Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?

(was not able, but I don't know the reason)

Answer:

Explanation:

we know that,

force = mass × acceleration

∴ since speed/velocity is constant, acceleration should be zero.

∴ f = m × 0

f = 0 N

∴ If we apply this to ,

work = force × displacement

we get ,

w = 0 × s

∴ we can say that the net work is zero.

and hence the answer is true!!!

Answer:

See explanation below

Explanation:

For two vectors A and B to have a positive resultant, they must both move in the positive direction i.e along the positive x axis

For two vectors A and B to have a zero resultant, they must both have the same magnitude but move in the opposite direction i.e one in the positive direction while the other in the negative direction.

Fie example if A = +5N, B = -5B

So that A+B = +5-5 = 0N

Answer:

allows for efficient transport of materials

Explanation:

Living cells relatively have a small size, so small that a microscope is needed to see their structure. However, this small size is for their benefitting as it increases their surface area to volume ratio (SA:V).

A large surface area to volume ratio (SA:V) enables easy movement of molecules to and fro the cell membrane via the process of DIFFUSION. Therefore, having a small size allows for efficient transport of materials needed in the cell.

Answer:

Explanation:

Given;

output power of the incandescent lightbulb , P = 120 W

input voltage, V = 200 V

The resistance of the filament is calculated as;

[tex]P = \frac{V^2}{R}[/tex]

Where;

R is the resistance of the filament

[tex]R = \frac{V^2}{P} \\\\R = \frac{200^2}{120} \\\\R = 333.33 \ ohms[/tex]

The rms current is given as;

[tex]P = I_{rms} V_{rms}\\\\I_{rms} = \frac{P}{V_{rms}} \\\\I_{rms} =\frac{120}{200}\\\\ I_{rms} = 0.6 \ A[/tex]

Answer: The red at the bottom represents the base of the tube, not the red liquid.

Explanation:

The densest materials have more weight per unit of volume.

This means that those elements will always flow to the bottom of the containers, like the one in the image.

The red liquid being the least dense one, can not go to the bottom by its own means.

There could be some cases, like:

The red liquid when solid, is way denser than in its liquid phase, and then the red at the bottom could be solid phase of the red liquid, but there is no mention of this in the question, then we can discard this idea.

Another trivial idea is that the red liquid at the bottom could be trapped by some kind of wall, but again, there is no mention of this, so again we can discard this idea.

The thing that makes sense is that the red at the bottom represents the base of the tube and not the red liquid.

Answer:

Low-temperature blackbody

Explanation:

There are 3 types of blackbody temperatures.

Low-temperature blackbody

High temperature extended area blackbody

High-temperature cavity blackbody

A Low-temperature blackbody is a type of black body radiation that has the range of -40° C to 175° C, typically between 233 K and 448 K. A perfect fit for the temperature range mentioned in the question, "a few hundred Kelvin". Therefore, it's the kind of blackbody temperature that the object would emit.

Answer:

You have a displacement of 5 units to the right.

Explanation:

First you go three to the right which lands on the 3 mark. Then you move it 4 to the left which substracts 4, landing the object at -1. Finally you move 6 to the right, and you finish at marker 5. Since displacement is not total distance but just final distance from the start point directly to end point, it is only a displacement of 5.

Answer:

I think it would be +5

Explanation:

I looked on a number line and started at 0.

I then moved the object 3 units to the right and got 3.

I then moved the object 4 units the the left and got -1.

I then moved the object 6 units to the right and my final answer was positive 5. (+5 or just 5)

I hope I was right and helped

A battery of 9 volts is connected to a resistor of 1000 ohms, then the value of current flowing through it will be 0.009 A.

Any flow of electrical charge carriers, such as ions, holes, or subatomic charged particles, is referred to as electrical current.

When electrons serve as the charge transfer in a wire, the amount of charge moving through any point of wire in a given amount of time is measured as electric current. Electric charges' motion is intermittently reversed in the alternating current, but not in direct current.

In many situations, the direction of movement in electric circuits is assumed to be the direction of positive charge flow, which is the way opposite of the real particle drift. When properly specified, the current is known as conventional current.

From the question,

V = ir

i = v/r

i = 9/1000

i = 0.009 A

Therefore, the value of the current flowing through it is  0.009 A.

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Answer:

Approximately [tex]1.69 \times 10^{30}\; \rm kg[/tex].

Explanation:

Let [tex]M[/tex] and [tex]m[/tex] denote the mass of the star and the planet, respectively.

Let [tex]G[/tex] denote the constant of universal gravitation ([tex]G \approx 6.67408 \times 10^{-11}\; \rm m^{3} \cdot kg^{-1}\cdot s^{-2}[/tex].)

