a water balloon is thrown a target at 18 m/s. if the water balloon has a mass of 0.4 kg, what is the momentum?

The above question requires a personal answer about your perception of corals, environmental conservation, and the activity you did in the classroom. In that case, I can't answer your question, but I'll show you how to answer it.

Please consider the following information to answer your questions:

In summary, coral reefs must be protected so that the natural marine balance is maintained.

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The relation of the kinetic energy allows to find the correct result for the energy stored in the electromagnet is:

B) 283 MJ

Kinetic energy is the energy due to the movement of bodies.

          K = ½ m v²

Where K is the kinetic energy, m is  the mass and v the spped.

They indicate that the weight of the bodye is W = 981 N and its final velocity is v = 7 [tex]v_s[/tex].

          W = m g

Since the projectile starts from rest, its initial velocity is zero, therefore the change in energy is

        ΔK = [tex]K_f - K_o = K_f[/tex]  

we substitute

        ΔK = ½ 981 / 9.8 (7 340) ²

         ΔK = 2.835 10⁸ J

They indicate that all the energy of the electromagnet is transformed into the energy of the projectile,

          Em = K

When reviewing the results, the correct one is:

       B) 283 MJ

In conclusion, using the relationship of kinetic energy we can find the correct result for the energy stored in the electromagnet is:

   B)  283 MJ

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When the speed of a car increases by a factor of 50%, the minimum braking distance will also be increased by a factor.

Braking distance will increase by a factor  2.25

Solution

From Newton's equation of motion, we can say that;

v² = u² + 2as

Where initial velocity is zero, we have;

v² = 2as

s = v²/2a

s is the distance and v is the final speed.

50% increase in speed means it has increased by a factor of 1.50

Hence we have

1.50²v² is directly proportional to the 1.50²d

Distance = 1.50²v² = 2.25d

Hence, breaking distance will increase by a factor 3.0625

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By Newton's second law, the net force acting on the block in the vertical direction is

∑ F [ver] = n - mg = 0

where n = magnitude of normal force and mg = weight of the block. It follows that n = mg.

When the block is at rest, the applied force X will not be enough to move the box until it can overcome the maximum mag. of static friction. If µ[s] is the coefficient of static friction, then the maximum mag. of the frictional force is

f = µ[s] n = µ[s] mg

The net horizontal force would be

∑ F [hor] = X - µ[s] mg = 0

so a minimum force of X = µ[s] mg is required to get the block moving. Any mag. smaller than this and the block stays at rest/in equilibrium.

Once the mag. of X exceeds µ[s] mg, the block will begin to move. At that point, if the coefficient of kinetic friction is µ[k], then the net force on the block is

∑ F [hor] = X - µ[k] mg = 0

so a minimum force of X = µ[k] mg would be needed to keep the block moving at constant speed, or otherwise X = µ[k] mg + ma if the block is accelerating with mag. a.

The principles here are captured in the attached plot.

Answer: 300N

Explanation:

Impulse= Mass * Velocity

F.T = M * V

F= MV/T

F= (0.15*20)/ 0.01

F= 300N

Newton's second law.

__________________

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The impulse-momentum relationship is a direct result of Newton's second law of motion.

The impulse-momentum theorem states that the impulse applied to an object is equal to the change in its momentum. It proves that the change in momentum of an object depends not only on the amount of force applied but also on the duration of force applied.

Newton's second law states that the force acting on an object is equal to the rate of change of its momentum.

This means that a force applied to an object will cause a change in its momentum. The impulse-momentum relationship describes the relationship between the force applied to an object and the resulting change in its momentum.

The impulse-momentum relationship states that the impulse acting on an object is equal to the change in its momentum.

Impulse is defined as the force applied to an object over a period of time, while momentum is the product of an object's mass and velocity. Therefore, the impulse-momentum relationship can be expressed as:

Impulse = Change in momentum

This relationship is important in understanding the behavior of objects in motion, particularly in collisions or other situations where forces act over a period of time.

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From Newton's second law of motion, the average force on the ball by the racket is 98 Newtons. The correct answer is option C

Given that a 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. And the time for contact with the racket is 0.04 s, that is,

mass m = 0.07 kg

velocity v = 56 m/s

time t = 0.04 s

force f = ?

To calculate the average force on the ball by the racket, let us apply Newton's second law of motion.

Impulse = change in momentum

ft = mv

Substitute all the parameters into the equation above

0.04f = 0.07 x 56

make f the subject of the formula

f = 3.92 / 0.04

f = 98 N

Therefore, the average force on the ball by the racket is 98 Newtons. The correct answer is option C

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The average force on the ball by the racket is 98 N. The correct option is the third option - 98 N

From the question, we are to determine the average force on the ball by the racket.

