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sladkih [1.3K]
3 years ago
15

A 2.7-kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0 m high cliff. At the instant the ball i

s thrown, a woman starts running away from the base of the cliff with a constant speed of 6.00 m/s. The woman runs in a straight line on level ground, and air resistance acting on the ball can be ignored.How far does the woman run before she catches the ball?
Physics
1 answer:
scoray [572]3 years ago
3 0

Answer:

The distance traveled by the woman is 34.1m

Explanation:

Given

The initial height of the cliff

yo = 45m final, positition y = 0m bottom of the cliff

y = yo + ut -1/2gt²

u = 20.0m/s initial speed

g = 9.80m/s²

0 = 45.0 + 20×t –1/2×9.8×t²

0 = 45 +20t –4.9t²

Solving quadratically or by using a calculator,

t = 5.69s and –1.61s byt time cannot be negative so t = 5.69s

So this is the total time it takes for the ball to reach the ground from the height it was thrown.

The distance traveled by the woman is

s = vt

Given the speed of the woman v = 6.00m/s

Therefore

s = 6.00×5.69 = 34.14m

Approximately 34.1m to 3 significant figures.

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A pure sound wave, generated by a tuning fork, is considered a periodic wave. Which statement is true for this tuning fork sound
soldier1979 [14.2K]

Answer : Every wave has the same wave pattern.

Explanation : A pure sound wave has a single frequency and its generated by a tuning fork . The sound wave is a  simple periodic wave.

When an object such as a guitar push and pull on the around it. When it pushes on the air, then the pressure increases and when it pull on the air,  then the pressure decreases and sound waves are formed.

Hence, sound wave has the same wave pattern.  

4 0
3 years ago
Read 2 more answers
9. A car driver brakes gently. Her car slows down front --
sleet_krkn [62]

Answer:

9) This is a case of deceleration

10)-0.8 ms-2

b) acceleration is the change in velocity with time

11)

a) 100 ms-1

b) 100 seconds

12) 10ms-1

13) more information is needed to answer the question

14) - 0.4 ms^-2

15) 0.8 ms^-2

Explanation:

The deceleration is;

v-u/t

v= final velocity

u= initial velocity

t= time taken

20-60/50 =- 40/50= -0.8 ms-2

11)

Since it starts from rest, u=0 hence

v= u + at

v= 10 ×10

v= 100 ms-1

b)

v= u + at but u=0

1000 = 10 t

t= 1000/10

t= 100 seconds

12) since the sprinter must have started from rest, u= 0

v= u + at

v= 5 × 2

v= 10ms-1

14)

v- u/t

10 - 20/ 25

10/25

=- 0.4 ms^-2

15)

a=v-u/t

From rest, u=0

8 - 0/10

a= 8/10

a= 0.8 ms^-2

7 0
3 years ago
un conductor de plata (p=1,6x10°-6ohm x m) tiene una seccion de 5x10°-6 m°2 y una longitud de 110 m . calcular la resistencia
Alexus [3.1K]

Responder:

35,2 ohm.

Explicación:

Dado:

La resistencia específica del conductor es,

La longitud del conductor es,

El área de la sección transversal del conductor es,

Sabemos que la resistencia de un conductor es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.

Por lo tanto, la resistencia se puede expresar como:

R=\frac{\rho\times l}{A}

Ahora, conecte los valores dados y resuelva para 'R'. Esto da,

R=\frac{1.6\times 10^{-6}\ ohm\cdot m\times 110\ m}{5\times 10^{-6}\ m^2}\\\\R=35.2\ ohm

Por lo tanto, la resistencia del conductor es de 35,2 ohm.

4 0
3 years ago
A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f
Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t

Now we need to find v_y as a function of v_0. We use the horizontal velocity, which is always the same as follow:

v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\

We know the angle at 3 seconds:

v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}

Substitute  v_{t=3} in  v_x and then solve for  v_y

\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0

With this expression we go back to the kinematic equation and solve it for initial speed

\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s

3 0
3 years ago
WHY ARE ALL GIRLS THE SAME, don't even say their not cause if you say that then I guess you don't have a life!!!!!​
Zinaida [17]
Because they are. it’s just how life works
6 0
2 years ago
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