a research submarine what is the maximum depth it can go

Answer:

Which feature is created when a block of rock dropped down in relation to the block of rock beside it?

Explanation:

Grabens drop down relative to adjacent blocks and create valleys. Horsts rise up relative to adjacent down-dropped blocks and become areas of higher topography.

Answer:

yes

Explanation:

Bouncing is a way to increase impulse. Because an object that bounces changes directions the force of impulse must be absorbed then generated by the target object. (Impulse is nearly doubled.)

Answer:

C

Explanation:

h= height = 19 m

g.e = gravity on earth = 9.8 m/s^2

g.u = gravity on Uranus = 9.8 x (88.9/100) = 8.71 m/s^2

Apply :

h = vot + 1/2 gt^2

h= 1/2 g t^2

t = √(2h/g)

Earth:

t= √(2(19)/9.8) = 1.97 s

Uranus

t= √(2(19)/8.71) = 2.09 s

Tu - Te = 2.09 - 1.97 = 0.12 s

Answer : 0.12 s

Answer:

The frequency of light can be calculated using the formula:

`c = λv`

Where `c` is the speed of light in a vacuum, `λ` is the wavelength of light, and `v` is the frequency of light.

The speed of light in a vacuum is `3.00 × 10^8 m/s`.

To convert the wavelength from nanometers to meters, we need to divide by `1 × 10^9`.

Thus, the frequency of light with a wavelength of 655 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(655 × 10^-9 m)`

`v = 4.58 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 655 nm is `4.58 × 10^14 Hz`.

Similarly, the frequency of light with a wavelength of 515 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(515 × 10^-9 m)`

`v = 5.83 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 515 nm is `5.83 × 10^14 Hz`.

Finally, the frequency of light with a wavelength of 475 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(475 × 10^-9 m)`

`v = 6.32 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 475 nm is `6.32 × 10^14 Hz`.

So, the frequency of light with a wavelength of 655 nm is `4.58 × 10^14 Hz`, the frequency of light with a wavelength of 515 nm is `5.83 × 10^14 Hz` and the frequency of light with a wavelength of 475 nm is `6.32 × 10^14 Hz`.

Answer:

9.09 s

Explanation:

If the sprinter ran the 100 meters at the constant speed of 11 m/s it would take him 9.09 s to cover the full distance.

We can find this number by dividing 100 meters (the distance covered) by 11 meters per second (the speed)

\(\frac{100}{11} =9.09\)

Answer:

10778292789403987593790

Explanation:

I am a Cow'

Answer:

0.102 kg

Explanation:

Work is measured in Joules, an amount of energy.  As the engine lifts the block, it will exert energy (doing work) to give that energy to the block in the form of gravitational potential energy.

Gravitational potential energy is given by the formula \(U=mgh\), where U is the gravitational potential energy, m is the mass of the object with the energy, g is the gravitational constant near the surface of the earth, and h is the height it was raised.

We can isolate "m" in the equation, substitute the known values into the equation, including g = 9.81 m/s^2, and calculate m.

\(\dfrac{U}{gh}=m\)

\(\dfrac{(1.00[J])}{(9.81[\frac{m}{s^2}])(1.00[m])}=m\)

Rewriting the units of Joules into its base SI units...

\(\dfrac{(1.00[\frac{kg \cdot m^2}{s^2}])}{(9.81[\frac{m}{s^2}])(1.00[m])}=m\)

Canceling units, and calculating...

\(0.1019367991845[kg]=m\)

Rounding to 3 sig figs...

\(0.102[kg]=m\)

Answer:

Vertical component: 161 m/s

Horizontal component: 192 m/s

Explanation:

Now the vertical component of velocity is found by

In our case

opposite = v_y

hypotenuse = v0 = 250 m/s

Therefore,

multiplying both sides by 250 gives

since Θ = 40, the above becomes

Evaluating the left-hand side gives

rounding to the nearest whole number

Hence, the vertical component of the velocity is 161 m/s.

Next, we find the horizontal component.

Now,

Now in our case

adajcent = v_x

hypotenuse = v0 = 250

therefore,

Multiplying both sides by 250 gives

since Θ = 40, the above becomes

The left-hand side evaluates to give

rounding to the nearest whole number gives

Hence, the horizontal component of the velocity is 192 m/s.

