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DanielleElmas [232]

2 years ago

13

A bullet starting from rest accelerates uniformly at a rate of 1,250 meters per square second. What is the bullet's speed after

it has traveled 100 meters?

1 answer:

Tatiana [17]2 years ago

8 0

Answer:

125,000

Explanation:

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2 years ago

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Suppose you are designing an amplifier and loudspeaker system to use at a rock concert. You want to make it as loud as possible.

OverLord2011 [107]

Answer is given below

Explanation:

Audio power amplifiers are found in all types of sound systems, including sound reinforcement, public address and home audio systems, as well as musical instrument amplifiers such as guitar amplifiers.
This is the last electronic step in the general audio playback series before sending the signal to the loudspeaker. So when we want maximum volume or loud sound, we have to get it with maximum output and high input and low output impedance

4 0

2 years ago

A major league baseball pitcher throws a pitch that follows these parametric equations: x(t) = 142t y(t) = –16t2 + 5t + 5. The t

azamat

Answer:

(a) $x'(t)= 142$

(b) 142

(c) $y'(t)= -32t+5$

(d) 96.8 mph

(e) 0.426 s

(f) 0.061 rad

Explanation:

Velocity is a time-derivative of position.

(a) $x(t) = 142t$

$x'(t)= 142$

(b) Since $x'(t)= 142$ is independent of $t$ , it follows it was constant throughout. Hence, at any point or time, the horizontal velocity is 142.

(c) $y(t) = - 16t^2+5t+5$

$y'(t)= -32t+5$

(d) When it passes the home plate, the ball has travelled 60.5 ft (from the question). This is horizontal, so it is equivalent to $x(t)$ .

$x(t)= 142t = 60.5$

$t=\dfrac{60.5}{142}= 0.426$ .

In this time, the vertical velocity, $y'(t)$ is

$y(t)= -32\times0.426+5 = -8.632$

The speed of the ball at thus point is $s=\sqrt{142^2+(-8.632)^2}=142$ ft/s

To convert this to mph, we multiply the factor 3600/5280

$s=142\times\dfrac{3600}{5280}=96.8 \text{ mph}$

(e) The time has been determined from (d) above.

$t= 0.426$

(f) This angle is given by

$\theta=\tan^{-1}\dfrac{y'(t)}{x'(t)}$

$\theta=\tan^{-1}\dfrac{-8.632}{142}=\tan^{-1}-0.0607=3.47$ (Note here we are considering the acute angle so we ignore the negative sign)

In radians, this is

$\theta=3.47\times\dfrac{\pi}{180}=0.061 \text{ rad}$

6 0

2 years ago

A hot air balloon of total mass M (including passengers and luggage) is moving with a downward acceleration of magnitude a. As i

LUCKY_DIMON [66]

Answer:

The fraction of mass that was thrown out is calculated by the following Formula:

M - m = (3a/2)/(g²- (a²/2) - (ag/2))

Explanation:

We know that Force on a moving object is equal to the product of its mass and acceleration given as:

F = ma

And there is gravitational force always acting on an object in the downward direction which is equal to g = 9.8 ms⁻²

Here as a convention we will use positive sign with acceleration to represent downward acceleration and negative sign with acceleration represent upward acceleration.

Case 1:

Hot balloon of mass = M

acceleration = a

Upward force due to hot air = F = constant

Gravitational force downwards = Mg

Net force on balloon is given as:

Ma = Gravitational force - Upward Force

Ma = Mg - F (balloon is moving downwards so Mg > F)

F = Mg - Ma

F = M (g-a)

M = F/(g-a)

Case 2:

After the ballast has thrown out,the new mass is m. The new acceleration is -a/2 in the upward direction:

Net Force is given as:

-m(a/2) = mg - F (Balloon is moving upwards so F > mg)

F = mg + m(a/2)

F = m(g + (a/2))

m = F/(g + (a/2))

Calculating the fraction of the initial mass dropped:

$M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]$

5 0

2 years ago

A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. Th

alina1380 [7]

A) Zero

The motion of the shot is a projectile's motion: this means that there is only one force acting on the projectile, which is gravity. However, gravity only acts in the vertical direction: so, there are no forces acting in the horizontal direction. Therefore, the x-component of the acceleration is zero.

B) -9.8 m/s^2

The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:

$F=mg$

where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:

$a_y = g = -9.8 m/s^2$

where the negative sign means it points downward.

C) 7.6 m/s

The x-component of the shot's velocity is given by:

$v_x = v_0 cos \theta$

where

$v_0 = 12.0 m/s$ is the initial velocity

$\theta=51.0^{\circ}$ is the angle of the shot

Substituting into the equation, we find

$v_x = (12.0 m/s)(cos 51^{\circ})=7.6 m/s$

D) 9.3 m/s

The y-component of the shot's velocity is given by:

$v_y = v_0 sin \theta$

where

$v_0 = 12.0 m/s$ is the initial velocity

$\theta=51.0^{\circ}$ is the angle of the shot

Substituting into the equation, we find

$v_y = (12.0 m/s)(sin 51^{\circ})=9.3 m/s$

E) 7.6 m/s

We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:

$v_x = 7.6 m/s$

F) -11.1 m/s

The y-component of the velocity at time t is given by:

$v_y(t) = v_y + at$

where

$v_y = 9.3 m/s$ is the initial y-velocity

a = g = -9.8 m/s^2 is the vertical acceleration

t is the time

Since the total time of the motion is t=2.08 s, we can substitute this value into the equation, and we find:

$v_y(2.08 s)=9.3 m/s + (-9.8 m/s^2)(2.08 s)=-11.1 m/s$

where the negative sign means the vertical velocity is now downward.

3 0

1 year ago