A shell is launched with an initial velocity at an angle of 40.0° above horizontal
from ground level. The shell needs to hit 1.5 km away. There is no appreciable air resistance,
and g = 9.80 m/s2
.
a. What should the initial velocity be?
b. What are the components of the shell’s velocity when it hits the ground? (horizontal
and vertical components)

Answer:

oa

Explanation:

it may be oa is the right answer for this question

but I don't know properly

Assuming players stick together immediately after the collision, their speed will be 1.3206 m/sec, and - ve indicates their direction.

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of 1st player= 101.0 kg

(u₁) is the initial velocity of 1st player (u₁) = 4.10 m/s

(m₂) is the mass of 2nd player = 117 kg

(u₂) is the initial velocity of 2nd player= -6  m/s

(v) is the velocity after collision =.?

According to the law of conservation of momentum;

Momentum before collision =Momentum after collision

m₁u₁+m₂u₂=(m₁+m₂)V

101.0 kg × 4.10 m/sec + 117 kg × (-6 m/sec) = (101.0 kg + 117 kg) × V

V= - 1.3206 m/sec

Hence, the velocity of players just after impact, if they cling together, will be - 1.3206 m/sec, and - ve shows the direction.

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The Sketch  of the figure will bear the parameters of the charge that lies   at (4,0,0)

Generally, the equation for is  mathematically given as

1)

Since both of the other points, "q1"<"q2," are on the x-axis, the third one must also be on the x-axis, and it must be close to the charge of lesser magnitude. off q1 by x cm

Now, forces on q3 due to q1

[tex]\vec{F}_{1}=\frac{k q_{1} q_{3}}{x^{2}} \hat{x}[/tex]

Force on q3 due to q2

[tex]\vec{F}_{2}=\frac{k q_{2} q_{3}}{(6+x)^{2}}(-\hat{x})[/tex]

[tex]\quad \vec{F}_{1}+\vec{G}_{2}=0$[/tex]

[tex]&\frac{k q_{1} q_{3}}{x^{2}}-\frac{k q_{2} q_{3}}{(x+6)^{2}}=0 \\&\frac{2}{x^{2}}=\frac{8}{(x+6)^{2}} \Rightarrow \frac{x+6}{x}=\sqrt{\frac{8}{2}}=2 \\&x+6=2 x \Rightarrow x=6 \mathrm{~cm}[/tex]

In conclusion, the charge lies at (4,0,0)

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The cost of gas is $1.27 per liter and the gallon of gas costs $4.80. where 1 gallon is equal to 3.78541 liters.

To convert the cost of gas from dollars per liter to dollars per gallon, The conversion factor between liters and gallons.

1 gallon is equal to 3.78541 liters (the exact conversion factor is 3.785411784).

Given the cost of gas is $1.27 per liter, let's calculate the cost per gallon:

Cost per gallon = Cost per liter × Liters per gallon

Cost per gallon = $1.27 × 3.78541

Cost per gallon = $4.80

So, a gallon of gas costs $4.80.

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The distance car moves with an average speed of 75 km/hr from town P to town Q in 2 hours is 149,760 m.

Given:
Speed = 75 km/hr

Time = 2 hr

The distance can be calculated from the product of velocity and time.

The standard unit of distance is a meter.

Convert speed and time into meters per second and seconds:

v = 75×5/18

v = 20.8 m/s

t = 3600×2

t = 7200 s

The distance is computed as:

v = d/t

d = vt

d = 20.8×7200

d = 149,760 m

Hence, the distance car moves with an average speed of 75 km/hr from town P to town Q in 2 hours is 149,760 m.

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Answer:

(a) The rms voltage across the resistor is given by V = (2.20 102 V) sin 2ft.

V = (2.20 102 V) sin 2ft

V = (2.20 102 V) (0.707)

V = 1.56 102 V

(b) The rms current flowing through the resistor is given by I = V/R.

