2}[/tex]), [tex]I[/tex] is the current, and [tex]r[/tex] is the distance from the segment to the point P. (I will get to the [tex]d\vec{l} \times \hat{r}[/tex] term in a bit)

This equation is fairly complicated. Luckily, it can be simplified by looking at the magnitude and direction separately.

The first thing to simplify is the cross product. Due to the fact that a cross product can be simplified from [tex]\vec{x} \times \vec{y}[/tex] to [tex]xy\sin(\theta)[/tex], where [tex]\theta[/tex] is the angle between the 2 vectors, and [tex]\hat{r}[/tex] is the unit vector of [tex]r[/tex] (i.e. [tex]\hat{r} = \vec{r}/r[/tex]) we can simplify [tex]d\vec{l} \times \hat{r}[/tex] to just [tex]dl \sin(\theta)[/tex].

Next, we will look at the integral. In this scenario, everything will function as a constant, so we can essentially ignore it.

Finally, [tex]\frac{\mu_{0}}{4\pi}[/tex] simplifies down to [tex]10^{-7}[/tex].

This gives us our new equation for the Magnetic Field produced by a single segment at a point:

[tex]B = \frac{Il\sin\theta}{r^{2}}*10^{-7}[/tex]

Now we need to find values for [tex]r[/tex] and [tex]\theta[/tex]. Luckily, we are dealing with a 45-45-90 triangle with sides of [tex]1.5 cm[/tex]. This means the distance [tex]r[/tex] is [tex](1.5\sqrt2)cm[/tex]! Similarly, because it is a 45-45-90 triangle, our [tex]\theta[/tex] is [tex]45\textdegree[/tex]!

Now we can start plugging things in:

[tex]B = \frac{(25A)(2*10^{-3}m)\sin(45\textdegree)}{(1.5\sqrt2*10^{-2}m)^2}*10^{-7}\frac{N}{A^2}[/tex]

[tex]B = 7.86^{-6} \frac{N}{A}[/tex] or [tex]T[/tex]

This is the magnitude due to only one single segment. In order to find the total field, we need to know the direction of the field due to each segment.

Finding the direction is really easy. Just use the right hand rule. Point your thumb in the direction of the current and curl the rest of your fingers around an imaginary pole. The direction your fingers point is the direction of the field. In this case, the field lines due to the segments point into the page in the 4th quadrant (the origin is the bend). This means that at point P, both segments induce the same field in the same direction. Therefore, we can take our value from before and double it, giving us our final answer:

[tex]B = 1.57*10^{-5} T[/tex]; into the page.

Answer:

 no

Explanation:

The angle subtended by the radius of the target at Emily's distance can be found using the tangent relation.

 tan(α) = opposite/adjacent = (1/4 ft)/(150 ft) = 1/600

The angle is found using the inverse relation -

α = arctan(1/600) ≈ 0.095°

If Emily's aim is off by 0.2°, she will miss the target by several inches.

Emily's projectile will miss her aiming point by ... (150 ft)tan(0.2°) ≈ 0.524 ft ≈ 6.28 in

The voltage between the plates is 3.9 × [tex]10^-^3[/tex] V.

The work-energy theorem states that the delta in the equation equals the change in kinetic energy plus the change in potential energy. Here, a charge's potential energy is expressed as qV, where V is the position's electric potential.  The greater the change in voltage per unit distance, the greater the electric field.

The kinetic energy of the electrons = 4.1 × [tex]10^-^1^5[/tex] J

Charge of the electron = 1.602 × [tex]10^-^1^9[/tex] coulomb

Using,

     ΔU = q × ΔV

4.1 × [tex]10^-^1^5[/tex] = 1.602 × [tex]10^-^1^9[/tex] × ΔV

      ΔV  = 3.9 × [tex]10^-^3[/tex] V

Therefore, the voltage between the plates is 3.9 × [tex]10^-^3[/tex] V.

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The initial momentum of the yellow and the orange train is 1000kgm/s.

Momentum is the product of the mass and velocity of any object.

Momentum is denoted by P.

Momentum P = mv , where m = mass and v = velocity.

Mass of the orange train = 200kg

Velocity of the orange train = 1m/s

So, the momentum of the orange train will be,

                            ∴    P = mv

                                  P = 200 x 1

                                  P = 200 kgm/s

∴   The initial momentum of the orange train is 200kgm/s.

Mass of the yellow train = 100kg

Velocity of the yellow train = 8m/s

So, the momentum of the yellow train will be,

                            ∴    P = mv

                                  P = 100 x 8

                                  P = 800 kgm/s

∴ The initial momentum of the yellow train is 800kgm/s.

Therefore, the initial momentum of the yellow and the orange train is 1000kgm/s.

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The answer is A. Decreasing the distance between the particles by a factor of 3.

The Universal Law of Gravitation is :

F = Gm₁m₂ / r² (where 'r' is the distance between them)

Since force is inversely proportional to the square of the distance between them, distance has to be decreased by a factor of 3 to increase the force to 9 times the original force.

