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2 years ago

a pitcher threw a baseball straight up at 35.8 meters per second. what is the ball's velocity after 2.50 seconds?

1 answer:

5 0

Ok the velocity of an object in free fall is given by the equation :

v=v0-gt, where v0 is the original velocity, g is the gravitational constant (9.8 m/s^2) and t is the time.

so, we substitute values into this equation. v=35.8-9.8*2.5; v=11.3 m/s

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Motion maps for two objects, Y and Z, are shown.

Vesna [10]

Answer:

The answer is B) 3 seconds

Explanation:

I just took the test on 2020 edge and got it right

5 0

2 years ago

Two sinusoidal waves are identical except for their phase. When these two waves travel along the same string, for which phase di

Kamila [148]

Answer:

zero or 2π is maximum

Explanation:

Sine waves can be written

x₁ = A sin (kx -wt + φ₁)

x₂ = A sin (kx- wt + φ₂)

When the wave travels in the same direction

Xt = x₁ + x₂

Xt = A [sin (kx-wt + φ₁) + sin (kx-wt + φ₂)]

We are going to develop trigonometric functions, let's call

a = kx + wt

Xt = A [sin (a + φ₁) + sin (a + φ₂)

We develop breasts of double angles

sin (a + φ₁) = sin a cos φ₁ + sin φ₁ cos a

sin (a + φ₂) = sin a cos φ₂ + sin φ₂ cos a

Let's make the sum

sin (a + φ₁) + sin (a + φ₂) = sin a (cos φ₁ + cos φ₂) + cos a (sin φ₁ + sinφ₂)

to have a maximum of the sine function, the cosine of fi must be maximum

cos φ₁ + cos φ₂ = 1 +1 = 2

the possible values of each phase are

φ1 = 0, π, 2π

φ2 = 0, π, 2π,

so that the phase difference of being zero or 2π is maximum

6 0

2 years ago

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

Bezzdna [24]

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q= $1.6\times 10^{-19} C$

a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

8 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

(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.

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

(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

Cylinder A is moving downward with a velocity of 3 m/s when the brake is suddenly applied to the drum. Knowing that the cylinder

Xelga [282]

Answer:

Incomplete question

Check attachment for the given diagram

Explanation:

Given that,

Initial Velocity of drum

u=3m/s

Distance travelled before coming to rest is 6m

Since it comes to rest, then, the final velocity is 0m/s

v=3m/s

Using equation of motion to calculate the linear acceleration or tangential acceleration

v²=u²+2as

0²=3²+2×a×6

0=9+12a

12a=-9

Then, a=-9/12

a=-0.75m/s²

The negative sign shows that the cylinder is decelerating.

Then, a=0.75m/s²

So, using the relationship between linear acceleration and angular acceleration.

a=αr

Where

a is linear acceleration

α is angular acceleration

And r is radius

α=a/r

From the diagram r=250mm=0.25m

Then,

α=0.75/0.25

α =3rad/sec²

The angular acceleration is =3rad/s²

b. Time take to come to rest

Using equation of motion

v=u+at

0=3-0.75t

0.75t=3

Then, t=3/0.75

t=4 secs

The time take to come to rest is 4s

7 0

2 years ago