From the formula,

[tex]F = \frac{mv}{t}[/tex]

Where F is the force

m is the mass

v is the velocity

and t is the time

From the given information

m = 0.07 kg

v = 56 m/s

t = 0.04 s

Putting the parameters into the formula,

we get

[tex]F = \frac{0.07 \times 56}{0.04}[/tex]

[tex]F = \frac{3.92}{0.04}[/tex]

F = 98 N

Hence, the average force on the ball by the racket is 98 N. The correct option is the third option - 98 N

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Answer:

mountains are limited in their theoretical height by several processes. First is isostasy: the bigger a mountain gets, the more it weighs down its tectonic plate, so it sinks lower. ... Bottom line: mountains can get taller than Mount Everest in earth gravity, like the Appalachians probably did—but not much taller.

Answer:

It probably won't get any taller, though. From a geological standpoint, the Appalachians haven't seen much growth in quite a while. Since the dawn of the dinosaurs about 225 million years ago, this range has been getting whittled down by weathering forces.

Explanation:

Answer:

4.80 mol N2O5/min

Explanation:

The balanced equation tells us that 2 moles of N2O5 are required for every 1 mole of O2.  Therefore:

(2.40 mol O2/min)*(2 mol N2O5/mol O2) = 4.80 mol N2O5/min

Answer:

0.1g to 0.0000001g hope it helps uu

Answer:

Forces come in pairs.

Explanation:

Distances are between two points. The distance from A to B is equal but opposite the distance from B to A.

Answer:

1kW = 1000W

600W = 0.6kW

Cost of electric bill = 0.6kWh × 24 × 30 × $0.05

                               = $21.60

Average velocity is defined as the ratio in change in position to change in time,

v[ave] = ∆x/∆t

which on its own doesn't have anything to do with acceleration.

If acceleration is constant, the average velocity is the literal average of the initial and final velocities,

v[ave] = (v[final] + v[initial]) / 2

If this constant acceleration has magnitude a, the final velocity can be expressed in terms of the initial velocity by

v[final] = v[initial] + a*t

and plugging this into the previous equation gives

v[ave] = (v[initial] + a*t + v[initial])/2

v[ave] = v[initial] + 1/2*a*t

If the body in consideration is initially at rest, then

v[ave] = 1/2*a*t

which might be the relation you're looking for. But bear in mind the conditions I've underlined.

If acceleration is not constant and changes over time, so that the acceleration is some function of time a(t), then you can determine the velocity function v(t) by using the fundamental theorem of calculus. You need to know a particular velocity for some time to completely characterize v(t), though. For example, if you're given the initial velocity v[initial] = v(0), then

[tex]\displaystyle v(t) = v(0) + \int_0^t a(u) \, du[/tex]

or if you know any other velocity for some time t₀ > 0,

[tex]\displaystyle v(t) = v(t_0) + \int_{t_0}^t a(u) \, du[/tex]

Answer:

D

Explanation:

Explanation:

hope it's useful for you knows

Answer:

Vertebrate zoology

Explanation:

Have a great day!

Answer:

Approximately [tex]5.11 \times 10^{-19}\; {\rm J}[/tex].

Explanation:

Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.

Look up the Rydberg constant for hydrogen: [tex]R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}[/tex].

Look up the speed of light in vacuum: [tex]c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}[/tex].

Look up Planck's constant: [tex]h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}[/tex].

Apply the Rydberg formula to find the wavelength [tex]\lambda[/tex] (in vacuum) of the photon in question:

[tex]\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}[/tex].

The frequency of that photon would be:

[tex]\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}[/tex].

Combine this expression with the Rydberg formula to find the frequency of this photon:

[tex]\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}[/tex].

Apply the Einstein-Planck equation to find the energy of this photon:

[tex]\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}[/tex].

(Rounded to three significant figures.)

Using Newton's second law

[tex]\\ \sf\Rrightarrow F=ma[/tex]

[tex]\\ \sf\Rrightarrow 300=2m[/tex]

[tex]\\ \sf\Rrightarrow m=150kg[/tex]

Hey there!

The formula of “mass” in physics is:

m = F/a

Whereas “f” is ‘force’, “a” is ‘acceleration’, & “m” is your ‘mass’ of course.

mass = 300 Net force/2 acceleration

300 Net force/2 acceleration = m

mass = 150

Therefore, your answer is: 150 kg

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

Answer:

18:equilibrium

19:10.8

20:separating the surfaces with a layer of air

21:it remains motionless

22:cm/s

Answer:

the slope of velocity-time graph represent an object acceleration

The magnitude of the magnetic field 0.01 m from the same wire is 0.2 T.

