What is the gravitational acceleration on the board of the Deepsea Challenger when it is in the Mariana Trench, 11 km below the surface of the Pacific Ocean?
It is slightly less than 9.81 m/s2.
It is very large due to the high pressure.
It is zero.
It is significantly less than 9.81 m/s2.

The hypothesis will be:

H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.

H₁ =  The amount of salt added to ice will affect the rate at which the ice will melt.

The independent variable which is can be changed by increasing the rate of salt added to the equation.

The dependent variable which is ice will change or melt in response as it will decrease if the rate of the salt added increases.

Salt does not really lower the temperature of an ice cubes, it is known to just lowers their freezing point, that is lowers their melting point.

Note that if salt is around, ice cubes are known to be colder to be solid, and they tend to melt at a temperature that is said to be lower than the freezing point of pure water.

If  the ionic compound salt is known to be added, it tends to lowers the freezing point of the water, which implies that the ice on the ground is not able to freeze that layer of water at all.

Hence, the  hypothesis will be:

H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.

H₁ =  The amount of salt added to ice will affect the rate at which the ice will melt.

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The voltage across the wire is 1.72 x [tex]10^{-3}[/tex] V

The resistance of a wire is the opposition to the flow of current. It depends on the following;

Given that the resistivity of a 1.0 m long wire is 1.72 × 10-8 ωm and its cross sectional area is 2.0 × 10-6 m2. if the wire carries a current of 0.20A

The given parameters are;

The formula to use to get R will be

R = ρL / A

Substitute all the necessary parameters into the formula

R = 1.72 x [tex]10^{-8}[/tex] x 1 / 2 x [tex]10^{-6}[/tex]

R = 8.6 x [tex]10^{-3}[/tex]  Ω

From Ohm's law, V = IR

Substitute all the necessary parameters into the formula

V = 0.2 x 8.6 × [tex]10^{-3}[/tex]

V = 1.72 x [tex]10^{-3}[/tex] V

Therefore, the voltage across the wire is 1.72 x [tex]10^{-3}[/tex] V

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Answer:  It simply means that a freely falling object would increase its velocity by 9.8 m/s per second

Answer:

Hello!

"The acceleration due to gravity on the surface of the earth is 9.8 m/s ²." What does it means that

That every second an object is in free fall, gravity will cause the velocity of the object to increase 9.8 m/s. So, after one second, the object is traveling at 9.8 m/s.

The option that is the correct one concerning the uncontrolled burn phase is:

Uncontrolled Combustion is known to be the the time and place in which a kind of an ignition will stop and it is said to be never  fixed by anything in regards to the compression ignition engine as seen in SI engines.

Note that the four Stages of combustion  are:

1.     Pre-flame combustion

2.     Uncontrolled combustion

3.     Controlled combustion and

4.     After burning

Hence, The uncontrolled burn phase is characterized by uncontrolled combustion in a cylinder until fuel accumulated during ignition delay is burned as all the fuel need to burn out.

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We push a car to the distance of 33.939m if we do 28000J work , exerting a force of 825N.

F = applied force

S = distance moved

=28000J/825N

=33.939meter

Thus, we can conclude that the distance moved by the car is 33.939m.

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Answer:

Gravity exist between two objects that have mass

The impulse shared by the object equals the difference in momentum of the object. In equation form,

F • t = m • Δ v. In a collision, objects experience an impulse; the impulse causes and is equal to the difference in momentum.

a)There is this impulse-momentum change equation.

[tex]where m$ is the mass of a body, $F$ is a force acting to the body, $t$ is time and $D E L A T A N\}=V_{2}-V_{1}$ is the change of velocity.We consider everything is happen along a straight line, and gravitation does not participate.So, the increase of momentum is $\mathrm{F}^{*} \mathrm{t}=10000 \mathrm{~N} * 60$ seconds $=600000 \mathrm{~N}^{*} \mathrm{~s}=600000\left(\mathrm{~kg}^{*} \mathrm{~m}\right)^{*} \mathrm{~s} / \mathrm{s}^{\wedge} 2=600000 \mathrm{~kg}{ }^{*} \mathrm{~m} / \mathrm{s}$.[/tex]

We consider everything exits happen along a straight line, and gravitation does not participate.

So, the increase of momentum is F×t = 10000 N × 60 seconds = 600000 N*s = 600000 (kg*m)*s/s^2 = 600000 kg*m/s.

[tex]$$\Delta(\mathrm{V})=\frac{\mathrm{F.t}}{\mathrm{m}}=\frac{600000}{1200}=500 \mathrm{~m} / \mathrm{s} .$$[/tex]

New velocity after  engine was firing during 60 seconds is 2000 + 500 = 2500 m/s.

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D. The minimum braking torque that must be applied is -11.2 Nm.

