Please solve the Problem.

The diver has the initial velocity, both (a) magnitude is 9.8 m/s and (b) direction is  73.5°.

When an object is released from rest in free air considering no friction, the motion is depend only on the acceleration due to gravity, g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy(P.E)

U =mgh

When the object strikes the ground, all the potential energy converted into kinetic energy.

K.E = 1/2mv²

where v is the speed just before hitting the ground.

A diver springs upward from a board that is 4.40 m above the water. At the instant, she contacts the water her speed is 13.5 m/s and her body makes an angle of 78.1 ° with respect to the horizontal surface of the water.

(a)

From energy conservation principle, initial and final mechanical energy are equal.

1/2mu² + mgh = 1/2mv²

where, u is the initial velocity of the diver.

u = sq rt  (v² - 2gh)

u = sq rt (13.5² - 2x9.81x4.4)

u = 9.798 m/s or 9.8 m/s

Thus, the velocity of the diver is 9.8 m/s.

(b)

The horizontal component of velocity will remain constant.

The horizontal component of acceleration is zero.

Then,

ucosθ = vcosΦ

θ = cos⁻¹ [ (13.5 x cos 78.1)/9.8 ]

θ = 73.5°

Thus, the direction of velocity is  73.5°.

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a)]An expression for the magnitude of charge transferred Q is Q = W/V.

b) The magnitude of the charge, in coulombs is  1.02 x 10⁻⁹ Coulomb.

c) Number of electrons in this are 63.75 x 10⁸ electrons.

The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.

A balloon is charged until it has a voltage of v=9800V. A student reaches out and is shocked by a spark. The electric force does W= 1x10⁻⁶ J of work transferring the charge.

a) Work done = Charge x potential

W = Q x V

Q = W/V

b)Substitute the values into the expression, we have

|Q| =  1x10⁻⁶/9800

|Q| =  1.02 x 10⁻⁹ Coulomb

c) No of electrons  = total charge /charge on electron

n = 1.02 x 10⁻⁹/1.6 x 10⁻¹⁹

n = 63.75 x 10⁸ electrons

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(a) The average force experienced by the person for 0.85 cm compression is 3,722,425.8 N.

(b) The average force experienced by the person for 13 cm compression is 243,389.4 N.

The kinetic energy of the person is calculated as follows;

K.E = ¹/₂mv²

K.E = ¹/₂ x 62.5 x 22.5²

K.E = 15,820.31 J

E = ¹/₂Fx

2E = Fx

F = (2E)/x

F = (2 x 15,820.31)/(0.0085)

F = 3,722,425.8 N

F = (2 x 15,820.31)/(0.13)

F = 243,389.4 N

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Answer:

A parallel circuit has two or more paths for current to flow through.

Voltage is the same across each component of the parallel circuit.

The sum of the currents through each path is equal to the total current that flows from the source.

Explanation:

Answer:

  (b)  0.29 m/s

Explanation:

The average speed is found by dividing the net displacement by the amount of time it took to achieve that displacement.

__

The net displacement is ...

  6 m -8 m = -2 m . . . . . where north is positive

The total time is ...

  2 s + 5 s = 7 s

Then the distance per unit time (speed) is ...

  (2 m)/(7 s) = 2/7 m/s ≈ 0.29 m/s

From the calculations, the Pd across the capacitors are 360V and 240V respectively.

The capacitor is a device used to store charges. Since the charge for capacitors in series is the same;

1/C = 1/20 + 1/30 = 0.05 + 0.033

C = 12 * 10^-3 F

Q = 12 * 10^-3 F * 600 V

Q = 7.2  C

Voltage across the 20mf  capacitor = 7.2  C/20 * 10^-3

= 360 V

Voltage across the 30mf capacitor =  7.2  C/30 * 10^-3

=240 V

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The angular velocity of the propeller is 136.1 rad/s; linear velocity is 153.1 m/s; centripetal acceleration 20835.2 m/s² and 2123.8 g.

The angular velocity of the propeller in rad/s is given as follows:

1300 rev/min = 1300 * 2π/60 = 136.1 rad/s.

b. The linear velocity, v = radius * angular velocity

Linear velocity, v = 2.25/2 * 136.1

v = 153.1 m/s

c. Centripetal acceleration, [tex]a = \frac{v^{2}}{r}[/tex]

[tex]a = \frac{(153.1)^{2}}{1.125} = 20835.2\:ms^{2}[/tex]

Centripetal acceleration in terms of g; [tex]g = \frac{20835.2}{9.81} = 2123.8 g[/tex]

Therefore, the angular velocity of the propeller is 136.1 rad/s; linear velocity is 153.1 m/s; centripetal acceleration 20835.2 m/s² and 2123.8 g.

