pressure exerted on a surface

The initial speed of the Snow Leopard was 35m/s if it jumped at an angle of 35 degrees.

When a leopard jumps. It's path will be a parabola which is corresponding to a projectile.

The angle at which the leopard jumped is 35°.

The horizontal range of the projectile is 96 meters.

We can use the formula for horizontal range R of the projectile,

R = U²sin2A/g

Where,

U is the initial velocity of leopard,

A is the angles at which the leopard jumped.

g is acceleration due to gravity having value 9.8m/s².

Putting all the values,

96 = U²sin(70°)/9.8

U² = 96×9.8/0.77

U² = 1222

U = 34.9 m/s

Nearly 35 m/s.

The initial speed of leopard is 35m/s.

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Given:

the mass of the weight-watcher is

The work off equivalent to

Required: height climbed by the person

Explanation:

first we need to change the work into calories.

it is given that

then the work done is

now change this work done from calories to joules.

we know that

Then the work done is ,

as the person climbs to the mountain the work done is stored as potential energy.

we assume that a person attained some height h,

then the work-energy relation,

Plugging all the values in the above relation and solve for h, we get

Thus, the height climbed by the person is

Answer:

what is the speed of the bird

Answer:

HOW TO DO IT…

Use the correct cutting oil on the tap when cutting threads.

Turn the tap clockwise one-quarter to one-half turn, then turn back three-quarters of a turn to break the chip.

When tapping a blind hole, use the taps. in the order starting, plug, and then bottoming.

Given:

Let's determine the preferred units used when measuring the velocity of the center of mass.

The center of mass can be said to be the mean position of matter in a given system.

This is the point in a body where the whole mass of the body is concentrated upon.

The SI unit for center of mass is meters (m).

When measuring the center of mass, the preferred unit is meters per second.

herefiore, when measuring the velocity of the center of mass, the preferred unit is meters/second.

• NSWER:

A. meters/second

The difference in the acceleration due to gravity on the moon and the Earth are different due to the difference in their masses and radius.

Yes, can shield a gravitational field in a vacuum by coming in between the mass and gravitational source.

Newton's law of universal gravitation states that every two objects in the universe attracts each other with a force directly proportional to the product of the masses and inversely proportional to the distance between the two objects.

F = GmM/R²

where;

From Newton's second law of motion,

F = mg

where;

mg  = GmM/R²

g = GM/R²

Thus, the acceleration due to gravity on the moon and the Earth are different because of difference in their masses and radius.

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Answer: i usually use the calendar on my phone to set my day or i make plans with other people.

The arrow be released at an angle of 18 degrees to hit the bull's-eye if its initial speed is 36.0 m/s. The arrow go over the branch.

The path of the arrow is projectile when it is launched by the archer.

The horizontal distance to be covered by the arrow is 77 meters.

The initial velocity is 36m/s while releasing.

(a) We can use the formula,

R = U²sin2A/g

Where,

U is the initial velocity of leopard,

A is the angles at which the leopard jumped.

g is acceleration due to gravity having value 9.8m/s².

Putting all the values,

77 = (36)²sin(2A)/9.8

Sin(2A) = 77×9.8/36×36

Sin(2A) = 0.58

2A = 36°

A = 18°

The angles at which the arrow should be launcher is 18°.

(b) The length of the branch of the tree is 3.5 meters.

Whether is should come it the way or not,

We have to find the height of the arrow.

H = U²Sin(A)/2g.

Where H is the height,

H = 36×36Sin(18°)/2×9.8

H = 36×36×0.3/2×9.8

H = 19.38 meters.

The height of arrow is more than of the height of branch of tree. So it will go over the branch.

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Drag is ignored in projectile predictions because projectiles usually have a relatively short time of flight.

The surface area of the object, the wind speed, as well as the relative velocity of the airplane will affect the drop of relief packages.

The drag force will cause the projectile to take a longer time to land and may cause to land far from its expected drop point.

Drag is a force that acts in opposition to the motion of an object moving through a fluid.

Drag can be thought of as friction in fluids because similar to friction, it acts in an opposite direction of the relative motion of a moving object.

For example, airplanes moving through air experience a drag; ships and boats moving through water experience drag too.

Drag also occurs in projectiles moving through the air. However, because of the relatively short time of flight, it is usually ignored in projectile motion.

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Complete question:

While studying projectile motion, we consider ideal scenarios, where the projectile travels along its trajectory only under the influence of gravity. In real-world situations, however, other forces act on the projectile.

