What is the x component of (+3m)ι^ And (+3m/s) i^?

A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen

Svetllana [295]

Answer:

$h_p = 30.46\ m$

Explanation:

<u>Free Fall Motion</u>

A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.

The speed vf of the object when a time t has passed is given by:

$v_f=g\cdot t$

Where $g = 9.8 m/s^2$

Similarly, the distance y the object has traveled is calculated as follows:

$\displaystyle y=\frac{g\cdot t^2}{2}$

If we know the height h from which the object was dropped, we can solve the above equation for t:

$\displaystyle t=\sqrt{\frac{2\cdot y}{g}}$

The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

$\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec$

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

$t_2=2.56 - 2 = 0.56\ sec$

Therefore, it has traveled down a distance:

$\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m$

Thus, the height of the pen is:

$h_p = 32 - 1.54\Rightarrow h_p=30.46\ m$

8 0

2 years ago

An ultrasound pulse has a wavelength of 1.0mm. Its speed in water is 1400m. What’s the frequency?

timama [110]

Answer:

$1.4\cdot 10^6 Hz$

Explanation:

The relationship between the frequency, the wavelength and the speed of a wave is given by the wave's equation:

$v=f \lambda$

where

v is the speed of the wave

f is the frequency

$\lambda$ is the wavelength

For the pulse in this problem,

$\lambda = 1.0 mm = 0.001 m\\v = 1400 m/s$

Solving for f, we find the frequency:

$f=\frac{v}{\lambda}=\frac{1400}{0.001}=1.4\cdot 10^6 Hz$

6 0

2 years ago

A point charge with charge q1 is held stationary at the origin. A second point charge with charge q2 moves from the point (x1, 0

Scilla [17]

Answer:

$W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$

Explanation:

Position of charge q₁ is (0,0)

Position of charge q₂ is (x₁,0)

So, the electric potential energy between the charges is given by :

$U_1=k\dfrac{q_1q_2}{x_1}$

Now the position of charge q₂ has been changes from (x₁,0) to (x₂,y₂). Now, electric potential energy between the charges is :

$U_2=k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}}$

We know form the work energy theorem that, the change in potential energy is equal to the work done. Mathematically, it is given by :

$W=-\Delta U$

$W=-(U_2-U_1)$

$W=(U_1-U_2)$

$W=(k\dfrac{q_1q_2}{x_1}-k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}})$

$W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$

Hence, the work done by the electrostatic force on the moving point charge is $kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$ . Hence, this is the required solution.

3 0

2 years ago

Alex throws a 0.15-kg rubber ball down onto the floor. The ball’s speed just before impact is 6.5 m/s, and just after is 3.5 m/s

Jet001 [13]

Answer: Change in ball's momentum is 1.5 kg-m/s.

Explanation: It is given that,

Mass of the ball, m = 0.15 kg

Speed before the impact, u = 6.5 m/s

Speed after the impact, v = -3.5 m/s (as it will rebound)

We need to find the change in the magnitude of the ball's momentum. It is given by :

So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.

Read more on Brainly.com - brainly.com/question/12946012#readmore

7 0

2 years ago

Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hourmiles/hour

slavikrds [6]

Answer:

The acceleration of the cheetahs is 10.1 m/s²

Explanation:

Hi there!

The equation of velocity of an object moving along a straight line with constant acceleration is the following:

v = v0 + a · t

Where:

v = velocity of the object at time t.

v0 = initial velocity.

a = acceleration.

t = time

We know that at t = 2.22 s, v = 50.0 mi/h. The initial velocity, v0, is zero.

Let's convert mi/h into m/s:

50.0 mi/h · (1609.3 m / 1 mi) · (1 h / 3600 s) = 22.4 m/s

Then, using the equation:

v = v0 + a · t

22.4 m/s = 0 m/s + a · 2.22 s

Solving for a:

22.4 m/s / 2.22 s = a

a = 10.1 m/s²

The acceleration of the cheetahs is 10.1 m/s²

5 0

2 years ago