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The force of the object with the given mass and acceleration is determined as 10 N.

The force of the given object is calculated from Newton's second law of motion as follows;

F = ma

where;

F = 5 x 2 = 10 N

Thus, the force of the object with the given mass and acceleration is determined as 10 N.

The complete question is below:

Rearranging the equation m = F/a and solve for F, m= 5Kg and a= 2 m\s2

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The most significant change in the comet's energy as it moves from point D to point A KE increases, GPE decreases.Option C is correct.

Comets are frozen leftovers from the formation of the solar system composed of dust, rock, and ice.

They range from a few miles to tens of miles wide, but as they orbit closer to the Sun, they heat up and spew gases and dust into a glowing head that can be larger than a plane.

As the comet transitions from point D to point A, the biggest energy shift is seen. GPE declines as KE rises. The right answer is C.

Hence option C is correct.

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The  power the swimmer must provide to overcome that drag force is 50 W.

Work done is equal to product of force applied and distance moved.

Work = Force x Distance

Given is  an athlete, swimming at a constant speed, covers a distance of 203 m in a time period of 5 minutes and 50 seconds = 350 seconds. The drag force exerted by the water on the swimmer is 86.0 N.

Work done by drag force is  W= F x d

So, W = 86 N x  203 m

W = 17.458 kJ

Power , a swimmer is provided P = Work done/ Time

P = 17458 / 350

P = 50 W

Thus, the power, the swimmer must provide to overcome that drag force is  50 W.

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The acceleration of each object will be 2.31 m/s².While the direction for mass 1 is down, mass 2 is left and mass 3 is up.

The tension force is described as the force transferred through a rope, string, or wire as it is pulled by opposing forces.

The tension force is applied along the whole length of the wire, pulling energy equally on both ends.

From Newton's second law, the net force is equal to the product of the mass and acceleration. Forces and tension acting upward are positive while the downward is taken as negative.

Foer mass m₁ only vertical force is considered.

[tex]\rm \sum F_y=ma_y \\\\ + T_{12}-m_1g =-m_1a \\\\ -T_{12}+39.2 \ N=(4.00 \ kg )a[/tex]

For mass m₂ both horizontal and verticle force is considered;

[tex]\rm \sum F_x=ma_x \\\\ -T_{12}+f+T_{23}\\\\ -T_{12}+\mu_kN+T_{23} \\\\ T_{12}-0.350(9.80 \ N)-T_{23}=(1.00 \ kg)a[/tex]

[tex]\rm \sum F_y =ma_y \\\\ N-m_2g=0[/tex]

Foer mass m₃ only vertical force is considered.

[tex]\rm \sum F_y=ma_y \\\\ T{23}-m_2g=+m_3a \\\\ T_{23}-19.6 \ N=(2.00 \ kg)a[/tex]

Adding  all the obtained forces we get;

[tex]\rm +39.2 \ N-3.43 \ N-19.6 N=(7.00 \kg) a\\\\ a=2.31 \ m/s^2[/tex]

Hence, the acceleration of each object will be 2.31 m/s².While the direction for mass 1 is down, mass 2 is left and mass 3 is up.

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Answer:

I don't now the answer

Explanation:

I'm so sorry

Answer:

CₙH₂ₙ

Explanation:

Alkenes have at least one double bond between 2 of the parent chain carbons. If the general formula for alkanes (no double bonds) is CₙH₂ₙ₊₂, it makes sense that the general formula for alkenes is CₙH₂ₙ. This means that alkenes have 2 less hydrogens than their alkane counterparts.

Answer:

kajsskskw जिम्मेवार द्वन्द्वका दिदीबहिनी दिनेश दिनेश दिनु सायदै दिनेश दिनु दिनु द्वन्द्वका दिनु

The oceanic lithosphere is a layer on the Earth's crust that consists of rocks of the mantle.

The lithosphere can be defined as the outer rock layer present on the Earth's surface.

The oceanic lithosphere is composed of different rocks of the mantle such as, among others, gabbro and/or basalt.

The structure of the oceanic lithosphere is similar around the world because it is slender compared to the continental lithosphere and always determined by the supply of magma.

In conclusion, the oceanic lithosphere is a layer on the Earth's crust that consists of rocks of the mantle.

