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3 years ago

Find the interest. A 105-day note for $14,200 at 8.5% interest.

Physics

1 answer:

mr Goodwill [35]3 years ago

7 0

<h3><u>Answer</u>;</h3>

$347.22

<h3><u>Explanation</u>;</h3>

Principal = $14,200

Rate = 8.5%

Time = 105 days = 105/365

Interest = Principal x Rate x Time

Interest = 14,200 x 0.085 x 105/365

Interest = 347.219

= $347.22

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Electrons flow from the negative terimal of a

gizmo_the_mogwai [7]

Answer:

Option B

Explanation:

The electrons flow from negative terminal of a battery to the positive terminal because as the charge of electron is negative, it will get repelled by the negative terminal of the battery

Conventional flow actually assumes that the current flows out of the positive terminal, through the circuit and into the negative terminal of the battery

It actually says that direction of flow of current is in opposite direction to the direction of flow of electrons

The direction of current will not change even a resistor is placed in the circuit and it generates a potential difference across the resistor

But the current tries to move in such a way in which there will be less resistance to the flow

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3 years ago

If the lily pads are spaced 2.4 m apart and Ferdinand jumps with a speed of 5 m/s taking 0.6 s to go from lily pad to lily pad,

Zina [86]

Answer:

$36.87^{\circ}$

Explanation:

v = Velocity of Ferdinand = 5 m/s

$\theta$ = Angle of jump

T = Time taken = 0.6 s

R = Distance between lily pads = 2.4 m

Horizontal range is given by

$R=v_xT\\\Rightarrow R=vcos\theta T\\\Rightarrow cos\theta=\dfrac{R}{vT}\\\Rightarrow \theta=cos^{-1}\dfrac{R}{vT}\\\Rightarrow \theta=cos^{-1}\dfrac{2.4}{5\times 0.6}\\\Rightarrow \theta=36.87^{\circ}$

The angle at which Ferdinand make each of his jumps is $36.87^{\circ}$

8 0

3 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

7 0

3 years ago

What happens when the crests of two waves overlap

gregori [183]

Energy of the waves are redistributed to form a resultant wave with amplitude given by the summation of individual wave's amplitude.
<span>If the two waves are of same frequency, speed and amplitude and travelling in opposite direction den stationary waves are form.</span>

4 0

3 years ago

A rocket has total mass Mi=360 kg , including Mf=330kg of fuel and oxidizer. In interstellar space, it starts from rest at the p

steposvetlana [31]

let the parameters are:

M=instantaneous mass of the rocket

v=velocity of the rocket

t=time

initial mass=Mi=360 kg

fuel mass=Mf=330 kg

relative speed=Ve=1500 m/s

rate of mass decay=k=2.5 kg/s

writing newton’s second law of motion:

M*du/dt=-Ve*dM/dt

==>du=-Ve*dM/M

Integration of both sides,

u=-Ve*ln(M)

using the limits with at t=0, u=0 and velocity being v(t),

at t=0, mass=Mi, at any time t, mass=Mi-k*t

v(t)-0=-Ve*ln((Mi-k*t)/Mi)

v(t)=-Ve*ln(1-(k*t/Mi))

<h3>What is needed for rocket propulsion?</h3>

Rocket propulsion is the strategy utilized to make the basic thrust to lift a rocket into the air. The force that the rocket employments to lift off from the soil is known as rocket propulsion. The third law of movement by Newton serves as the foundation for rocket propulsion. Here, the fuel is shot out from the exit mightily in arrange to cause an equal and converse reaction. The some sorts of rocket propulsion are fuelled by liquid, strong, cold gas, and particles. The rocket's mass, fuel burn rate, and weaken speed all impact speeding up.

To learn more about Rocket propulsion, visit;

brainly.com/question/15363207

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1 year ago