The net work done (in J) required to accelerate a 1500 kg car from 55 m/s to 65 m/s is 3127500 J

First, we shall obtain the initial kinetic energy. Details below:

KE₁ = ½mu²

= ½ × 1500 × 55²

= 41250 J

Next, we shall final kinetic energy. Details below:

KE₂ = ½mv²

= ½ × 1500 × 65²

= 3168750 J

Finally, we shall determine the net work done. Details below:

W = KE₂ - KE₁

= 3168750 - 41250

= 3127500 J

Thus, the net work done is 3127500 J

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Answer:Magnetic fields from two sources add up as vectors at each point, so the strength of the field is not necessarily the sum of the strengths1. Magnetic fields are vectors, which means they have direction as well as size. Therefore, the sum of two magnetic fields is not simply the sum of their magnitudes2.

Explanation:

Question:  Initially, the car travels along a straight road with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s, determine the constant deceleration of the car.

Answer:

1.29 m/s²

Explanation:

From the question,

a = (v-u)/t............................ Equation 1

Where a = deceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.

Given: v = 13 m/s, u = 35 m/s, t = 17 s.

a = (13-35)/17

a = -22/17

a = -1.29 m/s²

Hence the deceleration of the car is 1.29 m/s²

2.5 watt

Explanation:

we know

power = work done /time

work done gainst gravity is = mgh

as mg = weight = force = 5 N

and height = 2 m

so work done = mgh = 5*2 = 10N

therefore ,

Power = 10/4 = 2.5 watt

Answer:

(1.8 × 10^9) meters in one minute

Explanation:

To determine how far light can travel in one minute, we need to multiply its speed by the number of seconds in a minute.

The speed of light is 3 × 10^8 meters per second.

There are 60 seconds in a minute.

Therefore, the distance light can travel in one minute is:

Distance = Speed × Time

Distance = (3 × 10^8 meters per second) × (60 seconds)

Calculating this, we get:

Distance = 3 × 10^8 meters/second × 60 seconds

Distance = 18 × 10^8 meters

Distance = 1.8 × 10^9 meters

So, light can travel approximately 1.8 billion (1.8 × 10^9) meters in one minute.

Answer:

A

Explanation:

Answer: a or e

Explanation:

Answer:

B. 23 kg*m/s north

Explanation:

p_1 = p_2

(m*v)_1 + (m*v)_2 = (m*v)_1' + (m*v)_2'

(70kg*1.4m/s)+(75kg*-1.0m/s) = p_2

(98kg*m/s)+(-75kg*m/s) = 23kg*m/s north

B. 23 kg*m/s north

Two hockey players are about to collide on the ice, so the total momentum of the system when they collide is  23 kg m/s. Hence, option B is correct.

The result of a particle's mass and velocity is momentum. It has both magnitude and direction because momentum is a vector quantity. According to Isaac Newton's second law of motion, the force pushing a particle has an equal and opposite effect on the temporal rate at which its momentum changes.

\(Momentum (p)= Mass(m)*Velocity(v)\)

The given values according to the question is :

p₁ = p₂

(m × v)₁+ (m × v)₂ = (m × v)₁' + (m × v)₂'

(70 kg × 1.4 m/s) + (75 kg) × (-1.0 m/s) = p₂

(98 kg m/s) + (- 75 kg m/s) = 23 kg m/s North

Therefore, the total momentum after collision is 23 kg m/s North.

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From the histogram given, there were at approximately 7 downloads between 3pm and 4pm . This can be derived by counting the rows in that time period.

A histogram is a graph that depicts the frequency distribution of a few data points from a single variable.

Histograms frequently divide data into "bins" or "range groups" and count the number of data points that belong to each of those bins.

Histograms are frequently used to depict the key properties of data distribution in a handy format. It is especially beneficial when working with huge data sets (more than 100 observations). It can aid in the detection of uncommon observations (outliers) or gaps in the data.2

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The current in the series circuit  across a 9.0 V potential difference is determined as 0.333 A.

The current in the series circuit is determined by applying ohms law as shown below;

V = IR

where;

R = 12 Ω  + 15 Ω  = 27 Ω

I = V/R

I = 9/27

I = 0.333 A

Thus, the current in the series circuit  across a 9.0 V potential difference is determined as 0.333 A.

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The value of 'x' such that the mass то exerts a force of mg on the inclined surface at P. (N=mg)​ is given by

x = N / (mg cos θ)

Generally, To determine the value of 'x', you need to know the angle of inclination of the surface and the mass of the object. The force of gravity (N) acting on the object is equal to the mass of the object (m) multiplied by the acceleration due to gravity (g). The force of gravity is also equal to the normal force (N) exerted by the surface on the object, which is perpendicular to the surface.

