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Kazeer
17 days ago
15

A projectile is launched from the ground with an initial velocity of 12ms at an angle of 30° above the horizontal. The projectil

e lands on a hill 7.5m away. The height at which the projectile lands is most nearly
A- 1.78m
B- 3.10m
C- 5.34m
D- 6.68m
E- 12.0m
Physics
2 answers:
ValentinkaMS [2.4K]17 days ago
7 0

Answer:gggffdss

Explanation:cc bbhfewaz

Softa [2K]17 days ago
6 0

vi^{2}sin2thita/g =12^{2}sin2[30]/9.8=12.7Answer:

Explanation:

The range is specified as

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A ski lift is used to transport people from the base of a hill to the top. If the lift leaves the base at a velocity of 15.5 m/s
Yuliya22 [2438]
This can be determined using the principle of energy conservation. The ski lift begins with a velocity of v= 15.5 m/s, and all of its kinetic energy Ek converts into potential energy Ep, thus we set Ep equal to Ek.
Because Ek is given by (1/2)*m*v², where m denotes mass and v represents speed, while Ep equals m*g*h, where m is mass, g is 9.81 m/s², and h is height. Now:

Ek=Ep 

(1/2)*m*v²=m*g*h, canceling out the mass,

(1/2)*v²=g*h, rearranging for height by dividing by g,

(1/2*g)*v²=h and substituting the values:

h=12.245 m. The hill's height rounded to the nearest tenth is h=12.25 m.

5 0
18 days ago
Read 2 more answers
Feng and Isaac are riding on a merry-ground. Feng rides on a horse at the outer rim of the circular platform, twice as far from
Ostrovityanka [2208]
The question lacks complete details or specifications. Here is the missing information: 1. impossible to establish 2. half of Isaac's 3. identical to Isaac's 4. double Isaac's The angular velocity of Feng will match that of Isaac. Thus, the correct choice is option 3.
4 0
1 day ago
The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m
Sav [2230]

Answer:

The density comes out to be 10^{6} Mg/µL

Explanation:

Given data:

The density of nuclear matter is approximately 10^{18} kg/m³

1 ml corresponds to 1 cm³

To determine:

The density of nuclear matter in Mg/µL

Solution:

We recognize that:

1 Mg equals 1000 kg

Thus, 1 m³ is equal to 10^{6} cm³

Moreover, 1 cm³ is equivalent to 1 mL

Thus, we can conclude that 1 mL is equal to 10³ µL

With this, we convert the density as follows:

Density = 10^{18} kg/m³

Density = 10^{18} kg/m³ × \frac{10^{-3} }{10^{6} }  Mg/µL

Density = 10^{6} Mg/µL

8 0
23 days ago
Read 2 more answers
A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold
Maru [2360]

Answer:

(a). The coefficient of permeability is 8.6\times10^{-3}\ cm/s.

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

Q=\dfrac{V}{t}

Insert the values into the formula

Q=\dfrac{405}{90}

Q=4.5\ cm^3/s

(a). We will compute the coefficient of permeability

Applying the formula for permeability

Q=kiA

k=\dfrac{Q}{iA}

k=\dfrac{Ql}{Ah}

Where, Q=discharge

l = length

A = cross-sectional area

h=constant head that causes flow

Plugging the value into the formula

k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}

k=8.6\times10^{-3}\ cm/s

The coefficient of permeability measures as 8.6\times10^{-3}\ cm/s.

(c). To ascertain the discharge velocity during the testing phase

Utilizing the discharge velocity formula

v=ki

v=\dfrac{kh}{l}

Substituting the values into the equation

v=\dfrac{8.6\times10^{-3}\times38}{17.5}

v=0.0187\ cm/s

The discharge velocity during the test measures 0.0187 cm/s.

Thus, (a). The coefficient of permeability is 8.6\times10^{-3}\ cm/s.

(b). The seepage velocity computes at 0.0330 cm/s.

(c). The observed discharge velocity during the test equals 0.0187 cm/s.

8 0
8 days ago
The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
kicyunya [2264]

Answer:

tan \theta = \mu_s

Explanation:

Aby obiekt był w spoczynku na nachyleniu, wynikowa siła działająca na niego musi wynosić zero. Równanie sił działających w kierunku równoległym do nachylenia jest następujące:

mg sin \theta - \mu_s R =0 (1)

gdzie

mg sin \theta to składowa ciężaru równoległa do nachylenia, przy czym m oznacza masę obiektu, g oznacza przyspieszenie grawitacyjne, a \theta to kąt nachylenia

\mu_s R to siła tarcia, z \mu_s jako współczynnikiem tarcia oraz R jako reakcją normalną nachylenia

Równanie sił w kierunku prostopadłym do nachylenia to

R-mg cos \theta = 0

gdzie

R to reakcja normalna

mg cos \theta to składowa ciężaru prostopadła do nachylenia

Obliczając R,

R=mg cos \theta

I podstawiając do (1)

mg sin \theta - \mu_s mg cos \theta = 0

Rearanżując równanie,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

To jest warunek, przy którym równowaga jest zachowana: gdy tangens kąta staje się większy niż wartość \mu_s, siła tarcia nie jest w stanie zrównoważyć składowej ciężaru równoległej do nachylenia, dlatego obiekt zaczyna zsuwać się w dół.

4 0
16 days ago
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