Answer:
The ball land at 3.00 m.
Explanation:
Given that,
Speed = 40 m/s
Angle = 35°
Height h = 1 m
Height of fence h'= 12 m
We need to calculate the horizontal velocity
Using formula of horizontal velocity
![V_{x}=V_{i}\cos\theta](https://tex.z-dn.net/?f=V_%7Bx%7D%3DV_%7Bi%7D%5Ccos%5Ctheta)
![V_{x}=40\times\cos35](https://tex.z-dn.net/?f=V_%7Bx%7D%3D40%5Ctimes%5Ccos35)
![V_{x}=32.76\ m/s](https://tex.z-dn.net/?f=V_%7Bx%7D%3D32.76%5C%20m%2Fs)
We need to calculate the time
Using formula of time
![t = \dfrac{d}{v}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7Bd%7D%7Bv%7D)
![t=\dfrac{130}{32.76}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B130%7D%7B32.76%7D)
![t=3.96\ sec](https://tex.z-dn.net/?f=t%3D3.96%5C%20sec)
We need to calculate the vertical velocity
![v_{y}=v_{y}\sin\theta](https://tex.z-dn.net/?f=v_%7By%7D%3Dv_%7By%7D%5Csin%5Ctheta)
![v_{y}=40\times\sin35](https://tex.z-dn.net/?f=v_%7By%7D%3D40%5Ctimes%5Csin35)
![v_{y}=22.94\ m/s](https://tex.z-dn.net/?f=v_%7By%7D%3D22.94%5C%20m%2Fs)
We need to calculate the vertical position
Using formula of distance
![y(t)=y_{0}+V_{i}t+\dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=y%28t%29%3Dy_%7B0%7D%2BV_%7Bi%7Dt%2B%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
Put the value into the formula
![y(3.96)=1+22.94\times3.96+\dfrac{1}{2}\times(-9.8)\times(3.96)^2](https://tex.z-dn.net/?f=y%283.96%29%3D1%2B22.94%5Ctimes3.96%2B%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%28-9.8%29%5Ctimes%283.96%29%5E2)
![y(3.96)=15.00\ m](https://tex.z-dn.net/?f=y%283.96%29%3D15.00%5C%20m)
We need to calculate the distance
![s = y-h'](https://tex.z-dn.net/?f=s%20%3D%20y-h%27)
![s=15.00-12](https://tex.z-dn.net/?f=s%3D15.00-12)
![s=3.00\ m](https://tex.z-dn.net/?f=s%3D3.00%5C%20m)
Hence, The ball land at 3.00 m.
Density = Mass / Volume
A) Density = 888/800 = 1.11g/mL > 1.0g/mL
B) Density = 7.8 / 8.7 = 0.897g/mL < 1.0g/mL
C) density = 725 / 715 = 1.01 g/mL > 1.0g/mL
D)density = 1.3/1.1= 1.18 g/mL > than 1.0 g/mL
E) Density = 18/19 = 0.95g/mL < 1.0g/mL
F) Density = 1.25/1.78 = 0.7 g/mL < 1.0g/mL
SUMMARY:
A C D have densities greater than 1.0g/mL
Answer:
2.47 s
Explanation:
Convert the final velocity to m/s.
We have the acceleration of the gazelle, 4.5 m/s².
We can assume the gazelle starts at an initial velocity of 0 m/s in order to determine how much time it requires to reach a final velocity of 11.1111 m/s.
We want to find the time t.
Find the constant acceleration equation that contains all four of these variables.
Substitute the known values into the equation.
- 11.1111 = 0 + (4.5)t
- 11.1111 = 4.5t
- t = 2.469133333
The Thompson's gazelle requires a time of 2.47 s to reach a speed of 40 km/h (11.1111 m/s).
The oldest way ... the way we've been using as long as we've been
walking on the Earth ... has been to use plants. Plants sit out in the
sun all day, capturing its energy and using it to make chemical compounds.
Then we come along, cut the plants down, and eat them. Our bodies
rip the chemical compounds apart and suck the solar energy out of them,
and then we use the energy to walk around, sing, and play video games.
Another way to capture the sun's energy is to build a dam across a creek
or a river, so that the water can't flow past it. You see, it was the sun's
energy that evaporated the water from the ocean and lifted it high into
the sky, giving it a lot of potential energy. The rain falls on high ground,
up in the mountains, so the water still has most of that potential energy
as it drizzles down the river to the ocean. If we catch it on its way, we
can use some of that potential energy to turn wheels, grind our grain,
turn our hydroelectric turbines to get electrical energy ... all kinds of jobs.
A modern, recent new way to capture some of the sun's energy is to use
photovoltaic cells. Those are the flat blue things that you see on roofs
everywhere. When the sun shines on them, they convert some of its
energy into electrical energy. We use some of what they produce, and
we store the rest in giant batteries, to use when the sun is not there.
Answer:
Explanation:
The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.
So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.
force on it due to rest of the charges will be equal and opposite so
k3q Q / x² =k 8q Q / (L+x)²
8x² = 3 (L+x)²
2√2 x = √3 (L+x)
2√2 x - √3 x = √3 L
x(2√2 - √3 ) = √3 L
x = √3 L / (2√2 - √3 )
Let us consider the balancing force on 3q
force on it due to -Q and -8q will be equal
kQ . 3q / x² = k3q 8q / L²
Q = 8q (x² / L²)
so charge required = - 8q (x² / L²)
and its distance from x on negative x side = √3 L / (2√2 - √3 )