Let [tex]r[/tex] denote the orbital radius of this planet (assuming that [tex]r\![/tex] is constant.) The question states that [tex]r = 1.65 \times 10^{10}\; \rm m[/tex].

The size of gravitational attraction of the star on this planet would be:[tex]\displaystyle \frac{G \cdot M \cdot m}{r^{2}}[/tex].

If attraction from the star is the only force on this planet, the net force on this planet would be [tex]\displaystyle \frac{G \cdot M \cdot m}{r^{2}}[/tex].

Let [tex]\omega[/tex] denote the angular velocity of this planet as it travels along its circular orbit around the star. The size of [tex]\omega\![/tex] could be found from the period [tex]T[/tex] of each orbit: [tex]\omega = (2\, \pi) / T[/tex].

In other words, this planet of mass [tex]m[/tex] is in a circular motion with radius [tex]r[/tex] and angular velocity [tex]\omega[/tex]. Therefore, the net force on this planet should be equal to [tex]m \cdot \omega^2 \cdot r[/tex].

Hence, there are two expressions for the net force on this planet:

Equate the right-hand side of these two equations:

[tex]\displaystyle \frac{G \cdot M \cdot m}{r^2} = {\left(\frac{2\pi}{T}\right)}^{2}\, m \cdot r[/tex].

Simplify this equation and solve for [tex]M[/tex], the mass of the star:

[tex]\displaystyle M = \frac{{(2\pi / T)}^2 \cdot r^3}{G}[/tex].

Notice that [tex]m[/tex], the mass of the planet, was eliminated from the equation. That explains why this question could be solved without knowing the exact mass of the observed planet.

Convert the orbital period of this star to standard units:

[tex]\begin{aligned}T &= 14.5\; \text{day} \times \frac{24\; \text{hour}}{1\; \text{day}} \times \frac{3600\; \text{second}}{1\; \text{hour}} \\ & = 1.2528 \times 10^{6}\; \rm \text{second}\end{aligned}[/tex].

Calculate the mass of the star:

[tex]\begin{aligned} M &= \frac{{(2\pi / T)}^2 \cdot r^3}{G} \\ &\approx \frac{\displaystyle {\left(\frac{2\pi}{1.2528 \times 10^{6}\; \rm s}\right)}^{2} \times \left(1.65 \times 10^{10}\; \rm m\right)^{3}}{6.67408 \times 10^{-11}\; \rm m^{3}\cdot kg^{-1} \cdot s^{-2}}\\ &\approx 1.69 \times 10^{30}\; \rm kg\end{aligned}[/tex].

Answer:

d. 1.4 m/s

Explanation:

Given;

initial velocity of the `basketball, u = 4.0 m/s

acceleration of the basketball, a = -0.5 m/s² (leftward)

distance the basketball traveled, d = 14 m

the final velocity of the basketball, v = ?

(this final velocity should be less than the initial velocity since the ball is slowing down at a constant rate)

Apply the following the kinematic equation;

v² = u² + 2ad

v² = 4² + 2(-0.5)14

v² = 16 - 14

v² = 2

v = √2

v = 1.41 m/s

Therefore, the final velocity of the basketball is 1.4 m/s.

Answer:

20m

Explanation:

The two tens cancel each other out, as they are in opposite directions. Now we only care about the 20m, which if we have no 10's, will end up 20m away.

The distance covered by man is 40 m and the displacement is 20 m towards the east.

The distance can be defined as the measurement of the total ground covered by an object during its motion. It is a scalar quantity.

The displacement can be defined as the measurement of the shortest path over the ground is covered by an object during its motion. It is a vector quantity.

Given that a man starts walking towards the north and covered 10 m. he turns east and walks for 20 m then turns right and walks for 10 m. The attachment shows the total ground area covered by the man.

The total distance covered by the man is given below.

[tex]D = 10 + 20 + 10[/tex]

[tex]D = 40 \;\rm m[/tex]

The displacement of the man is the difference between its final position and initial position.

[tex]d = 20 - 0[/tex]

[tex]d = 20 \;\rm m[/tex]

Hence we can conclude that the distance covered by man is 40 m and the displacement is 20 m towards the east.

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Answer:

if the angular momentum is constant, the skater is closer, going faster.  

  w₂ = r₁² / r₂² w₁

Explanation:

For the skaters to perform in the turn in l, the angular number of the same must be equal, so we stammer the concept of conservation of angular momentum

             L₁ = L₂

the angular momentum is

           L = I w

we substitute

             I₁ w₁ = I₂ w₂

             w₂ = [tex]\frac{I_{1} }{I_{2} }[/tex] w₁

if we consider the skater as a particle, his moment of inertia is

             I = m r²

we substitute

              w₂ = r₁² / r₂² w₁

therefore, if the angular momentum is constant, the skater is closer, going faster.