If the surface is inclined at an angle θ with respect to the horizontal, the normal force will be equal to the component of the force of gravity along the surface. This component can be calculated using the formula N = mg cos θ, where θ is the angle of inclination and m is the mass of the object.

Therefore, to determine the value of 'x', you need to solve the equation N = mg cos θ for 'x'. To do this, you can rearrange the equation to solve for x:

x = N / (mg cos θ)

Substituting the values of N, m, and θ into this equation will give you the value of 'x' that you are looking for.

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For the first 6 kg block, the speed after the collision will be 2.5 m/s. For the second 6 kg block, the speed after the collision will be zero since it sticks to the first 6 kg block.

Speed is a measure of how quickly an object moves or how quickly a task is performed. It is usually measured in units of distance per unit of time, such as metres per second, miles per hour, or kilometres per hour. Speed is an important concept in physics and is used to describe motion of all kinds, from elementary particles to everyday objects such as cars, planes, and rockets.

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Answer:  it turns into energy

Explanation: In a nuclear fusion reaction, two reactants combine, and form new products. But the mass of the products is less than total mass of the reactants. This is because a part of mass is lost as energy. This can be seen in an example shown below. You can see that mass is of the products is less than that of the reactants.

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Answer:

a

  \(t = 3.798 \  s \)

b

\(S = 67.91 \ m \)

Explanation:

From the question we are told that

The height of the helicopter is \(h= 235 ft\)

The height of the bed of the truck is \(h_b = 3 \ ft\)

The speed of the truck is \(v = 40 \ miles / hour = \frac{40 * 1609.34}{3600} = 17.88 \ m/s\)

Generally the distance between the truck bed and the helicopter is mathematically represented as

\(H = h - h_b\)

=> \(H = 235 -3\)

=> \(H = 232 \ ft \)

Converting to meters

\(H = \frac{ 232}{3.281} = 70.7 \ m \)

Generally the time at which the helicopter should drop the package is mathematically represented as

\(t = \sqrt{\frac{2 * H}{g} }\)

\(t = \sqrt{\frac{2 * 70.7}{9.8} }\)

\(t = 3.798 \ s \)

Generally distance of the helicopter from the drop site at time t = 0 s is

\(S = v * t\)

=>     \(S =  17.88   *  3.798\)

=> \(S = 67.91 \ m \)

For the pulley of radius 7.0 cm connected to a motor that rotates at a rate of 6000 rad/s and then decelerates at 1000 rad/s within 5 seconds, we have:

a) Its angular acceleration is -1000 rad/s².

b) The number of rotations within the time range is 2785.2 revolutions.

c) The length of the string that winds it within the time range is 1225.5 meters.

d) The tangential acceleration of the string is -70 m/s².

e) The time taken for the pulley to stop from 1000 rad/s is 1 second.

a) The angular acceleration is given by:

\( \omega_{f} = \omega_{i} + \alpha t \)    (1)

Where:

\( \omega_{f} \): is the final angular velocity = 1000 rad/s

\( \omega_{i} \): is the initial angular velocity = 6000 rad/s

\( \alpha \): is the angular acceleration =?

t: is the time = 5 s

Solving equation (1) for α we have:

\( \alpha = \frac{\omega_{f} - \omega_{i} }{t} = \frac{1000 rad/s - 6000 rad/s}{5 s} = -1000 rad/s^{2} \)

Hence, the angular acceleration is -1000 rad/s². Its negative value is because the motor is decelerating.

b) The number of rotations can be calculated with the following equation:

\( \theta_{f} = \theta_{i} + \omega_{i}t + \frac{1}{2}\alpha t^{2} \)

Where:

\(\theta_{i}\): is the initial displacement or initial number of rotations = 0

\(\theta_{f}\): is the final displacement or the final number of rotations =?

\( \theta_{f} = 6000 rad/s*5 s - \frac{1}{2}1000 rad/s^{2}*(5 s)^{2} = 17500 rad*\frac{1 rev}{2\pi rad} = 2785.2 rev \)

Then, the number of rotations within the time range is 2785.2 revolutions.

c) The length of a rope that is winded within the time range, can be calculated knowing the perimeter of the pulley (circumference):

\( C = 2\pi r = 2\pi*0.070 m = 0.44 m \)

Now, since the pulley rotates 2785.2 revolutions in the interval time of 5 seconds, the length of the string is:

\(l = \frac{0.44m \: length}{1 \: rev}*2785.2 \: rev = 1225.5 m \: length\)

Hence, the length of the string is 1225.5 meters.

d) The tangential acceleration of the string is given by:

\( a = \alpha r \)

Where:

a: is the tangential acceleration =?

