which is the hottest and coldest planet in solar system ?​

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:

Vertebrate zoology

Explanation:

Have a great day!

Answer: ??? what the

Explanation:

Answer:

18:equilibrium

19:10.8

20:separating the surfaces with a layer of air

21:it remains motionless

22:cm/s

Answer:

incompressible and expandable

Answer:

30 m/s.

Explanation:

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.)

Answer:

No.

Explanation:

In the question, no mathematical angle is shown.

Answer:

So correct sentence is:

So correct sentence is:

Answer:

Hey There!

Refer to the attachment...

or in this way also you can solve shown in second attachment

Radiation is the only method of heat transfer that does not require matter.

Facts about Radiation :

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:

y-component = 45.75N

x-component = 0N

Explanation:

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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Answer:

D

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:

Let L be the length of the wire.

N λ = M L      where N is the number of wavelengths λ. Also, M and are integers and L is the length of the wire

There must be nodes at the ends of the wire and anti-nodes at the points of zero displacement - λ/2, λ, 3λ/ 2, etc.

The simplest would be 2 λ = L   or L = λ / 2

In such a situation one would have

N-A-N    or node-antinode-node

There is a 180 degree phase change upon reflection of the wave creating the standing wave

The distance between nodes or anti-nodes must be λ/2 and there must  be nodes at the ends of the wire

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:

i can

Explanation:

i know what's the answer

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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The car is accelerating at 3 m/s² in the positive direction (to the right). By Newton's second law, the net force on the car in this direction is

∑ F = F[a] - F[f] - F[air] = ma

3100 N - 200 N - F[air] = (650 kg) (3 m/s²)

Solve for F[air] :

F[air] = 3100 N - 200 N - (650 kg) (3 m/s²)

F[air] = 3100 N - 200 N - 1950 N

F[air] = 950 N

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:

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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Answer: 300N

Explanation:

Impulse= Mass * Velocity

F.T = M * V

F= MV/T

F= (0.15*20)/ 0.01

F= 300N

The distance traveled by the egg after 2 seconds is 19.6 m.

The given parameters:

The distance traveled by the egg after 2 seconds is calculated as follows;

[tex]h = v_0_y t + \frac{1}{2} gt^2\\\\[/tex]

where;

[tex]h = 0 \ + \ \frac{1}{2} \times 9.8 \times (2)^2\\\\h = 19.6 \ m[/tex]

Thus, the distance traveled by the egg after 2 seconds is 19.6 m.

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

[tex]\\ \sf\Rrightarrow \dfrac{F}{a}=m[/tex]

Or

[tex]\\ \sf\Rrightarrow m=\dfrac{F}{a}[/tex]

As the spring is compressed, it performs

-1/2 (252 N/m) (0.118 m)² ≈ -1.75 J

of work on the clay, while gravity does

(0.263 kg) (9.80 m/s²) (0.118 m) ≈ 0.304 J

on it. Then the total work W performed on the clay ball is approximately -1.45 J.

By the work-energy theorem, W is equal to the change in the clay ball's kinetic energy ∆K. If v is the speed with which it hits the spring, then

W = ∆K

-1.45 J = 0 - 1/2 mv²

Solve for v :

-1.45 J = -1/2 (0.263 kg) v²

v² ≈ 11.0 m²/s²

v ≈ 3.32 m/s