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Work done is potential energy here

[tex]\\ \rm\Rrightarrow PE=mgh[/tex]

[tex]\\ \rm\Rrightarrow 244292=1248(10)h[/tex]

[tex]\\ \rm\Rrightarrow 244292=12480h[/tex]

[tex]\\ \rm\Rrightarrow h=19.6m[/tex]

Answer:

We'll be using the formula here i.e.,

PE = mgh

Now, putting the values,

244292 = 1248(10) h

244292 = 12480h

h = 12480/244292

h = 19.6 m

Answer:x=20

Explanation:

The speed does the block move after it is hit by the bullet that remains stuck inside the block will be 23.7 m/sec and it takes 12.07 seconds to stop.

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

Apply the law of conservation of momentum principle;

m₁v₁+m₂v₂cosΘ =(m₁+m₂)V

3 kg ×  12 m/s +  0,1 kg × 400 m/s cos 20° = (3+0.1)V

36 + 40 cos 20° = 3.1 V

V=23.7 m/sec

The time it takes to stop when the friction coefficient between the block and the surface is 0.2 is found as;

V = u +at

V = 0+ μgt

23..7=0.2× 9.81 ×t

t=12.07 sec

Hence, it takes 12.07 seconds to stop.

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Answer:

(a) [tex]W\approx18834.7 \ J[/tex]

(b) [tex]P \approx 50.1 \ W[/tex]

(c) [tex]\text{Food Calories} \approx 4.5 \ cal}[/tex]

Explanation:

Part (a): To calculate the work done by the person, we can use the formula:

[tex]\rightarrow W= F d[/tex]

The force can be calculated using the person's mass and the acceleration due to gravity:

[tex]\rightarrow F=mg[/tex]

Given:

Number of steps, N = 133

Height of each step, h = 16.2 cm = 0.162 m

Time taken to walk, t = 6 minutes and 16 seconds = 376 seconds

Mass of the person, m = 89.2 kg

First, let's calculate the vertical distance traveled:

[tex]d=Nh\\\\\\\\\Longrightarrow d=(133)(0.162)\\\\\\\\\therefore d=21.546 \ m[/tex]

Now, we can calculate the work done:

[tex]W= F d\\\\\\\\\Longrightarrow W=mgd\\\\\\\\\Longrightarrow W=(89.2)(9.8)(21.546)\\\\\\\\\therefore \boxed{\boxed{W\approx18834.7 \ J}}[/tex]

Therefore, the person did approximately 18,515.7 joules of work.

Part (b): The average power performed by the person can be calculated using the formula:

[tex]\rightarrow P=\dfrac{W}{t}[/tex]

Calculating the average power by plugging in the appropriate values into the above formula:

[tex]P=\dfrac{W}{t}\\\\\\\\\Longrightarrow P= \dfrac{18834.7}{376} \\\\\\\\\therefore \boxed{\boxed{P \approx 50.1 \ W}}[/tex]

Therefore, the average power performed by the person during the walk was approximately 50.1 watts.

Part (c): To calculate the food Calories burned during the walk, we need to convert the work done from joules to kilojoules (kJ) and then to food Calories:

[tex]\rightarrow W \ \text{(in} \ kJ)=\dfrac{W \ \text{(in} \ J)}{1000}\\\\\\\\\Longrightarrow W \ \text{(in} \ kJ)=\dfrac{18834.7}{1000}\\\\\\\\\therefore W=18.8347 \ kJ[/tex]

[tex]\rightarrow\text{Food Calories} = \dfrac{ W \ \text{(in} \ kJ)}{4.1868} \\\\\\\\\Longrightarrow \text{Food Calories} = \dfrac{ 18.8347}{4.1868}\\\\\\\\\therefore \boxed{\boxed{ \text{Food Calories} \approx 4.5 \ cal}}[/tex]

Therefore, the person burned approximately 4.5 food Calories during the walk.

1- Selfishness

2- Bad Government

3- Segregation

From the definition of half life, the half life is when the number of daughter nuclei equals the number of parent nuclei.

The half life is the time required for only about half of the number of radioactive isotopes to  remain.

We know that, the half life of an isotope shows when the number of daughter nuclei equals the number of parent nuclei.

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If a wire lies within a magnetic field and an electric current is produced in the wire, the true statement about the magnetic field is option D.

The magnetic field is the region of space where a charged object experiences magnetic force when it is moving.

When a charged object moves in the magnetic field of other object, the magnetic force is experienced.

If a wire lies within a magnetic field and an electric current is produced in the wire, the magnetic field must be moving.

Thus, the correct option is D.

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Answer:

My answer is Drowsiness

Explanation:

i hope this helps if not I am sorry

The regions of Great Lakes that are having a localized maximum in snowfall are called: B. snowbelts.