Answer:

As you know, the denser objects have more weight per unit of volume, this will mean that the force that pulls down these objects is a bit larger.

This will mean that the denser objects will always go to the bottom.

This clearly implies that the red liquid, the one with one of the smaller densities, can not be at the bottom.

There are some cases where a liquid with a small density may become a lot denser as the temperature or pressure changes, and in a case like that, we could see the red liquid at the bottom, but for this case, there is no mention of changes in the temperature nor in the pressure, so this can be discarded.

The only thing that makes sense is that the red part at the bottom is the base of the tube, and has nothing to do with the red liquid.

Answer:

The value is  [tex]f_1 = 828 \ Hz[/tex]

Explanation:

From the question we are told that

    The speed is [tex]v = 25 \ m/s[/tex]

     The speed of the ambulance is [tex]u = 35 \ m/ s[/tex]

     The frequency of the siren is  [tex]f = 850 \ Hz[/tex]

Generally from Doppler effect equation we have that

        [tex]f_1 = \frac{v_s - v }{ v_s + u } * f[/tex]

Here [tex]v_s[/tex] is the velocity of sound with the value  [tex]v_s = 343 \ m/s[/tex]

  =>   [tex]f_1 = \frac{343 - 25 }{ 343 + 35 } * 850[/tex]

  =>   [tex]f_1 = 828 \ Hz[/tex]

Answer:

the correct one is the a,       U₂ = 2 U₁

Explanation:

The gravitational potential energy is

            U = m g h

if the stones reach the same height, the energy of the first stone is

            U₁ = m₁ g h

The second stone is twice the mass and reaches the same height

             m₂ = 2 m₁

potential energy is

           U₂ = m₂ g h

           U₂ = 2 m₁ g h

           U₂ = 2 U₁

therefore the energy is double.

When reviewing the statements, the correct one is the a

Answer:

W = 0

Explanation:

As the satellite moves in a circle the force is perpendicular to the path, therefore the work that is defined by

      W = F. r = f r cos θ

Since the force and the radius are perpendicular, the angle θ = 90º and the cosine 90 = 0, therefore there is no work for the circular motion.

W = 0

Answer:

a)   α = 0.0334 rad / s² ,  b) w = 2.14  rad/s see that the angular velocity doubles.

Explanation:

This is a magular kinematics exercise

Let's reduce the magnitudes to the SI system

   w = 0.17 rev /s (2π rad / 1rev) = 1.07 rad / s

a) as part of rest its initial velocity is zero w or = 0

        w = w₀ + α t

        α = [tex]\frac{\omega -\omega_{o} }{t}[/tex]

        α = [tex]\frac{1.07-0}{32}[/tex]

         α = 0.0334 rad / s²

b) If we double the angular relation what will be the final velocity

       w = w₀ + (2α) t

       w = 0 + 2 0.0334 32

       w = 2.14  rad/s

We see that the angular velocity doubles.

Answer:

v₀ = 677.94 m / s ,   θ = 286º

Explanation:

We can solve this exercise using the kinematic expressions, let's work on each axis separately.

X axis

has a relation of aₓ = 5.10 m / s², the motor is on for a time of t = 675 s, reaching the speed vₓ = 3630 m / s, let's use the relation

         vₓ = v₀ₓ + aₓ t

         v₀ₓ = vₓ - aₓ t

let's calculate

         v₀ₓ = 3630 - 5.10 675

         v₀ₓ = 187.5 m / s

Y Axis

        [tex]v_{y}[/tex] = v_{oy} - a_{y} t

         v_{oy} = v_{y} - a_{y} t

let's calculate

        v_{oy}  = 4276 - 7.30 675

         v_{oy} = -651.5 m / s

we can give the speed starts in two ways

a)   v₀ = (187.5 i ^ - 651.5 j ^) m / s

b) in the form of module and angle

Let's use the Pythagorean theorem

            v₀ = [tex]\sqrt{v_{ox}^{2} + v_{oy}^{2} }[/tex]

            v₀ = [tex]\sqrt{187.5^{2} +651.5^{2} }[/tex]

            v₀ = 677.94 m / s

we use trigonometry

            tan θ = [tex]\frac{v_{oy} }{v_{ox} }[/tex]

            θ = tan⁻¹ \frac{v_{oy} }{v_{ox} }

            θ = tan⁻¹ ([tex]\frac{-651.5}{187.5}[/tex])

            θ = -73.94º

This angle measured from the positive side of the x-axis is

            θ‘ = 360 - 73.94

            θ = 286º

Answer:

a. Potential energy of the sled is increased when height of sled is increased.

b. Potential energy of the sled is increased when height of sled is increased.

c. P.E₂₀ = 2 P.E₁₀

d. P.E₂₀₀ = 2 P.E₁₀₀

Explanation:

The potential energy of the sled can be given by the following:

[tex]Potential\ Energy = P.E = mgh\\[/tex]

where,

m = mass of sled

g = acceleration due to gravity

h = height of sled

a.

