Answer:
Explanation:
range of projectile = u² sin2θ / g
u = 30 m /s
θ = 40°
range = 30² x sin 80 / 9.8
= 90.44 m
b )
maximum altitude H = u² sin²θ / 2 x g
= 30² sin²40 / 9.8
= 37.94 m .
Freida wants to model the way atoms move when a substance changes its state. To do this, Freida makes a pyramid of marshmallows. Then, she knocked down the pyramid causing the marshmallows to fall. If the marshmallows represent the atoms in the substance, which change of state is Freida modeling?
Answer:melting 2020 edge
Explanation:
:)
Answer:
Melting
Explanation:
A person picking apples stand on a ladder 3.0 m above the ground. He throws them into
a basket 2.0 m away. How fast must the person throw the apple in order for it to land in
the basket?
Answer:
The speed the apple must be thrown in order for it to land in the basket is 2.554 m/s.
Explanation:
Given;
height above the ground, h = 3.0 m
horizontal distance, X = 2.0 m
The time to drop from the given height;
h = ¹/₂gt²
[tex]t = \sqrt{\frac{2h}{g} }\\\\t = \sqrt{\frac{2*3}{9.8} }\\\\t = 0.783 \ s[/tex]
The horizontal speed traveled by the apple is given by;
vₓ = X / t
vₓ = 2 / 0.783
vₓ = 2.554 m/s
Therefore, the speed the apple must be thrown in order for it to land in the basket is 2.554 m/s.
To remove a stain using a solvent the stain has to become dissolved in the solvent
True
False
Answer:
True
Explanation:
have a good day:)
Answer: This statement is True
go to his profile and roast the mess out of him plzz 403665fl 50 points
Answer:
ok
Explanation:
Please select the word from the list that best fits the definition The skeletons of tiny ocean animals grow together to form ______, structures that are found only in warm, clear ocean water.
Answer:
coral reef
Explanation:
what are fundamental quantities
Answer: length, luminous intensity,mass, time, temperature, electric current, amount of a substance.
Explanation:
find the mass of an object with a density of 1.5 g/cm^3 and had a volume of 8cm^3
Answer:
12 gExplanation:
The mass of a substance when given the density and volume can be found by using the formula
mass = Density × volume
From the question we have
mass = 1.5 × 8
We have the final answer as
12 gHope this helps you
One particle has mass m and a second particle has mass 2m. The second particle is moving with speed v and the first with speed 2v. How do their kinetic energies compare?
Answer:
Explanation:
The formula for kinetic energy to be used here is 1/2mv².
If the first particle is "particle a" and the second particle is "particle n"; there kinetic energies (K.E) will be
K.Eₐ = 1/2.m2v² = mv²
K.Eₙ = 1/2.2mv² = mv²
From the above, it can be said that there kinetic energies are the same.
NOTE that the m and v used in the question means mass and velocity respectively.
During a medieval siege of a castle, the attacking army uses a trebuchet to hurl heavy stones at the castle walls. If the trebuchet launches the stones with a velocity of +48.5" m"/s at an angle of 42.0°, how long does it take the stone to hit the ground? For those settings, what is the maximum range? How high will the stones go? Show all your work
Answer:
a) t = 6.62 s
b) x = 238.6 m
c) H = 53.7 m
Explanation:
a) We can find the time of flight as follows:
[tex] y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} [/tex]
Where:
[tex]y_{f}[/tex] is the final height = 0
[tex]y_{0}[/tex] is the initial height = 0
[tex]v_{0_{y}}[/tex] is the initial vertical velocity of the stone
t: is the time
g: is the gravity = 9.81 m/s²
[tex] v_{0}sin(42)t - \frac{1}{2}gt^{2} = 0 [/tex]
[tex] 48.5 m/s*sin(42)*t - \frac{1}{2}9.81 m/s^{2}*t^{2} = 0 [/tex]
By solving the above quadratic equation we have:
t = 6.62 s
b) The maximum range is:
[tex] x = v_{0_{x}}t = 48.5 m/s*cos(42)*6.62 s = 238.6 m [/tex]
c) The maximum height (H) can be found knowing that at this height the final vertical velocity of the stone is zero:
[tex] v_{f_{y}}^{2} = v_{0_{y}}^{2} - 2gH [/tex]
[tex] H = \frac{v_{0_{y}}^{2} - v_{f_{y}}^{2}}{2g} = \frac{(48.5 m/s*sin(42))^{2} - 0}{2*9.81 m/s^{2}} = 53.7 m [/tex]
I hope it helps you!
Have you ever written a bio-data or an application letter? Share your experience in the
space below. (e.g. How did you find the experience? What was hard? What was easy?)
Leave the space blank if you have never written any of these.
Answer:
I found the experience tasking
Explanation:
I wouldn't say it was hard, neither was it easy. I'd rather go for something like it being tasking. It's worthy of note that it was my first time, and I think it's very normal especially when one hasn't been doing something of that nature previously. Of course I did my draft, which unsurprisingly happened to be not good enough, and I had to look for templates to guide me through the acceptable way.
I still did it in my own way, but in the right way. Ever since then though, I have never stuttered when writing application letters, as it had since then seem inborn
The experience of writing a bio-data or an application letter was quite a tasking goal, where proper structure is required.
The given problem is based on the fundamentals of bio-data. Biodata is a document that is used to display the biographical data about the work experience in any organization.
As per my experience it was neither hard nor easy. I'd rather go for something like it being tasking. It's worthy of note that it was my first time, and I think it's very normal especially when one hasn't been doing something of that nature previously. I did my draft, which unsurprisingly happened to be not good enough, and I had to look for templates to guide me through the acceptable way. I still did it in my own way, but in the right way. Ever since then though, I have never stuttered when writing application letters, as it had since then seem inborn.Thus, it is concluded that writing a bio-data or an application letter was quite a tasking goal, where proper structure is required.
Learn more about the biodata here:
https://brainly.com/question/22965812
Upon being struck by 240-nm photons, a metal ejects electrons with a maximum kinetic energy of 2.97 eV. What is the work function of this metal?
Answer:
Work Function = 3.53 x 10⁻¹⁹ J = 2.2 eV
Explanation:
The work function of the metal metal can be found as follows:
Energy of Photon = Work Function + K.E
hc/λ = Work Function + K.E
Work Function = hc/λ - K.E
where,
h = Plank's Constant = 6.625 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of photons = 240 nm = 2.4 x 10⁻⁷ m
K.E = Maximum Kinetic Energy = (2.97 eV)(1.6 x 10⁻¹⁹ J/1 eV) = 4.752 x 10⁻¹⁹ J
Therefore,
Work Function = (6.625 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.4 x 10⁻⁷ m) - 4.752 x 10⁻¹⁹ J
Work Function = 8.281 x 10⁻¹⁹ J - 4.752 x 10⁻¹⁹ J
Work Function = 3.53 x 10⁻¹⁹ J = 2.2 eV