Given to us:

Magnetic field, [tex]B_1 = 0.1\ T[/tex]

Radius of wire, [tex]R_1 = 0.02\ m[/tex]

To find out the magnitude of the magnetic field 0. 01 m from the same wire, we need to find out current first. we will use the formula,

[tex]B = \dfrac{\mu_oI }{2\pi R},\\\rn\\where,\\B= magnetic\ field\\\mu_o = 4\pi\times 10^{-7} m\cdot kg\cdot s^{-2} A^{-2}\ is\ the\ magnetic\ constant\\I= current\\R= radius\ of\ the\ wire[/tex]

Putting the values,

[tex]B_1 = \dfrac{\mu_oI }{2\pi R_1},\\\rn\\\\0.1= \dfrac{4\times \pi \times 10^{-7}\times I}{2\times \pi\times0.02}\\\\I=10,000\ A[/tex]

Now, for [tex]B_2[/tex]

[tex]B_2 = \dfrac{\mu_oI }{2\pi R_2},\\\rn\\\\B_2= \dfrac{4\times \pi \times 10^{-7}\times 10,000}{2\times \pi\times0.01}\\\\B_2= 0.2\ T[/tex]

Hence, the magnitude of the magnetic field 0. 01 m from the same wire is 0.2 T.

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Hi there!

We can use the same equation for Gravitational Force:

[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]

Fg = force due to gravity (N)

G = gravitational constant

m1,m2 = masses of objects (kg)

r = distance between objects (m)

Plug in the values provided:

[tex]F_g = (6.67*10^{-11})\frac{(2500)(3000)}{8^2} = \large\boxed{7.814 * 10^{-6}N }[/tex]

[tex]\huge\bf\underline{\underline{\pink{A}\orange{N}\blue{S}\green{W}\red{E}\purple{R:-}}}[/tex]

Here we've been given,

We have to find the gravitational attraction force (Fg) = ?

The standard formula to solve is given by,

[tex]:\implies\tt{F_g = g \frac{m_1m_2}{ {r}^{2} } } [/tex]

[tex]:\implies\tt{F_g = 6.67 \times {10}^{ - 11} \times \frac{(2500 )(3000)}{ {8}^{2} } }[/tex]

[tex]:\implies\tt{F_g = 6.67 \times {10}^{ - 11} \times \frac{7500000}{64} }[/tex]

[tex]:\implies\tt{F_g = 7.814 \times {10}^{ - 6} }[/tex]

Answer:

i can

Explanation:

i know what's the answer

Answer:

You should try your best to answer the question.

Answer:

There are many forces involved in the game of football. These are: Force of Gravity, Normal Force, Force of Friction, and Applied Force. Force of Gravity applies to football when the football is thrown or kicked, when a player jumps in the air to avoid a tackle or catch a ball, and is constantly being applied.

Explanation:

How physics is used in football?

When you throw a football across the yard to your friend, you are using physics. You make adjustments for all the factors, such as distance, wind and the weight of the ball. The farther away your friend is, the harder you have to throw the ball, or the steeper the angle of your throw.

Answer:

So correct sentence is:

So correct sentence is:

The motion of the package can be described as the motion of a projectile,

given that it has an horizontal velocity and it is acted on by gravity.

Reasons:

Velocity of the aircraft = 70 m/s

Direction of flight of the aircraft = Eastward

Height from which the aircraft drops the package, h = 800 m

a) The initial velocity of the package, [tex]\vec{v}[/tex] = 70·i

b) The time it will take the package to reach the ground, t, is given by the formula;

[tex]\displaystyle h = \mathbf{\frac{1}{2} \cdot g \cdot t^2}[/tex]

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

[tex]\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}[/tex]

Which gives;

[tex]\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}[/tex]

The time it will take the package to reach the ground, t ≈ 12.77 seconds

c) The vertical velocity just before the package reaches the ground, [tex]v_y[/tex], is given as follows;

[tex]v_y^2[/tex] = 2·g·h

Therefore;

[tex]v_y[/tex] = √(2·g·h)

Which gives;

[tex]v_y[/tex] = √(2 × 9.81 × 800) ≈ 125.28

[tex]v_y[/tex] ≈ 125.28 m/s

Which gives; [tex]\vec{v}[/tex] = 70·i - 125.28·j

Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51

The speed of the package as it lands, |v| ≈ 143.51 m/s

d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.

Therefore;

The path of the package is the path of a projectile, which is a parabolic shape.

e) As seen by someone on the aeroplane, the horizontal velocity will be

zero, therefore, the package will appear as accelerating directly vertical

downwards.

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