Consider a cargo plane that is dropping relief packages to flood victims. In predicting and studying this motion, we might consider gravity, but ignore the horizontal and vertical forces associated with drag (or air friction). Discuss this simplification. Specifically address these questions:

•Why do we often ignore drag in projectile predictions?

•What conditions (of the object, its surroundings, and its launch) do you think might make drag a significant factor in the relief package drop?

•How would drag affect the projectile's motion if it really were a significant factor in the relief package drop?

Given that the time taken to reach the ground is t = 6 s

The final velocity is v = 0m/s

We have to find the height of the monument, h.

The initial velocity can be calculated by the formula

Here, g is the acceleration is due to gravity whose value is 9.81 m/s^2

Substituting the values, the initial velocity will be

The formula to find height is

Substituting the values, the height will be

Thus, the correct option is 177 m.

The potential energy of an object of mass m at a height h from the ground on the Earth, where the gravitational acceleration is g, is:

As we can see, the velocity of the object is not a variable that needs to be taken into account to find the potential energy,

Then, the answer is:

Answer:

U = 41.65 J

Explanation:

We have just to calculate it with the following expression of gravitational potential energy:

U = m*g*h

Where:

m: mass

g: gravity acceleration;

h: height.

So:

U = m * 9.8 * 0.85

U = 8.33 * m

Then, it is just putting the mass value of the dog in that equation to obtain the value of potential energy acquired by the dog when jumping.

The mass of this kind of dog is between 3.6kg - 5kg, so, the maximum value of the potential energy can be like this:

U = 5 * 9.8 * 0.85

U = 41.65 J

The mass of the Sun is 1.98 * 10³⁰ kg.

The mass of the Sun is calculated from the data given about Venus as follows:

radius of Venus from the Sun = 1.08 x 10¹¹ meters

Period of rotation of Venus around the Sun, T = 224.7 days.

T in seconds = 224.7 * 24 * 3600 = 1.94 * 10⁷ s

Force of gravity = Centripetal force

GMm/r₂ = mv²/r

M = mass of Sun

m = mss of Venus

M = rv²/G

G = 6.67 * 10⁻¹¹ m³kg⁻¹s⁻²

But v = 2πr/T

v = 2 * 3.14 * 1.08 x 10¹¹ / 1.94 * 10⁷

v = 3.496 * 10⁴ m/s

Hence;

M =  1.08 x 10¹¹ * (3.496 * 10⁴)² / 6.67 * 10⁻¹¹

M = 1.98 * 10³⁰ kg

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The final velocity of the plane at 250 m distance is 50 m/s.

To calculate the final velocity of plane we need to use the formula from equation of motion,

In equation of motion we will use the formula,

v² = u² + 2as

where,

v is final velocity

u is initial velocity

a is the acceleration of plane

s is the distance

In this given data are,

s = 250 m

a = 5 m/s²

u = 0 m/s ( as the initial of plane is resting )

v = ?

Now, since we know the values, putting them into the formula,

v² = ( 0 )² m/s + 2 x (5 m/s²) x 250 m

v² = 2 x (5 m/s²) x 250 m

v² = 2,500 m/s

v = √2,500 m/s

v = 50 m/s

Hence, the final velocity of the plane at 250 m is 50 m/s.

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Answer:

mrs.casts

Explanation:

she's married

According to the give statement:

a)Work done on the gas is 567 J.

b)The change in its internal energy is 167J.

The temperature and condition of a substance are examples of internal energy. For instance, the internal energy of water is influenced by its temperature as well as whether it is in a solid, liquid, or gas state. Due to its condition, liquid water has greater stored energy than copper metal at the same temperatures.

The work done on the gas is evaluated using formulas:

W = -p * ΔV

Only the large volumes which is signified by large data Substituting the known values we obtain

W= 0.8 * 1.01325 * 10⁵ Pa * -1 (-7) * 10⁻³ m³

W = 567 J

Change in Internal energy is given by

ΔU=W - Q

=567−400)

=167J

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The magnitude of the walrus's acceleration that accelerates from 7.0 km/h to 34.5 km/h over a distance of 95m is 6 m / s²

( v + u ) / 2 = d / t

v = Final velocity

u = Initial velocity

t = Time

d = Distance

u = 7 km / h

v = 34.5 km / h

d = 95 m

( 34.5 + 7 ) / 2 = 95 / t

t = 95 / 20.75

t = 4.58 s

a = v - u / t

a = Acceleration

a = ( 34.5 - 7 ) / 4.58

a = 6 m / s²

Therefore, the magnitude of the walrus's acceleration is 6 m / s²

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Solar system formation began approximately 4.5 billion years ago, when gravity pulled a cloud of dust and gas together to form our solar system.