The moment of inertia of the uniform cylinder is calculated as follows;

I = ¹/₂MR²

where;

I = (0.5)(8)(0.1²)

I = 0.04 kgm²

τ = -Iα

where;

α = ω/t

α = (80,000 x 2π/rev x 1 min/60s) / (30 s)

α =  (80,000 x 2π)/(60 x 30)

α = 279.25 rad/s²

τ = - ( 0.04 kgm²) x (279.25)

τ =  -11.2 Nm

Thus, the minimum braking torque that must be applied is -11.2 Nm.

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Answer:C is the answer

Explanation:

It is the most reasonable and the answer that makes most sense

Answer:

The angle at which Earth's axis tilts changes.

Answer:

Anticlockwise directions

Please mark me Brainliest to help me

The magnitude of the Coulomb force (in N) on one of the spheres, given the data is 4.59×10⁻⁴ N

Sphere 1 losses 2.40×10¹² electrons

But

1 electron = 1.6x10¯¹⁹ C

Thus,

Charge on sphere 1 = +1.6x10¯¹⁹ × 2.40×10¹² = +3.84×10¯⁷ C

Sphere 2 gains 2.40×10¹² electrons

But

1 electron = 1.6x10¯¹⁹ C

Thus,

Charge on sphere 2 = -1.6x10¯¹⁹ × 2.40×10¹² = -3.84×10¯⁷ C

Using the Coulomb's law equation, the force can be obtained as illustrated below:

F = Kq₁q₂ / r²

F = (9×10⁹ × 3.84×10¯⁷ × 3.84×10¯⁷) / (1.7)²

F = 4.59×10⁻⁴ N

Thus, the magnitude of the Coulomb's force is 4.59×10⁻⁴ N

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(a) The angular acceleration will be 26.035 rev/[tex]min^{2}[/tex].

(b) The final angular velocity is expected to be 33.846 rev/min.

Given.

t=1.3 min, Θ=22 rev, [tex]ω_{i}[/tex]=0

We know, Θ= [tex]ω_{i}[/tex]t+[tex]\frac{1}{2}[/tex][tex]\alpha[/tex][tex]t^{2}[/tex]

22=0+[tex]\frac{1}{2}[/tex][tex]\alpha[/tex][tex]1.3^{2}[/tex]

[tex]\alpha[/tex]=26.035 rev/[tex]min^{2}[/tex]

[tex]ω_{f} =ω_{i}+\alpha t[/tex]=0+26.035*1.3=33.846 rev/min

An object's rate of change in angular position or orientation over time is depicted by its angular velocity, rotational velocity, or both ( or ), also known as the angular frequency vector (i.e. how quickly an object rotates or revolves relative to a point or axis). The direction of the pseudovector is normal to the instantaneous plane of rotation or angular displacement, and its magnitude denotes the angular speed, or the rate at which the item rotates or revolves. It is customary to use the right-hand rule to specify the direction of angular motion. A general definition of angular velocity is "angle per unit time" (angle replacing distance from linear velocity with time in common). Radians per second is how angles are measured in the SI.

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Since the displacement is completely perpendicular to the direction of the applied force, the work done on the bag is zero.

The work is done on an object when the force applied is multiply by the distance moved by the object in the direction of the force applied.

Given that a man carries a hand bag by hanging on his hand moves horizontally where the bag does not up or down.

What is work if the displacement is not in the direction of force ?

The work done can only be zero if the displacement is perpendicular to the direction of force. otherwise, it will not be equal to zero.

Also, the work done will be zero, if the displacement is zero.

In the question above, the displacement is completely perpendicular to the direction of the applied force.

Therefore, the work done on the bag is zero.

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The mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387x 10^3 kg.

We need to be aware of the Bernoulli principle in order to determine the solution.

                        [tex]P+\frac{1}{2}dv^2+ dgh = constant.[/tex]    We substitute d for to represent density.

                         [tex]V_1=259m/s\\V_2=288m/s\\A=395m^2\\d=1.21kg/m^3[/tex]

                      [tex]P_1+\frac{1}{2}dV_1^2+dgh=P_2+ \frac{1}{2}dV_2^2+dgh\\P_1-P_2=\frac{1}{2}d(V_2^2-V_1^2)\\\frac{F}{A}= \frac{1}{2}d(V_2^2-V_1^2)\\[/tex]    

                       [tex]m=\frac{\frac{1}{2}d(V_2^2-V_1^2)A\\}{g} \\m=387*10^3kg.[/tex]

Consequently, we can say that the mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387 x 10^3 kg.

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When a piece of elastic material between two pieces of wood is being stretched beyond its limit, the elastic material or the object does not return to its original length when the force is removed

However, in physics, the type of stress applied when an elastic material is stretched is tensile stress.