It is clear from the formula that potential energy of sled is directly proportional  to the height of sled.

Therefore, potential energy of the sled is increased when height of sled is increased.

b.

It is clear from the formula that potential energy of sled is directly proportional to the mass of sled.

Therefore, potential energy of the sled is increased when mass of sled is increased.

c.

[tex]P.E\ at\ 10\ m:\\P.E_{10} = 10mg\\P.E\ at\ 20\ m:\\P.E_{20} = 20mg\\\frac{P.E_{20}}{P.E_{10}} = \frac{20mg}{10mg}[/tex]

P.E₂₀ = 2 P.E₁₀

d.

[tex]P.E\ at\ 100\ kg:\\P.E_{100} = 100gh\\P.E\ at\ 200\ m:\\P.E_{200} = 200gh\\\frac{P.E_{200}}{P.E_{100}} = \frac{200gh}{100gh}[/tex]

P.E₂₀₀ = 2 P.E₁₀₀

Answer:

Where is the picture

Explanation:

WHERE IS THE PICTURE

Answer:

The total power is  [tex]P = 6.665 *10^{4} \ W[/tex]        

Explanation:

From the question we are told that

     The distance is  [tex]r = 10 \ km = 1000 \ m[/tex]

      The amplitude of the electric field is   [tex]E = 0.20 \ volt/meter[/tex]

Generally the average intensity of the electromagnetic field from the radio transmitter is mathematically represented  as

           [tex]I = \frac{E^2}{ 2 \mu_o * c }[/tex]

Here c is the speed of light with value  [tex]c = 3.0 *10^{8} \ m/s[/tex]

         [tex]\mu_o[/tex] is the permeability of free space with value  [tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]

So

           [tex]I = \frac{0.2^2}{ 2 * 4\pi *10^{-7} * 3.0*10^{8} }[/tex]      

=>        [tex]I = 5.307 *10^{-5} \ W/m^2[/tex]

Generally this intensity can also be mathematically represented as

               [tex]I = \frac{P }{ 4 \pi r^2 }[/tex]

=>           [tex]P = I ( 4 \pi r^2 )[/tex]        

=>           [tex]P = 5.307 *10^{-5} ( 4 * 3.142 * 1000^2 )[/tex]        

=>           [tex]P = 6.665 *10^{4} \ W[/tex]        

The total power emitted by the radio transmitter is [tex]6.67\times 10^4 \ W[/tex].

The given parameters;

The intensity of the radio wave from the transmitter is calculated as follows;

[tex]I = \frac{E^2}{2\mu_0 c} \\\\I = \frac{0.2^2 }{2\times 4\pi \times 10^{-7} \times 3\times 10^8} \\\\I = 5.305 \times 10^{-5} \ W/m^2[/tex]

The total power emitted by the radio transmitter is calculated as follows;

[tex]I = \frac{P}{A} \\\\P = IA\\\\P = I \times 4\pi r^2\\\\P = (5.305\times 10^{-5} )\times 4\pi \times (10,000)^2\\\\P = 6.67\times 10^{4} \ W[/tex]

Thus, the total power emitted by the radio transmitter is [tex]6.67\times 10^4 \ W[/tex].

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Answer: Depends on the context.

Explanation: While simply throwing a football won’t breach the sound barrier, a specialized jet can breach it easily. It all just depends on your definition of an object. Hope this helps!