\( a = \alpha r = -1000 rad/s^{2}*0.070 m = -70 m/s^{2} \)                    

Therefore, the tangential acceleration of the spring is -70 m/s².

e) The time taken for the pulley to stop from 1000 rad/s can be calculated with equation (1):

\( t = \frac{\omega_{f} - \omega_{i}}{\alpha} \)

Where:

\( \omega_{f} \): is the final angular velocity = 0 (when the pulley stops)

\( \omega_{i} \): is the initial angular velocity = 1000 rad/s

\( t = \frac{\omega_{f} - \omega_{i}}{\alpha} = \frac{0 - 1000 rad/s}{-1000 rad/s^{2}} = 1 s \)

Then, the time taken for the pulley to stop is 1 second.

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The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

\(\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\\)

\(\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\\)

Substitute the given parameters and solve for the angular speed;

\(\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s\)

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

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The true statement about the electric potential for the equipotential surface is  \(V_A = V_B = V_C\)

A surface with an equipotential potential is one where all points on the surface have the same electric potential. .

That is an equipotential surface is that surface at every point of which, the electric potential is the same.

The formula for the potential across every point on the surface is given as;

V = F/Q x R

V = ER

where;

Since the surface is equipotential with uniform electric across the surface, the electric potential at any point across the surface will be the same.

So \(V_A = V_B = V_C\)

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Answer:b\(4.29\times 10^6\ N/C\)

Explanation:

Given

The radius of the ring is \(a=7.55\ cm\)

The charge on the ring is \(Q=7.06\times 10^{-6}\ C\)

Electric field along the axis at a distance \(x\) is given by

\(E=\dfrac{kxQ}{(x^2+a^2)^{\frac{3}{2}}}\)

To get the maximum value, differentiate \(E\) w.r.t \(x\) we get

\(E_{max}=\dfrac{2kQ}{3\sqrt{3}a^2}\quad [\text{at}\ x=\dfrac{a}{\sqrt{2}}]\\\\E_{max}=\dfrac{2\times 9\times 10^9\times 7.06\times 10^{-6}}{3\sqrt{3}(7.55\times 10^{-2})^2}\\\\E_{max}=\dfrac{24.456\times 10^3}{57\times 10^{-4}}=4.29\times 10^6\ N/C\)

At the end of 20 ns, the electron's velocity will be approximately 5.2 × 10⁶ m/s.

The force per unit charge exerted on a positive test charge that is at rest at a given position is the force per unit charge that is used to define the electric field analytically.

Electric charge or magnetic fields with variable amplitudes can produce an electric field. The attraction forces that keep together atomic nuclei and electrons at the atomic scale are brought on by the electric field.

Given that:

the electric field is 1.25 kN/C

So, acceleration of the electron: a = eE/m

= (1.6 × 10⁻¹⁹×1.25×10³)/(9.1×10⁻³¹) m/s²

= 2.6 × 10¹⁴  m/s².

Hence, after  the end of 20 ns the electron's velocity will be = at

= 2.6 × 10¹⁴ × 20 × 10⁻⁹ second

= 5.2 × 10⁶ m/s.

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Answer:

they are in imaginary world in this every thing is possible

Answer:

1.000153 T

Explanation:

The energy change ΔE = hc(1/λb - 1/λa)

= 6.626069 ✕ 10⁻³⁴ J · s 2.997925 × 10⁸ m/s(1/6.544750 × 10⁻⁷ m - 1/6.544550 × 10⁻⁷ m)

= 19.864457907 × 10⁻²⁶(1527942.2438 - 1527988.9374) = 19.864457907 × 10⁻²⁶(-46.6936)

= 927.543052 × 10⁻²⁶

= -9.275431× 10⁻²⁴ J.

This energy change ΔE = 2μBB. So the magnetic field, B is

B = ΔE/2μB where μB = 9.274009 ✕ 10⁻²⁴ J/T

B = -9.275431× 10⁻²⁴ J/9.274009 ✕ 10⁻²⁴ J/T = -1.000153 T

The magnitude of the magnetic field B = 1.000153 T