The planets, moons, asteroids and the whole lot else in the solar device shaped from the small fraction of material in the region that wasn't included within the younger sun.

The nebular hypothesis is the most extensively regularly occurring version within the field of cosmogony to explain the formation and evolution of the sun gadget. The protoplanetary disk is an accretion disk that feeds the significant superstar.

Protostar are the stars are thought to shape internal giant clouds of cold molecular hydrogen massive molecular clouds kind of 300,000 instances the mass of the solar and 20 parsecs in diameter.

Beneath positive circumstances the disk, that could now be called protoplanetary, may additionally deliver beginning to a planetary device. Large planet core formation is notion to proceed more or less alongside the lines of the terrestrial planet formation.

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The acceleration of the Camila on the turn with a radius of curvature of 42.4 m at a speed of 7.17 m/s is 1.2 m / s / s

a = v² / R

a = Centripetal acceleration

v = Linear velocity

R = Radius

v = 7.17 m / s

R = 42.4 m

a = 7.17² / 42.4

a = 1.2 m / s / s

Centripetal acceleration is the rate of change of tangential velocity of the body moving in a circular path.

Therefore, the acceleration of the car is 1.2 m / s / s

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ANSWER

Wave A = Lowest volume with the lowest pitch

Wave B = Highest volume with the lowest pitch

Wave C = Lowest volume with the highest pitch

Wave D = Highest volume with the highest pitch

EXPLANATION

We want to attach the correct option to the correct figure.

From the question, the amplitude is proportional to volume, and frequency is given to be perceived as pitch.

On a wave graph, the amplitude is represented on the y-axis while the period is represented on the x-axis.

The relationship between period and frequency is an inverted relationship, which means that for a wave, the higher the period, the lower the frequency, and vice versa.

This means that from the figures given, we can conclude that:

Wave A = Lowest volume with the lowest pitch

Wave B = Highest volume with the lowest pitch

Wave C = Lowest volume with the highest pitch

Wave D = Highest volume with the highest pitch

That is the answer.

Answer:

the mouse will overshoot and will not land in the nest.

a ) The tension in a vertical strand = 68.6 * [tex]10^{-5}[/tex] N

b ) The tension in a horizontal strand = 149.13 * [tex]10^{-5}[/tex] N

c ) Tension in horizontal strand / Tension in vertical strand = 2.17

a ) The tension in a vertical strand,

m = 7 * [tex]10^{-5}[/tex] kg

g = 9.8 m / s²

Since the strand is vertical and the spider is motionless,

∑ [tex]F_{y}[/tex] = 0

T - mg = 0

T = 7 * [tex]10^{-5}[/tex] * 9.8

T = 68.6 * [tex]10^{-5}[/tex] N

b ) The tension in a horizontal strand,

θ = 13.0°

The strands left and right have same magnitude and makes the same angle with the horizontal. Resolving,

sin θ = [tex]T_{y}[/tex] / T

[tex]T_{y}[/tex] = T sin 13.0°

∑ [tex]F_{y}[/tex] = 0

[tex]T_{y}[/tex] + [tex]T_{y}[/tex] - mg = 0

T sin 13.0° + T sin 13.0° = 7 * [tex]10^{-5}[/tex] * 9.8

2 T sin 13.0° = 68.6 * [tex]10^{-5}[/tex]

T = 68.6 * [tex]10^{-5}[/tex] / 0.46

T = 149.13 * [tex]10^{-5}[/tex] N

c ) Ratio of strands,

Tension in horizontal strand / Tension in vertical strand = 149.13 * [tex]10^{-5}[/tex] / 68.6 * [tex]10^{-5}[/tex]

Tension in horizontal strand / Tension in vertical strand = 2.17

Therefore,

a ) The tension in a vertical strand = 68.6 * [tex]10^{-5}[/tex] N

b ) The tension in a horizontal strand = 149.13 * [tex]10^{-5}[/tex] N

c ) Tension in horizontal strand / Tension in vertical strand = 2.17

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a ) The acceleration in a vertical takeoff from the Moon = 0.81 m / s²

b ) No, it could not lift off from Earth. Because, the thrust of the module's engines is less than its weight on Earth.