I = V/R

I = (1.56 102 V)/(1.90 102 Ω)

I = 8.21 A

Review procedures and confirm results to avoid possible errors.

Explanation:

The angular acceleration of the cylinder is mathematically given as

[tex]\alpha = 60.2175 rad/s^2[/tex]

Question parameters

Generally, the equation for Torque generated in the cylinder is  mathematically given as

[tex]\sigma[/tex] = w×r

for circular motion torque is

[tex]\sigma = × \alpha[/tex]

t[tex]=0.5 × m ×r^2 × \alpha\\\\ =0.5 × 2.15 ×0.111 ^2 × \alpha[/tex]

t = 0.0132 ×[tex]\alpha[/tex]

Therefore

0.0132 × [tex]\alpha[/tex] = w × r

[tex]\\\\\alpha =\frac{ 7.161 0.111}{(0.0132)}[/tex]

[tex]\alpha = 60.2175 rad/s^2[/tex]

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The fact that the layers of graphite are held together by only weak Van der Walls forces implies that they can slide over each other.

We know that graphite is composed of layers. These hexagonal layers are held together by weak Van Der Walls forces and as such are able to slide over each other. The carbon atom in each layer are held together by strong covalent bonds.

The fact that the layers of graphite are held together by only weak Van der Walls forces implies that they can slide over each other and as such make the graphite fluid.

Thus, the image that shows these layers of graphite is attached to this an answer

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Answer:it’s B

Explanation:

The density of the mixture is determined as 1,013.75 kg/m³.

The density of the mixture is calculated as follows;

density = total mass of the mixture / total volume of the mixture

mass of the fresh water = density x volume

                                        = 1000 kg/m³ x 0.0018 m³

                                        = 1.8 kg

mass of the sea water = 1025 kg/m³ x 0.0022 m³

                                     = 2.255 kg

Total mass = 1.8 kg + 2.255 kg = 4.055 kg

Total volume = 0.0018 m³  + 0.0022 m³ = 0.004 m³

Density = 4.055 kg/0.004 m³ = 1,013.75 kg/m³

Thus, the density of the mixture is determined as 1,013.75 kg/m³.

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c. Sodium and potassium

Perspiration, water given off by the intact skin, either as vapor by simple evaporation from the epidermis or as sweat, a form of cooling in which liquid actively secreted from sweat glands evaporates from the body surface.

When our body is sweating sodium and potassium is lost from the body along with water.

In order to maintain the integrity of the cells in the body ,Sodium and potassium are very important.

Sweat is also known as perspiration.

So,

By maintaining  the proper cell functioning and cell vitality optimum level of sodium and potassium should be maintained in the body.

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The acceleration is  1.55 m/s^2 while is component that is parallel to the ground is 0.775  m/s^2.

The term acceleration is the rate of change of speed. Given that the distance covered is 35.0 m and the final velocity is 10.43 m/s it the follows that;

v^2 = u^2 + 2as

u = 0 because he started from rest

v^2 = 2as

a = v^2/2s

a = (10.43 m/s)^2/2 * 35.0 m

a = 1.55 m/s^2

The component of this acceleration which is parallel to the ground is;

ax = a sin θ

ax =  1.55 m/s^2 cos 60

ax = 0.775  m/s^2

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The concept of a black hole seem reasonable to me because of its valid and authentic evidences.

A black hole is a region in space where light is unable to escape due to strong gravitation force. The strong gravity is formed because matter has been pressed into a tiny space. This compression occurs at the end of a star's life. Some black holes are formed when a stars die.

So we can conclude that the concept of a black hole seem reasonable to me because of its valid and authentic evidences.

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The length of the filament is 0.2 m.

To find the answer, we need to know about the Stefan - Boltzmann law.

L= power/(π×d×e×σ×T⁴)

= 40/(π×0.0001×0.35×5.67×10^(-8)×2500⁴)

= 0.2m

Thus, we can conclude that the length of the filament is 0.2 m.

