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tekilochka [14]

2 years ago

7

Nate throws a ball straight up to Kayla, who is standing on a balcony 3.8 m above Nate. When she catches it, the ball is still m

oving upward at a speed of 2.8 m/s. Required:With what initial speed did Nate throw the ball?

1 answer:

marin [14]2 years ago

8 0

Answer:

9.1m/s

Explanation:

Nate throws a straight ball to Kayla who is standing at a balcony 3.8m above Nate

When she catches the ball, it is still moving upward with a speed of 2.8m/s

v = 2.8m/s

u = ?

s = 3.8m

a= -9.8(The acceleration has a negative sign because the speed of the ball is declining)

Therefore the initial speed at which Nate threw the ball can be calculated as follows

v^2= u^2 + 2as

2.8^2= u^2 + 2(-9.8)(3.8)

7.84= u^2 + (-74.48)

7.84= u^2 - 74.48

u^2= 7.84 + 74.48

<h3 />

u^2= 82.32

u= √82.32

u = 9.1m/s

Hence the initial speed at which Nate threw the ball is 9.1m/s

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A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration

wel

Answer:

V=20cm/s

Explanation:

The average speed is the distance total divided the time total:

$V=X/T$

First stage:

T1=5s

$v_{f} =v_{o} - at$

But, $v_{f} =0$ (decelerates to rest)

then: $a =v_{o} /t=0.3/5=0.06m/s^{2}$

on the other hand:

$x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m$

X1=75cm

Second stage:

T2=5s

$x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m$

X2=125cm

Finally:

X=X1+X2=200cm

T=T1+T2=10s

V=X/T=20cm/s

8 0

2 years ago

The graph indicates Linda’s walk.

Sedaia [141]

I think the right answer is the first one. If she stops moving her Position does not change any more-and the Graph Shows that after 6 seconds she stays at the Position of 5 m. If she Went Back to the start point the Graph would have Developed Back to 0m(decreased).

3 0

2 years ago

Read 2 more answers

A girl is shown at position A on a swing when the seat is directly below the support bar. The seat is then at height A as shown

MrRa [10]

Answer:

<u></u>

<u>1. The potential energy of the swing is the greatest at the position B.</u>

<u>2. As the swing moves from point B to point A, the kinetic energy is increasing.</u>

Explanation:

Even though the syntax of the text is not completely clear, likely because it accompanies a drawing that is not included, it results clear that the posittion A is where the seat is at the lowest position, and the position B is upper.

The gravitational <em>potential energy </em>is directly proportional to the height of the objects with respect to some reference altitude. Thus, when the seat is at the position A the swing has the smallest potential energy and when the seat is at the <em>position B the swing has the greatest potential energy.</em>

Regarding the forms of energy, as the swing moves from point B to point A, it is going downward, gaining kinetic energy (speed) at the expense of the potential energy (losing altitude). When the seat passes by the position A, the kinetic energy is maximum and the potential energy is miminum. Then the seat starts to gain altitude again, losing the kinetic energy and gaining potential energy, up to it gets to the other end,

7 0

2 years ago

Read 2 more answers

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

DENIUS [597]

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle $\theta$ with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

Where $y_o$ is the initial height above the ground level, $v_{oy}$ is the vertical component of the initial velocity and t is the time

The y-component of the speed is

$v_y=v_{oy}-gt$

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of $v_o$

The object will reach the maximum height when $v_y=0$ . It allows us to compute the time to reach that point

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum heigh is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Solving for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Replacing the known values

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

5 0

2 years ago

When you urinate, you increase pressure in your bladder to produce the flow. For an elephant, gravity does the work. An elephant

weqwewe [10]

Answer:

a) v = 1.19 m / s , b) P₁ = 0.922 10⁵ Pa

Explanation:

1) Let's use the fluid continuity equation

Q = A v

The area of a circle is

A = π r2 = π d²/4

v = Q / A = Q 4 / pi d²

v = 0.006 4/π 0.08²

v = 1.19 m / s

2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point

P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rho g y₂

The exercise tell us

P₂ = 1.0013 105 Pa

v₁ = 0

y₁ = 1 m

y₂=0

Rho (water) = 1000 kg / m³

P₁ + rho y₁ = P₂ + ½ rho v₂²

P₁ = P₂ + ½ rho v₂² - rho g y₁

P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1

P₁ = 1.013 10⁵ +708.5 - 9800

P₁ = 92208.5Pa

P₁ = 0.922 10⁵ Pa

8 0

2 years ago