Answer:

While there is no text here are some climate conditions you could go off of, tropical, dry, temperate, cold, and polar, hope this helps you with whatever your trying to do

Explanation:

Answer:

54.5 kmph

Explanation:

From work-kinetic energy principles, work done by friction on both pavement and gravel shoulder = kinetic energy change of vehicle

ΔK = W = -(f₁d₁ + f₂d₂) where f₁ = frictional force due to pavement = μ₁mg where μ₁ = coefficient of friction of pavement = 0.35, m = mass of vehicle and g = acceleration due to gravity = 9.8 m/s² and d₁ = distance moved by vehicle across pavement = 30 m and

f₂ = frictional force due to gravel shoulder = μ₂mg where μ₂ = coefficient of friction of pavement = 0.50, m = mass of vehicle and g = acceleration due to gravity = 9.8 m/s² and d₂ = distance moved by vehicle across gravel shoulder = 60 m

ΔK = 1/2m(v₁² - v₀²) where v₀ = initial velocity of vehicle, v₁ = final velocity of vehicle = 20 kmph = 20 × 1000/3600 = 5.56 m/s and m = mass of vehicle

So,

ΔK = -(f₁d₁ + f₂d₂)

1/2m(v₁² - v₀²) = -(μ₁mgd₁ + μ₂mgd₂)

1/2(v₁² - v₀²) = -(μ₁gd₁ + μ₂gd₂)

v₁² - v₀² = -2g(μ₁d₁ + μ₂d₂)

v₀² = v₁² + 2g(μ₁d₁ + μ₂d₂)

v₀ = √[v₁² + 2g(μ₁d₁ + μ₂d₂)]

substituting the values of the variables into the equation, we have

v₀ = √[(5.56 m/s)² + 2 × 9.8 m/s²(0.35 × 30 m + 0.5 × 60 m]

v₀ = √[30.91 (m/s)² + 4.9 m/s²(10.5 m + 30 m]

v₀ = √[30.91 (m/s)² + 4.9 m/s²(40.5 m]

v₀ = √[30.91 (m/s)² + 198.45 (m/s)²]

v₀ = √[229.36 (m/s)²

v₀ = 15.14 m/s

v₀ = 15.14 × 3600/1000

v₀ = 54.5 kmph

So, the initial speed of the vehicle is 54.5 kmph

Answer:

1 µC

Explanation:

From the question given above, the following data were obtained:

Electric field intensity (E) = 9000 N/C

Distance (r) = 1 m

Charge (Q) =?

NOTE: Electric force constant (K) = 9×10⁹ Nm²C¯²

The magnitude of the point charge can be obtained as follow:

E = KQ/r²

9000 = 9×10⁹ × Q / 1²

9000 = 9×10⁹ × Q

Divide both side by 9×10⁹

Q = 9000 / 9×10⁹

Q = 1×10¯⁶ C

Recall:

1 micro charge (µC) = 1×10¯⁶ C

Hence, the magnitude of the point charge is 1 µC

Answer:

Explanation:

The magnitude of the point charge is expressed as;

V = kQ/r

k is the coulombs constant = 9.0×10^9 N⋅m²/C²

V is the electric field = 9000N/C

Q is the charge

r is the distance = 1.00m

Get the required charge

Substitute the given parameters into the formula;

9000 = 9.0×10^9Q/1.0

9000 = 9.0×10^9Q

Q = 9000/9.0×10^9

Q = 9.0×10^3/9.0×10^9

Q = 9.0/9.0×10^(3-9)

Q = 1.0×10^6C

Q = 1ηC

hence the magnitude of the point charge is 1ηC

The mass m of the object = 5.25 kg

Given

k = spring constant = 3.5 N/cm

Δx= 30 cm - 15 cm = 15 cm

Required

the mass m

Solution

F=m.g

Hooke's Law

F = k.Δx

[tex]\tt m.g=k.\Delta x\\\\m.10=3.5\times 15\\\\m=5.25~kg[/tex]

Answer:

The value is  [tex]F = 1.568 *10^{-9} \ N[/tex]

Explanation:

From the question we are told that

     The mass  of the first lead sphere is [tex]m = 1.60 \ kg[/tex]

      The mass of the second lead sphere is  [tex]M = 16 \ g = 0.016 \ kg[/tex]

      The separation between masses is  [tex]r = 3.30 \ cm = 0.033 \ m[/tex]

Generally the gravitational force between each sphere is mathematically represented as

          [tex]F = \frac{G * m * M }{r^2 }[/tex]

Here G is the gravitational constant with value  [tex]G = 6.67 *10^{-11 } \ m^3 \cdot kg^{-1} \cdot s^{-2}[/tex]

         [tex]F = \frac{6.67 *10^{-11 } * 1.60 * 0.016 }{0.033^2 }[/tex]

=>       [tex]F = 1.568 *10^{-9} \ N[/tex]

Answer:

D. There are fewer younger adults than older adults.

Explanation:

Failed the assignment but got this answer correct.

Answer:

D is correct

Explanation:

Answer:

at the highest point the velocity and acceleration are perpendicular but when projectile starts to move are said to be parallel to one another

Explanation:

the clear explanation is shown above.