Recall:

Stess is defined as force per unit area

Mathematically; Stress = F/A

Elasticity can be defined as the ability of a deformed elastic material or body to return to its original size and shape when the forces causing the deformation are removed.

So therefore, when a piece of elastic material between two pieces of wood is being stretched beyond its limit, the elastic material or the object does not return to its original length when the force is removed

Complete question:

What happens when a piece of elastic material between two pieces of wood is being stretched beyond its limit?

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The orbital speed of an ice cube in the rings of saturn is approximately 3.56 * 10^6 m/s

The law of gravitation states that the force of gravitation is directly proportional to the product of the masses and inversely proportional to the distance between the masses. Mathematically;

F = GMm/r²

where

M and m are the mass of ice cube and

Recall that;

s = Gm1/r^2

Also;

F = sm²

Substitute to have;

s = m²/F

For the centripetal acceleration

a = v²/r

Such that;

v²/r = Gm/r²

v² = Gm/r

v = √Gm/r

Substitute the given parameters into the formula to have:

V =  √6.67×10^-11 *  5.68 x 10^26 / 3.00 x 10^5

V = 355358.97m/s = 3.56 * 10^6 m/s

Therefore the orbital speed of an ice cube in the rings of saturn is approximately 3.56 * 10^6 m/s

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Assuming point K is held in equilibrium, by Newton's second law we have

• net horizontal force

[tex]F_C \cos\left(\tan^{-1}\left(\dfrac57\right)\right) - F_A = 0[/tex]

• net vertical force

[tex]F_C \sin\left(\tan^{-1}\left(\dfrac57\right)\right) - F_B = 0[/tex]

where the angle [tex]\theta[/tex] that rope C makes with the horizontal axis satisfies

[tex]\tan(\theta) = \dfrac{9-4}{11-4} = \dfrac57[/tex]

Solve the first equation for [tex]F_C[/tex].

[tex]F_C = F_A \sec\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

(Recall that [tex]\sec(x)=\frac1{\cos(x)}[/tex].)

Substitute this into the second equation and solve for [tex]F_B[/tex].

[tex]F_B = F_C \sin\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

[tex]F_B = F_A \sec\left(\tan^{-1}\left(\dfrac57\right)\right) \sin\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

[tex]F_B = F_A \tan\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]

(Recall that [tex]\tan(x)=\frac{\sin(x)}{\cos(x)}[/tex].)

[tex]F_B = \dfrac57 F_A[/tex]

[tex]\boxed{F_B \approx 36.1\,\rm N}[/tex]

Answer:

s^ -1    ( or    1/sec)

Explanation:

Velocity is given in units of displacement / sec

like feet /sec   or   m/sec    

so b would have units of   s^-1

(or perhaps a more general term would be   time^-1)

The required orbital speed of the ice cube is 355,358m/s

The force of gravitation is directly proportional to the product of the masses and inversely proportional to the distance between them. This can be expressed mathematically as;

Fr = GMm/r²

The distance is calculated as;

s = Gm/r²

Solving both equation, we will have:

v²/r = Gm/r²

v² = Gm/r

Take the square root of both sides

v = √Gm/r

Solve the required orbital speed

V =  √6.67×10^-11 * 5.68 x 10^26 / 3.00 x 10^5

V = 355358.97m/s

Hence the required orbital speed of the ice cube is 355,358m/s

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The correct answer is 5828.675 J.

Given combined mass 4kg and mass of bullet 150gm=0.150kg.

Total mass= 4+0.150=4.150kg

Velocity=53 m/s

Kinetic energy = [tex]\frac{1}{2} *m*v^{2}[/tex] =0.5*4.150*[tex]53^{2}[/tex] =5828.675 J

Kinetic energy is a type of power that an item or particle possesses as a result of motion. When an item undergoes work—the transfer of energy—by being subjected to a net force, it accelerates and consequently obtains kinetic energy. A moving object or particle's kinetic energy, which depends on both mass and speed, is one of its characteristics. Any combination of motions, including translation (or travel along a path from one location to another), rotation about an axis, and vibration, may be used as the type of motion.

A body's translational kinetic energy is equal to  [tex]\frac{1}{2} *m*v^{2}[/tex] , or one-half of the product of its mass, m, and square of its velocity, v.

a trolley and a sandbag have a combined mass of 4 kg. a bullet with a mass of 150 gram is fired towards the trolley and is lodged in the sand bag immediately after the Collision, on the trolley and sandbag in combination with the bullet, move backwards at 53 metre per second calculate the kinetic energy of the trolley , with the sandbag and bullet ,directly after the Collision ​

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Answer:

A i think...

Explanation:

Sorry if its wrong

(a) The electric field strength at a point 1.00 cm to the left of the middle is  2.0 x 10⁷ N/C.

(b) The magnitude of the force is 94.4 N and direction of the force on it towards the left.