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B. The meaning of 'atomos' according to Democritus in 450 BCE is indivisible.

A Greek philosopher  known as Democritus first thought of the existence of tiny particles that compose everything around us.

Democritus of Abdera, named the building blocks of matter atomos, meaning literally “indivisible.

Thus, the meaning of 'atomos' according to Democritus in 450 BCE is indivisible.

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The time taken for the bus to travel 95.1 m is 2.8 s

v² = u² + 2as

v² = 26.8² + (2 × 4.73 × 95.1)

v² = 1617.886

Take the square root of both sides

v² = √1617.886

v = 40.2 m/s

v = u + at

40.2 = 26.8 + (4.73 × t)

Collect like terms

4.73 × t = 40.2 - 26.8

4.73 × t = 13.4

Divide both side by 4.73

t = 13.4 / 4.73

t = 2.8 s

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Due to density, the two bats have the same size, shape, and mass.

The two bats could be the same size, shape, and mass if they have the same density because density is directly proportional to mass so if the density same, the mass will also be the same.

So we can conclude that due to density, the two bats have the same size, shape, and mass.

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Use the fundamental theorem of calculus.

[tex]f(t) = f(0) + \displaystyle \int_0^t f'(u) \, du[/tex]

In order to find velocity and position exactly, you need to know the initial velocity and position.

1) Integrate the acceleration function to get the velocity function. By the FTC,

[tex]v(t) = v(0) + \displaystyle \int_0^t a(u) \, du[/tex]

[tex]v(t) = v(0) + \displaystyle \int_0^t (3u^3+2u^2) \, du[/tex]

[tex]v(t) = \dfrac34 t^4 + \dfrac23 t^3 + v(0)[/tex]

At [tex]t=2\,\rm s[/tex], the velocity is

[tex]v(2) = \dfrac34 2^4 + \dfrac23 2^3 + v(0) = \boxed{\dfrac{52}3 + v(0)}[/tex]

2) Integrate the velocity function to the get the position function. By the FTC again,

[tex]x(t) = x(0) + \displaystyle \int_0^t v(u) \, du[/tex]

[tex]x(t) = x(0) + \displaystyle \int_0^t \left(\frac34 u^4 + \frac23 u^3 + v(0)\right) \, du[/tex]

[tex]x(t) = \displaystyle \frac3{20} t^5 + \frac16 t^4 + v(0) t + x(0)[/tex]

At [tex]t=4\,\rm s[/tex], the object's position is

[tex]x(4) = \dfrac3{20}4^5 + \dfrac16 4^4 + 4v(0) + x(0) = \dfrac{2944}{15} + 4v(0) + x(0)[/tex]

Fortunately, you don't need the initial position to find the displacement, since [tex]x(0)[/tex] cancels out in the end.

[tex]\Delta x = x(4) - x(0) = \boxed{\dfrac{2944}{15} + 4v(0)}[/tex]

Answer:

A. the center of mass

Explanation:

[tex]C_m[/tex] is usually used to designate the center of mass of an object or system of objects.

B. is incorrect because the combined mass of a system is usually denoted by ∑m.

C. is incorrect because the momentum of a system is denoted by the letter p.

The magnitude of the tension in the string marked A and B is mathematically given as

Generally, the equation for is mathematically given as

The angle at A is...

[tex]tan\theta = 3/8[/tex]

When below the negative x.

B= tan\phi

[tex]tan\phi = 5/4[/tex]

When below the negative x

C=

[tex]tan \rho = 1/6[/tex]

Hence, considering the Horizontal components

74.9cos(9.46) = A*cos(20.6) + B*cos(51.3)

A = 78.9 - 0.668B

Hence, considering the Vertical components

74.9*sin(9.46) + Asin(20.6) = Bsin(51.3)

40.07 = 1.015B

B = 39.5 N

In conclusion, Sub the value of B is the equation of A

A = 78.9 - 0.668B

Sub

A = 78.9 - 0.668( 39.5 N)

A = 52.5 N

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The depth of the well at the given air temperature is 456.3 m.