a ) Vertical takeoff from the Moon,

F = 31000 N

m = 12500 kg

[tex]g_{m}[/tex] = 1.67 m / s²

W = m g

[tex]W_{m}[/tex] = 12500 * 1.67

[tex]W_{m}[/tex] = 20875 N

∑ [tex]F_{y}[/tex] = m [tex]a_{y}[/tex]

F -  [tex]W_{m}[/tex] = m [tex]a_{y}[/tex]

31000 - 20875 = 12500 [tex]a_{y}[/tex]

[tex]a_{y}[/tex] = 0.81 m / s²

b ) Lift off from the Earth,

Force required for a 12500 kg object to lift off from Earth,

F = m g

F = 12500 * 9.8

F = 122500 N

Since the force required to lift off from Earth is much higher than the actual force, the module cannot lift off from Earth

Therefore,

a ) The acceleration in a vertical takeoff from the Moon = 0.81 m / s²

b ) No, it could not lift off from Earth. Because, the thrust of the module's engines is less than its weight on Earth.

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The initial velocity of the student as he leaves the ground is 3.57 m/s.

The velocity of the student as he leaves the ground can be calculated using the following kinematic equation for upward motion.

v² = u² - 2gh

where;

At the maximum height, the final velocity of the student = 0

0 = u² - 2gh

u²  =  2gh

u = √2gh

u = √(2 x 9.8 x 0.65)

u = 3.57 m/s

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THE MASS OF AN OBJECT IS 1100

a) there are 2 and 3 significant figures in the numbers 99.0 and 100.0

b)for the uncertainity of 1 in each number, the percentage uncertainity for in each is 1 an 1.01

Here the problem we are dealing with is related to significant numbers which are the number of digits in a value, frequently an estimation, that contributes to the degree of accuracy of the value. We begin counting significant figures at the primary non-zero digit. in the first case, we are given two numbers 99 and 100, where in  99  we have 2 significant numbers and in 100, there are 3 significant numbers. For the second case, the formula for calculating the uncertainity percentage is percentage = uncertainty /value x 100

So,the uncertainty is 1, so the  certaintiy percentage for three significant numbers are which is 100 , the uncertainty percentage = 1/100 *100 = 1

For 99 , the uncertainty percentage = 1/99* 100 = 1.01

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A parallel plate capacitor is connected to a battery that maintains a constant potential difference.

As we know,

Here, Q is the charge which is inversely proportional to the separation between the plates.

That means if the separation between the plates increases the magnitude of the charge on the plates decreases.

Thus, if the separation between the plates is doubled the magnitude of the charge will be cut in half.

Answer:

A

Explanation:

What is the Ksp expression for Ni3(PO4)2(s) in water? A) Ksp = [Ni2+]3[PO4]2(s) B) Ksp = [Ni2+]2[PO43-13 C) Ksp= (3x[Ni2+])(2x[PO43-]) C) Ks

Answer 0 votes

Answe

A) Ksp = [Ni2+]3[PO4]2(s)

The oscillation frequency of the spring is equal to 1.23 s⁻¹.

Gravitational potential energy can be described as the energy contained by the object because of its displacement of some height above the surface.

Given the height of the block before dropping, h = 3cm = 0.03 m

The amplitude of oscillation, A = 20 cm = 0.2m

The expression for the oscillation frequency is:

[tex]f = \frac{1}{2\pi }\sqrt{\frac{k}{m} }[/tex]

From the equation of motion: v²= u² + 2gh

v = √2gh

Because of the equilibrium between the block and the spring, the spring force is equal to the weight of the block.

[tex]kx = mg\\k =\frac{mg}{x}[/tex]

From the law of the conservation f energy, find the value of 'x' displacement:

[tex]m\frac{v^2}{2} + mgx = \frac{mg}{2x}(x+a )^2-mgA[/tex]

[tex]x^2 +2hx-A^2 =0[/tex]

[tex]x^2 + 2\times x \times 0.03- (0.2)^2 =0\\x^2 +0.0.6x -0.04 =0\\x = 0.172 \;m[/tex]

The oscillation frequency is equal to:

[tex]f = \frac{1}{2\pi }\sqrt{\frac{k}{m} }[/tex]

[tex]f = \frac{1}{2\pi }\sqrt{\frac{g}{x} }[/tex]

[tex]f = \frac{1}{2\pi }\sqrt{\frac{9.81}{0.172} }[/tex]

f = 1.20 s⁻¹

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