PLEASE HELP QUICKLY!A volleyball player jumps to hit a ball horizontally at 7.0 m/s straight on. If the
height at which the ball was hit is 3.0 m tall, how far did the ball go horizontally
before it hit the ground?
5.5 m
3.6 m
O 4.3 m
4.2 m

Answers

Answer 1

Answer:

Answer:

5.5 is the correct answer

please keeps as Brainly list

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A trumpet player hears 5 beats per second when she plays a note and simultaneously sounds a 440 Hz tuning fork. After pulling her tuning valve out to slightly increase the length of her trumpet, she hears 3 beats per second against the tuning fork. Was her initial frequency 435 Hz or 445 Hz? Explain.

Answers

Answer:

her initial frequency is 445 Hz

Explanation:

Given;

initial beat frequency, $F_B$ = 5

observed frequency, F = 440 Hz

let the initial frequency = F₁

F₁ = F ± 5 Hz

F₁ = 440 Hz ± 5 Hz

F₁ = 435 or 445 Hz

This result obtained shows that her initial frequency can either be 435 Hz or 445 Hz

The last beat frequency will be used to determine the actual initial frequency.

F = v/λ

Frequency (F) is inversely proportional to wavelength. That is an increase in length will cause a proportional decrease in frequency.

This shows that the final frequency is smaller than the initial frequency because of the increase in length.

Initial frequency - frequency of tuning fork = 5 beat frequency

Reduced initial frequency - frequency of tuning fork = 3 beat frequency

Initial frequency = 5Hz + 440 Hz = 445 Hz

Final frequency (Reduced initial frequency) = 440 + 3 = 443 Hz

Check: 445 Hz - 440 Hz = 5 Hz

443 Hz - 440 Hz = 3 Hz

An 800 kHz radio signal is detected at a point 2.1 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 800 mV/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. The intensity of the radio signal at that point is closest to

Answers

Answer:

$I=8.48* 10^(-4)\ W/m^2$

Explanation:

Given that,

Frequency of the radio signal, $f=800\ kHz=8* 10^5\ Hz$

It is detected at a pint 2.1 km from the transmitter tower, x = 2.1 km

The amplitude of the electric field is, E = 800 mV/m

Let I is the intensity of the radio signal at that point. Mathematically, it is given by :

$I=(E^2_(rms))/(c\mu_o)$

$E_(rms)$ is the rms value of electric field, $E_(rms)=(E)/(√(2) )$

$I=(E^2)/(2c\mu_o)$

$I=((800* 10^(-3))^2)/(2* 3* 10^8* 4\pi * 10^(-7))$

$I=8.48* 10^(-4)\ W/m^2$

So, the intensity of the radio signal at that point is $8.48* 10^(-4)\ W/m^2$ . Hence, this is the required solution.

. Using your knowledge of circular (centripetal) motion, derive an equation for the radius r of the circular path that electrons follow in terms of the magnetic field B, the electrons' velocity v, charge e, and mass m. You may assume that the electrons move at right angles to the magnetic field.2. Recall from electrostatics, that an electron obtains kinetic energy when accelerated across a potential difference V. Since we can directly measure the accelerating voltage V in this expierment, but not the electrons' velocity v, replace velocity in your previous equation with an expression containing voltage. The electron starts at rest. Now solve this equation for e/m.

You should obtain e/m = 2V/(B^2)(r^2)

3. The magnetic field on the axis of a circular current loop a distance z away is given by

B = mu I R^2 / 2(R^2 + z^2)^ (3/2)

where R is the radius of the loops and I is the current. Using this result , calculate the magnetic field at the midpoint along the axis between the centers of the two current loops that make up the Helmholtz coils, in terms of their number of turns N, current I, and raidus R.Helmholtz coils are separated by a distance equal to their raidus R. You should obtain:

|B| = (4/5)^(3/2) *mu *NI/R = 9.0 x 10^-7 NI/R

where B is magnetic field in tesla, I is in current in amps, N is number of turns in each coil, and R is the radius of the coils in meters

Answers

Answer:

Explanation:

Magnetic field creates a force perpendicular to a moving charge in its field which is equal to Bev where B is magnetic field , e is amount of charge on the moving charge and v is the velocity of charge particle .

This force provides centripetal force for creation of circular motion. If r be the radius of the circular path

Bev = mv² / r

r = mv / Be

2 ) If an electron is accelerated by an electric field created by potential difference V then electric field

= V / d where d is distance between two points having potential difference v .

force on charged particle

electric field x charge

= V /d x e

work done by field

= force x distance

= V /d x e x d

V e

This is equal to kinetic energy created

V e = 1/2 mv²

= 1/2 m (r²B²e² / m² )

V = r²B²e/ 2 m

e / m = 2 V/ r²B²

3 )

B = $(\mu* I* R^2)/(2(R^2+Z^2)^(3)/(2) )$

In Helmholtz coils , distance between coil is equal to R so Z = R/2

B = $(\mu* I* R^2)/(2(R^2+(R^2)/(4) )^(3)/(2) )$

For N turns of coil and total field due to two coils

B = $(\mu* I* N)/(R*((5)/(4))^(3)/(2) )$

= $(\mu* I* N)/(R)* ((4)/(5))^(3)/(2)$

= 9.0 x 10^-7 NI/R

A 1200-kg cannon suddenly fires a 100-kg cannonball at 35 m/s. what is the recoil speed of the cannon? assume that frictional forces are negligible and the cannon is fired horizontally.