The depth of the well is calculated from the principle of echo.

v = 2d/t

2d =  vt

d = vt/2

where;

d = (338 x 2.7)/2

d = 456.3 m

Thus, the depth of the well at the given air temperature is 456.3 m.

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Answer:it's 132.2m

Explanation:

Hello !

Because first recall parabolic peak height equation and

given that v=47.4m/s , h=10.5m

secondly recall parabolic total distance equation

The distance from the base of the wall that the cannonball land will be 138.7 m.

Consider that a body of mass 'm' is projected with a velocity of projection 'v' and angle of projection 'θ'. Assume that 'h' be the maximum height attained by the projectile.

2gh = v²sin²θ

This expression can be used to calculate the maximum height of the projectile.

Given, the height of the castle wall, h = 10.5 m

The velocity of the cannonball, v = 47.4m/s

The peak height equation can be used to find the angle θ:

v²sin²θ = 2gh

(47.4)² sin²θ = 2(9.81)(10.5)

Sinθ = 0.3028

θ = 17.63°

Therefore, the distance (d) from the base of the wall that the cannonball land:

d =  (47.4)² × 2 sin(17.63) / 9.81

d = 138.7m

Therefore the distance from the base of the wall that the cannonball land is 138.7m.

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Hi there!

1.

Let's use the rotational equivalent of Newton's Second Law.

[tex]\Sigma \tau = I\alpha[/tex]

τ = Torque (Nm)

I = Moment of Inertia (1/2mR² for solid cylinder)

α = Angular acceleration (rad/sec²)

The torque is equivalent to:
[tex]\tau = r \times F[/tex]

r = distance from lever arm to pivot point, aka the radius for this cylinder (0.111 m)
F = applied force (7.161 N)

We are already given various values, so let's create a working equation and solve.

[tex]rF = I\alpha \\\\\alpha = \frac{rF}{I}[/tex]
[tex]\alpha = \frac{(0.111)(7.161)}{\frac{1}{2}(2.15)(0.111^2)} = \boxed{60.013 \frac{rad}{sec^2}}[/tex]

2.

So, let's begin by doing a summation of forces acting on the block.

We have its force of gravity downward(+), and the tension from the string upward. (-), assigning signs based on the path of the falling block.

[tex]\Sigma F = F_g - T\\\\\Sigma F = m_b g - T\\\\m_b a = m_b g - T[/tex]

Solving for 'T':

[tex]T = m_b g - m_b a[/tex]

Let's now do a summation of torques on the cylinder. We only have the torque caused by the tension in the string connected to the block.

[tex]\Sigma \tau = rT\\\\I\alpha = r(T = m_b g - m_b a)[/tex]

Using the relationship between angular and translational acceleration:

[tex]a = \alpha r\\\\alpha = \frac{a}{r}[/tex]

And the equation for the moment of inertia:
[tex]I = \frac{1}{2}Mr^2[/tex]

Simplify the expression.

[tex]\frac{1}{2}Mr^2(\frac{a}{r}) = r( m_b g - m_b a)\\\\\frac{1}{2}Ma = ( m_b g - m_b a)[/tex]

Solve for 'a':
[tex]\frac{1}{2}ma + m_b a= m_b g \\\\a = \frac{m_b g}{\frac{1}{2}M + m_b} = \frac{0.730(9.81)}{\frac{1}{2}(2.15) + 0.730} = \boxed{3.97 \frac{m}{s^2}}[/tex]

3.

This is just a case of kinematics since we have a constant acceleration. To solve for this instance, however, let's use calculus.

We know that:
[tex]a(t) = 3.97[/tex]

v(t) is the integral. There is no initial velocity (block starts from rest), so:
[tex]v(t) = 3.97t + C\\C = 0 m/s \\\\v(t) = 3.97t[/tex]

Now, we can take the integral of the velocity/time equation to find the displacement between the interval.