The primary means by which heat is transferred through fluids is convection currents (option D).

Convection is the transmission of heat in a fluid by the circulation of currents.

Heat can be transferred by different methods depending on the medium. Fluids like gases and liquids transfer heat through the process of convection.

Therefore, the primary means by which heat is transferred through fluids is convection currents.

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From the calculation, the normal force is 6161.2 N.

The normal force is given by the expression;

N - mg = ma

Then;

N = mg + ma

m = 84.4 kg

g = 9.8 m/s^2

a = 63.2 m/s2

Now we have;

N = m(g + a)

N = 84.4 (9.8 + 63.2)

N = 6161.2 N

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The final speed and angle of the first puck are 1.17 m/s and 54.5° respectively.

In an elastic collision, both momentum and energy are conserved. But only momentum is conserved in inelastic collision.

Given that two hockey pucks are moving towards each other. The first one is 0.13 kg and moving at a speed of 1.11 m/s, while the second puck is 0.16 kg and moving at 1.21 m/s, and they collide. (Assume elastic collision.) After collision, the second puck ends up with a speed of 1.16m/s at an angle of 42 degrees below its original path, while the first puck ends up with an unknown speed at an angle above its original path.

The given parameters are;

The mathematical representation of the above question will be in two components.

Horizontal component

M1U1 - M2U2 = M1V1cosФ - M2V2cosФ

Substitute all the parameters

0.13 x 1.11 - 0.16 x 1.21 = 0.13 x V1 cosФ - 0.16 x 1.16cos42

0.1443 - 0.1936 = 0.13V1cosФ - 0.1379

0.13V1cosФ = 0.0886

V1cosФ = 0.0886/0.13

V1cosФ = 0.6815 ........ (1)

Vertical component

0 = M1V1sinФ - M2V2sinФ

M1V1sinФ = M2V2sinФ

Substitute all the parameters

0.13 x V1 sinФ = 0.16 x 1.16sin42

V1 sinФ = 0.1242/0.13

V1 sinФ = 0.9553 ......... (2)

Divide equation 2 by 1

V1 sinФ / V1 cosФ = 0.9553/  0.6815

Tan Ф = 1.40

Ф = [tex]Tan^{-1}[/tex](1.4)

Ф = 54.5°

Substitute Ф into equation 2

V1 sin54.5 = 0.9553

V1 = 0.9553 / 0.8141

V1 = 1.17 m/s

Therefore, the final speed and angle of the first puck are 1.17 m/s and 54.5° respectively.

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As, per the buoyancy force,the volume that the balloon should have is 2863[tex]m^{3}[/tex]

What is buoyancy force?

Air buoyancy is the upward force exerted on an object by the air that is displaced by object. Air buoyancy is responsible for the buoyancy created by the displaced air.

[tex]F_{b}[/tex] = -Vρg ,       where V= volume of the object

                                   ρ = density of the object

                                   g = acceleration due to gravity

                                   [tex]F_{b}[/tex] = buoyant force

The buoyancy force must be equal to the total load lifted

ρ[tex]_{He}[/tex] × V × g + 269 + 2910 = ρ[tex]_{air}[/tex] × V × g

0.179 × V + 3179 = 1.29 V

0.179V + 3179 = 1.29V

0.179V- 1.29V = - 3179

1.11V = 3179

On solving , we get

V = 2863 [tex]m^{3[/tex]

Therefore, the volume that the balloon should have is 2863[tex]m^{3}[/tex].

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The horizontal distance XQ traveled by the bomb is 250 m.

Let the XQ be the horizontal distance traveled by the bomb.

h = vt + ¹/₂gt

Let the vertical velocity = 0

h = 0 + ¹/₂gt²

h = ¹/₂gt²

2h = gt²

t² = 2h/g

t = √(2h/g)

t = √(2 x 100 / 9.8)

t = 4.5 s

XQ = vx(t)

where;

vx is horizontal speed, = 200 km/hr = 55.56 m/s

XQ = 55.56 x 4.5

XQ = 250 m

Thus, the horizontal distance XQ traveled by the bomb is 250 m.

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The point where m3 experiences a zero net gravitational force due to M1 and m2 is 57.42 m.

m1<------(x)------> m3 <-----------(94.8 m - x)-------->m2

a point, x, where m3 experiences a zero net gravitational force due to M1 and m2;

Force on m3 due to m1 = Force on m3 due to m2

Gm1m3/d² = Gm2m3/r²

m1/d² = m2/r²

where;

m1/(x²) = m2/(94.8 - x)²

m1(94.8 - x)² = m2x²

(94.8 - x)² = (m2/m1)x²

(94.8 - x)² = (10.6/25)x²

(94.8 - x)² = 0.424x²

(94.8 - x)² = (0.651)²x²

94.8 - x = 0.651x

94.8 = 1.651x

x = 94.8/1.651

x = 57.42 m

Thus, the point where m3 experiences a zero net gravitational force due to M1 and m2 is 57.42 m.

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