The electric field strength at a point 1.00 cm to the left of the middle is calculated as follows;

E = kq/r²

E1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²

E1 = 1.35 x 10⁸ N/C

E2 =  -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²

E2 = - 1.35 x 10⁸ N/C

E3 = - (9 x 10⁹ x 2 x 10⁻⁶)/(0.03)²

E3 = -2.0 x 10⁷ N/C

E = E1 + E2 + E3

E = +1.35 x 10⁸ N/C - 1.35 x 10⁸ N/C - (-2.0 x 10⁷ N/C)

E = +2.0 x 10⁷ N/C

F = Eq

F = 2.0 x 10⁷ x -4.72 x 10⁻⁶

F = -94.4 N

Thus, the direction of the force will be towards the left.

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The reason behind the heavier hydrometer bulb is the sinking of hydrometer is inversely proportional to the density of hydrometer, hence hydrometers is made heavier.

We all know that hydrometers float in liquid hence to maintain the centre of gravity while floating the hydrometer is made heavier using lead shots.

Question: 5

The length of the pendulum is 7.6 m.

So, L = T²g/4π²

= 7.6 m

Thus, we can conclude that the length of the pendulum is 7.6 m i.e option C is correct.

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Question: 6

The object takes 2.55 seconds to reach the ground.

t= √(2h/g)

t= √(2×31/9.8)

= 2.55s

Thus, we can conclude that the option A is correct.

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Kinetic energy of pieces A and B are 2724 Joule and 5176 Joule respectively.

Mass of piece B = Mb

Ma×Va = Mb×Vb

So, 1.9Mb × Va = Mb×Vb

=> 1.9Va = Vb

=>1/2 × Ma × Va² + 1/2 × Mb × Vb² = 7900

=> 1/2 × Ma × Va² + 1/2 × (Ma/1.9) × (1.9Va)² = 7900

=> 1/2 × Ma × Va² ×(1+1.9) = 7900 j

=> 1/2 × Ma × Va² = 7900/2.9 = 2724 Joule

Thus, we can conclude that the kinetic energy of piece A and B are 2724 Joule and 5176 Joule respectively.

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(1) The effective spring constant of the system is 7.14 N/m.

(2) The maximum x-acceleration of the glider is 0.9 m/s².

(3) The x-coordinate of the glider at time t= 0.650T is 0.28 m.

(4) The kinetic energy of the glider at x=0.00 m is zero.

The effective spring constant of the system is calculated as follows;

F = kx

where;

k = F/x

k = 0.5/0.07

k = 7.14 N/m

a = ω²x

where;

ω = √k/m

ω = √(7.14/0.55)

ω = 3.6 rad/s

a =  (3.6)² x 0.07

a = 0.9 m/s²

T = 2πx/v

T = 2πx/(ωx)

T = 2π/ω

T = 2π/(3.6)

T = 1.75 seconds

t = 0.65T

t = 0.65 x 1.75

t = 1.14 seconds

x = vt

x = (ωx)t

x = (3.6 x 0.07) x 1.14

x = 0.28 m

At position x = 0, the glider is at rest, the velocity is zero and the kinetic energy will be zero.

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Answer:

work done = ( force × displacement)

(a)The force acting on the block is it's self weight and displacement is equal to height of the tower.

work done by gravity = (250 × 10) = 2500 joule

(b) The work done by gravity 2500 joule is transferred to the object in the form of it's kinetic energy.

The maximum force constant of the spring Kmax is 2337.9 N/m.

The force constant or spring constant is defined as the force required to stretch or compress a spring such that the displacement in the spring is 1 meter.

Force constant is denoted by K and its unit is N/m.

Where;

K = spring constant

x = displacement

The work done by the spring is given below as follows:

Work done = Fx/2

Kinetic Energy = mv²/2

Force on an inclined plane = mgsinθ

Total force, F = mgsinθ + frictional force

F = 1390 * sin 22° + 515

F = 1035.7 N

Work done = change in KE

Fx/2 = mv²/2

Fx = mv²

m  = 1390/9.81 = 141.692

Solving for x;

x = mv²/F

x = 141.692 * 1.8²/1035.7

x = 0.443 m

The maximum force constant of the spring Kmax = 1035.7/0.443

Kmax = 2337.9 N/m

In conclusion, the maximum force constant of the spring  is the ratio of the total force and displacement.

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Note that the complete question is given below:

You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1490 N will move with speed 2.0 m/s at the top of a ramp that slopes downward at an angle 21.0 ∘. The ramp will exert a 533 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 8.0 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp. Calculate the maximum force constant of the spring Kmax that can be used in order to meet the design criteria

Answer:

M V R = constant      angular momentum is constant because  no forces act in the direction of V

Since M (mass) = constant

V R = constant

The force is directed along the gravitational force vector (towards the center of rotation)