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3 years ago

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A book is pushed with an initial horizontal velocity of 5.0 meters per second off the top of a desk. What is the initial vertica

l velocity of the book, in m/s?

1 answer:

PSYCHO15rus [73]3 years ago

7 0

Answer:

The Initial vertical velocity is zero

Explanation:

Given that,

A book is pushed with an initial horizontal velocity, Vx = 5 m/s

According to projectile motion, the horizontal component of velocity i.e., in X-axis remains the same though out its path.

The vertical velocity varies i.e., the velocity towards the Y- axis.

The velocity of the falling book in its trajectory is the resultant velocity. Due to the action of gravity, it keeps on increasing with time.

The resultant velocity is given by relation

$V_{R} = \sqrt{{V_{x}^{2}} +[ {gt}]^{2} }$

where

gt = Vy - vertical velocity

g - acceleration due to gravity

t - time

Since, the initial horizontal velocity at that instant is equal to the resultant velocity.

Vx = 5 m/s

$V_{R}$ = 5 m/s

So, on substituting in the above equation

Vy = 0 m/s

Therefore, the initial vertical velocity is equal to zero

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A simple experiment to measure the speed of sound doesn't involve a stopwatch. You can fill up along tube with water and put a t

Serhud [2]

Answer:

Explanation:

In order to answer this problem you have to know the depth of the column, we say R, this information is important because allows you to compute some harmonic of the tube. With this information you can compute the depth of the colum of air, by taking tino account that the new depth is R-L.

To find the fundamental mode you use:

$f_n=\frac{nv_s}{4L}$

n: mode of the sound

vs: sound speed

L: length of the column of air in the tube.

A) The fundamental mode id obtained for n=1:

$f_1=\frac{v_s}{4L}$

B) For the 3rd harmonic you have:

$f_3=\frac{3v_s}{4L}$

C) For the 2nd harmonic:

$f_2=\frac{2v_s}{4L}$

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3 years ago

While in a car at 4.47 meters per second a passenger drops a ball from a height of 0.70 meters above the top of a bucket how far

viktelen [127]

Answer:

1.7 m

Explanation:

$v_x$ = Velocity of ball in x direction = 4.47 m/s

$u_y$ = Velocity of ball in y direction = 0

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

t = Time taken

$s_y$ = Vertical displacement = 0.7 m

$s_y=u_yt+\dfrac{1}{2}gt^2\\\Rightarrow 0.7=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{0.7\times 2}{9.81}}\\\Rightarrow t=0.38\ \text{s}$

Horizontal displacement is given by

$s_x=v_xt\\\Rightarrow s_x=4.47\times 0.38\\\Rightarrow s_x=1.7\ \text{m}$

The passenger should throw the ball 1.7 m in front of the bucket.

5 0

3 years ago

A ball of 0.1kg is thrown straight up into the air with

Contact [7]

a) The momentum of the ball at the maximum height is zero

b) The momentum of the ball at halfway is 1.06 kg m/s

Explanation:

a)

There is only one force acting on the ball during its motion: the force of gravity, acting downward. As a result, the ball hasa constant acceleration downward ( $g=9.8 m/s^2$ , acceleration of gravity), and therefore it is in free fall motion.

This means that the velocity of the ball is given by the equation:

$v=u-gt$

where

v is the velocity at time t

u = 15 m/s is the initial velocity

$g=9.8 m/s^2$

As we see from the equation, the term $-gt$ is negative: this means that as the ball goes higher and higher, its velocity decreases. The point of maximum height is reached when the velocity has become zero (after that point, the velocity changes direction and points downward), so when

$v=0$

The momentum of the ball at any point is given by

$p=mv$

where m = 0.1 kg is its mass. Therefore, at the maximum height, since $v=0$ , the momentum is also zero:

$p=0$

b)

We can use the following suvat equation to find the maximum height reached by the ball:

$v^2-u^2=2as$

where

v = 0 is the velocity at the point of maximum height

u = 15 m/s is the initial velocity

$a=-g=-9.8 m/s^2$ is the acceleration (downward)

s is the vertical displacement (the maximum height)

Solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{0-(15)^2}{2(-9.8)}=11.48 m$

This means that the ball is halfway when its height is

$h'=\frac{s}{2}=\frac{11.48}{2}=5.74 m$

We can find the velocity of the ball v' when it is at h' by using again the same equation:

$v'^2-u^2=2ah'\\v=\sqrt{u^2+2ah'}=\sqrt{(15)^2+2(-9.8)(5.74)}=10.6 m/s$

And therefore, the momentum of the ball here is

$p'=mv'=(0.1)(10.6)=1.06 kg m/s$

Learn more about momentum:

brainly.com/question/7973509

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3 years ago

Why do you think Saturn and Jupiter have more moons than the other planets<br> in our solar system?

Aneli [31]

Answer:

yeah

Explanation:

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3 years ago

A group of hikers hear an echo 3.3 s after they shout. The temperature is 20◦C.

romanna [79]

Answer:

560 m

Explanation:

The speed of sound in air is approximately:

v ≈ v₀ + 0.6T

where v₀ is the speed of sound at 0°C (273 K) in m/s, and T is the temperature in Celsius.

The speed of sound at 20°C at that altitude is:

v ≈ 327 + 0.6(20)

v ≈ 339 m/s

The sound travels from the hikers to the mountain and back again, so it travels twice the distance.

339 m/s = 2d / 3.3 s

2d = 1118.7 m

d = 559.35 m

Rounding, the mountain is approximately 560 m away.

4 0

3 years ago