A snowbelt can be defined as a geographical region (area) near the Great Lakes in North America that is typically characterized by heavy localized maximum in snowfall, especially off the southern and eastern shores.

This ultimately implies that, snowbelts are the regions of Great Lakes that are having a localized maximum in snowfall.

In conclusion, hurricanes are more likely to form in the tropical Atlantic during a La Nina year because it weakens the wind shear.

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The number of times that the weight must be lifted is 9 times.

The term calories refers to the energy value that is derived from food. Following the principle of energy conversion, we know that the energy in the food is converted to the work done as the weight is lifted.

Since work done = mgh = 30.0-kg * 9.8ms^2 * 0.550 m = 161.7 J or 38.6 Cal

Thus, for the number of  times the weight must be lifted;

38.6 n = 3.35 * 10^2

n =  3.35 * 10^2/38.6

n = 9 times

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The effort distance​ will be 160 cm.Applying the moment at the center as follows will provide the effort distance:

Mechanical advantage is a measure of the ratio of output force to input force in a system, it is used to obtain the efficiency of forces in levers and pulleys.

Given data;

Effort,[tex]\rm F_e=00 N[/tex]

Load,[tex]\rm P= 400 \ N[/tex]

Distance from the fulcrum,[tex]\rm d=40 \ cm[/tex]

The effort distance​ is found by applying the moment at the center as;

[tex]\rm F_e \times d= P \times d' \\\\ 200 \ N \times d'=800 \ N \times 40 \ cm \\\\ d'=160 \ cm[/tex]

Hence, the effort distance​ will be 160 cm

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The direction of the magnetic force is to the right.

The magnetic field is the region in space where the influence of the magnet is felt. The magnetic force is always in the direction of the magnetic field.

We can see from the image, that the direction of the magnetic force is to the right.

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Answer:

1

Explanation:

4.34 + 5.66 = 10

number of significant figures in 10: 1

(the significant figure here is 1)

here's a clear explanation from Northern Kentucky University (NKU)

[To determine the number of significant figures in a number use the following 3 rules:

Non-zero digits are always significant

Any zeros between two significant digits are significant

A final zero or trailing zeros in the decimal portion ONLY are significant

Example:  .500 or .632000 the zeros are significant

                .006  or .000968 the zeros are NOT significant ]

[For addition and subtraction use the following rules:

Count the number of significant figures in the decimal portion ONLY of each number in the problem

Add or subtract in the normal fashion

Your final answer may have no more significant figures to the right of the decimal than the LEAST number of significant figures in any number in the problem.]

(https://www.nku.edu/~intsci/sci110/worksheets/rules_for_significant_figures.html#:~:text=To%20determine%20the%20number%20of,decimal%20portion%20ONLY%20are%20significant)

Hi there!

A)
Since the ball's velocity is initially only in the HORIZONTAL direction, there is NO vertical component of its velocity. Therefore, we can treat this like a free-fall scenario. (Dropped from rest).

We can rearrange the following kinematic equation to make solving for the time taken for the balls to hit the ground easier.

[tex]d_y = v_yt + \frac{1}{2}at^2[/tex]

dy = vertical displacement (height of table, 0.85 m)

vy = initial vertical velocity (0 m/s, only horizontal)
a = acceleration (due to gravity in this situation, 9.8 m/s²)

t = time (? sec)
Simplify and rearrange the variables for 't'.

[tex]d_y = 0(t) + \frac{1}{2}at^2\\\\d_y = \frac{1}{2}at^2\\\\t^2 = \frac{2d_y}{a}\\\\t = \sqrt{\frac{2d_y}{a}}}[/tex]

Plug in the given values.

[tex]t = \sqrt{\frac{2(0.85)}{9.8}} = \boxed{0.4165 s}[/tex]

B)

Now, remember that the ball's acceleration in the vertical direction due to gravity does NOT impact its horizontal velocity. Its vertical and horizontal velocities are COMPLETELY independent. Thus, we can simply use the time solved for above and each ball's respective velocities in the following kinematic equation:
[tex]d_x = v_xt[/tex]

dₓ= horizontal displacement (? m)

vₓ = horizontal component of velocity (3.5 and 5.3 m/s for A and B respectively)

t = time (0.4165 s)

Solve for the distance traveled by each ball:
Ball A:
[tex]d_x = 3.5 * 0.4165 = \boxed{1.458 m}[/tex]

Ball B:
[tex]d_x = 5.3* 0.4165 = \boxed{2.207 m}[/tex]