[tex]\int\limits^{0.590}_{0.390} {3.97t} \, dt = \boxed{0.389 m}[/tex]

4.

So, let's now find the translational acceleration produced by the new cylinder. Use the kinematic equation:
[tex]\Delta x = v_0 t + \frac{1}{2}at^2[/tex]

Since the block starts from rest, we can do some simplifying and rearranging:
[tex]2\Delta x = at^2\\\\a = \frac{2\Delta x}{t^2} = \frac{2(0.450)}{0.490^2} = 3.748 \frac{m}{s}^2[/tex]

So, let's go back to our summation of torques equation. This time, however, we cannot simplify 'I'.

[tex]I(\frac{a}{r}) = r( m_b g - m_b a)[/tex]

Solving for 'I':
[tex]I = \frac{r^2( m_b g - m_b a)}{a} = \frac{(0.111^2)(7.161 - .730(3.748))}{3.748}\\\\ = \boxed{0.0145 kgm^2}[/tex]

The design that would be used to make an efficient commercial vehicle chassis that has many realistic truck components is:

This refers to the main vehicle structure of a car that bears all the main stress that other components are attached to.

Hence, we can see that to make a good design for an efficient commercial vehicle chassis that has many realistic truck components, we would have to consider various factors such as handling maximum stress, maximum equilateral stress, and deflection.

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Answer:

Fruit Punch

Explanation:

A pure substance is whereby there is only one type of element or a compound in the periodic table in the substance.

Carbon: Well its just C in the periodic table, so this is definitely pure.

Water Molecule: What makes water? H2O right? Contains Hydrogen and Oxygen, and as we all know H2O is a compound, therefore this is a pure substance.

Fruit Punch: What makes fruit punch, water and fruits. Fruits may contain citric acid(a compound itself), and is mixed with water with already has a compound, so having 2 different compounds will result in a mixture and therefore it will not be pure.

The maximum height reached by the human cannonball is 4.74 m.

The height reached is calculated as follows;

H = u²sin²θ/2g

where;

H = (15² x [sin(40)]² )/(2 x 9.8)

H = 4.74 m

Thus, the maximum height reached by the human cannonball is 4.74 m.

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Answer:

35000 KJ

Explanation:

The equation for the kinetic energy is given by the formula :

[tex]E_{k} = \frac{1}{2} mv^{2}[/tex]

[tex]E_{k} = \frac{1}{2} (700)(10)^{2}[/tex]

[tex]E_{k} = \frac{1}{2} (700)(100)[/tex]

[tex]E_{k} = (350)(100)[/tex]  OR [tex]E_{k} = \frac{1}{2} (70000)[/tex]

[tex]E_{k} = 35000[/tex]

Units will be kilojoules since the units of mass was kilograms .

Our final answer is 35000 KJ

Hope this helped and have a good day

The answers are as follows;

a) the inductive reactance is 322 ohm

b) The maximum voltage is 387.5 V

c) The rms and maximum currents in the inductor are 1.2 A and 0.85 A.

The reactance is obtained from;

XL = 2πfL

XL = 2 * 3.14 * 57.0 * 0.900

XL = 322 ohm

The maximum voltage is obtained as;

Vo = Vrms * √2

Vo =  274 V * √2

Vo = 387.5 V

Io = Vo/XL

Io =  387.5 V/ 322 ohm

Io = 1.2 A

Irms = 274 V/322 ohm

Irms = 0.85 A

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For a ball that starts rolling down a ramp the value of the change in its kinetic energy, ΔEk is: Positive

Kinetic energy can be defined as the ability assumed by an object in motion.

Kinetic energy is also said to be a form of mechanical energy that is possessed by an object to be in motion.

In conclusion, For a ball that starts rolling down a ramp the value of the change in its kinetic energy, ΔEk is: Positive

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