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A bird flying straight upward at 5 m/s drops a berry when it is 300 m above the ground. How fast is the berry going when it hits

Nikitich [7]

Answer:

v=77.62 m/s

Explanation:

Given that

h= - 300 m

speed of the bird ,u= 5 m/s

Lets take Speed of the berry when it hit the ground = v m/s

we know that ,if object is moving upward

v² = u² - 2 g h

u=Initial speed

v=Final speed

h=Height

Now by putting the values

v² = u² - 2 g h

v² = 5² - 2 x 10 x (-300) ( take g = 10 m/s²)

v² =25 + 20 x 300

v² ==25 + 6000

v² =6025

v=77.62 m/s

Therefore the final speed of the berry will be 77.62 m/s.

5 0

3 years ago

Which law of motion accounts for the following statement?

strojnjashka [21]

I know that its not the second law. I'm almost positive its the first one. Please let me know if I'm wrong. This sentence makes no sense when you put it with the third law. So, the first law is my guess...

3 0

3 years ago

Read 2 more answers

A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision

Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is $a=2,800m/s^{2}$

Explanation:

Known Data

We will asume initial speed has a negative direction, $v_{i}=-30m/s$ , final speed has a positive direction, $v_{f}=26m/s$ , $\Delta t=20ms=0.020s$ and mass $m_{b}$ .

Initial momentum

$p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s$

final momentum

$p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s$

Impulse

$I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s$

Average Force

$F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}$

Average acceleration

$F=ma$ , so $a=\frac{F}{m_{b}}$ .

Therefore, $a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}$

8 0

3 years ago

You have a motion security light by your door cat runs across the in the light turns on what is the input of the system and what

Neporo4naja [7]

A motion security line is a system that is used to detect motion.
The input for the system is MOTION and the output is LIGHT. That is, when the system detects motion it switch on light.
Remember, an Input is the information that was inserted into a system while the output is the result of the processed information.

5 0

3 years ago

a) Calculate the magnitude of displacement of the car in 40 seconds. b) During which part of the journey was the car acceleratin

Zielflug [23.3K]

Answer:

a) 600 meters

b) between 0 and 10 seconds, and between 30 and 40 seconds.

c) the average of the magnitude of the velocity function is 15 m/s

Explanation:

a) In order to find the magnitude of the car's displacement in 40 seconds,we need to find the area under the curve (integral of the depicted velocity function) between 0 and 40 seconds. Since the area is that of a trapezoid, we can calculate it directly from geometry:

$Area \,\,Trapezoid=(\left[B+b]\,(H/2)\\displacement= \left[(40-0)+(30-10)\right] \,(20/2)=600\,\,m$

b) The car is accelerating when the velocity is changing, so we see that the velocity is changing (increasing) between 0 and 10 seconds, and we also see the velocity decreasing between 30 and 40 seconds.

Notice that between 10 and 30 seconds the velocity is constant (doesn't change) of magnitude 20 m/s, so in this section of the trip there is NO acceleration.

c) To calculate the average of a function that is changing over time, we do it through calculus, using the formula for average of a function:

$Average\,of\,f(x)=\frac{1}{b-a} \int\limits^b_a {f(x)} \, dx$

Notice that the limits of integration for our case are 0 and 40 seconds, and that we have already calculated the area under the velocity function (the integral) in step a), so the average velocity becomes:

$Avearage=\frac{600\,\,m}{40\,\,s}= 15\,\,\frac{m}s}$

7 0

3 years ago