Answers

Answer:

Recoil velocity of cannon = 2.92 m/s

Explanation:

By law of conservation of momentum, we have momentum of cannon = momentum of cannonball.

Mass of cannon = 1200 kg

Mass of cannon ball = 100 kg

Velocity of cannon ball = 35 m/s

We have, Momentum of cannon = momentum of cannon ball

1200 x v = 100 x 35

v =3500/1200 = 2.92 m/s

Recoil velocity of cannon = 2.92 m/s

Final answer:

The recoil speed of the cannon is 2.92 m/s.

Explanation:

To find the recoil speed of the cannon, we can use the conservation of momentum. The initial momentum of the cannon and cannonball system is zero since the cannon is at rest before firing. The final momentum is the sum of the momenta of the cannon and cannonball after firing. Using the equation:

Initial momentum = Final momentum

(mass of cannon) x (recoil speed of cannon) = (mass of cannonball) x (velocity of cannonball)

Plugging in the given values:

(1200 kg) x (recoil speed of cannon) = (100 kg) x (35 m/s)

Solving for the recoil speed of the cannon:

recoil speed of cannon = (100 kg x 35 m/s) / 1200 kg = 2.92 m/s

Learn more about recoil speed of cannon here:

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A lead ball is dropped into a lake from a diving board 6.10 mm above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 4.50 ss after it is released. How deep is the lake?

Answers

Answer:

D=1.54489 m

Explanation:

Given data

S=6.10 mm= 0.0061 m

To find

Depth of lake

Solution

To find the depth of lake first we need to find the initial time ball takes to hit the water.To get the value of time use below equation

$S=v_(1)t+(1/2)gt^(2) \n 0.0061m=(0m/s)t+(1/2)(9.8m/s^(2) )t^(2)\n t^(2)=(0.0061m)/(4.9m/s^(2) )\n t=\sqrt{1.245*10^(-3) }\n t=0.035s$

So ball takes 0.035sec to hit the water

As we have found time Now we need to find the final velocity of ball when it enters the lake.So final velocity is given as

$v_(f)=v_(i)+gt\nv_(f)=0+(9.8m/s^(2) )(0.035s)\n v_(f)=0.346m/s$

Since there are (4.50-0.035) seconds left for (ball) it to reach the bottom of the lake

So the depth of lake given as:

$D=|vt|\nD=|0.346m/s*4.465s|\nD=1.54489m$

Answer: d = 1.54m

The depth of the lake is 1.54m

Explanation:

The final velocity of the ball just before it hit the water can be derived using the equation below;

v^2 = u^2 + 2as ......1

Where ;

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled.

Since the initial velocity is zero, and the acceleration is due to gravity, the equation becomes:

v^2 = 2gs

v = √2gs ......2

g = 9.8m/s^2

s = 6.10mm = 0.0061m

substituting into equation 2

v = √(2 × 9.8× 0.0061)

v = 0.346m/s

The time taken for the ball to hit water from the time of release can be given as:

d = ut + 0.5gt^2

Since u = 0

d = 0.5gt^2

Making t the subject of formula.

t = √(2d/g)

t = √( 2×0.0061/9.8)

t = 0.035s

The time taken for the ball to reach the bottom of the lake from the when it hits water is:

t2 = 4.5s - 0.035s = 4.465s

And since the ball falls for 4.465s to the bottom of the lake at the same velocity as v = 0.346m/s. The depth of the lake can be calculated as;

depth d = velocity × time = 0.346m/s × 4.465s

d = 1.54m

The depth of the lake is 1.54m

A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106 N/C. If the test charge is replaced with another test charge of 23 mC, what happens to the external electric field at P

Answers

Answer:

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.

Explanation:

From coulomb's law, F = Eq

Thus,

F = E₁q₁

F = E₂q₂

Then

E₂q₂ = E₁q₁

$E_2 = (E_1q_1)/(q_2)$

where;

E₂ is the external electric field due to second test charge = ?

E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C

q₁ is the first test charge = 13 mC

q₂ is the second test charge = 23 mC

Substitute in these values in the equation above and calculate E₂.

$E_2 = (4*10^6*13)/(23) = 2.26 *10^6 \ N/C$

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.

However, the direction of the external field is still to the right.

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