An object is launched from the top of a building which is 100 m tall (relative to the ground) at a speed of 22 m/s at an angle of 23 degrees BELOW the horizontal. How long, in seconds, does it take to reach the ground

Answer:

D Venus

Explanation:

in terms of size, average, density, mass and surface gravity venus is similar to earth

Answer:

the correct answers is D

Explanation:

Thermal energy can be transferred by three methods: conduction, convention, and radiation.

Radiation transfer occurs when there is no movement of matter for the exchange of energy.

In this case, checking the correct answers is D

since in this case the transfer is between light and sand without matter exchange

Answer:

Explanation:

The energy of a photon is given by the Planck relation

          E = h f

the speed of light is related to wavelength and frequency

          c = λ f

           f- c /λ

we substitute

          E = h c /λ

let's calculate

          E = 6.63 10-34 3 10⁸ / 550 10-9

          E = 3.616 10-19 J

let's reduce to eV

          E = 3.616 10-19 J (1 eV / 1.6 10-19)

          E = 2.26 eV

Answer:

c

Explanation:

Superconductors perform best at very cold temperatures because resistivity of this kind of materials decays drastically with temperature. Chromium is most likely to be the best conductor of electricity because it belongs to the Transition Metal group of the periodic table

Answer:

a)    F = 35.7 N, b)   W = 846.7 J, c)   W = - 846.9 J, d) W=0

Explanation:

a) For this exercise let's use Newton's second law, let's set a reference frame with the x-axis horizontally

let's break down the pushing force.

        cos (-23,7) = Fₓ / F

        sin (-237) = F_y / F

        Fₓ = F cos 23.7 = F 0.916

        F_y = F sin (-23.7) = - F 0.402

Y axis  

       N- W - F_y = 0

       N = W + F 0.402

X axis

       Fₓ - fr = 0

       F 0.916 = fr

       F = fr / 0.916

       F = 32.7 / 0.916

       F = 35.7 N

It is asked to calculate several jobs

b) the work of the pushing force

       W = fx x

       W = 35.7 cos 23.7 25.9

       W = 846.7 J

c) friction force work

        W = F x cos tea

friction force opposes movement

        W = - fr x

         W = - 32.7 25.9

         W = - 846.9 J

d) The work of the force would gravitate, as the displacement and the force of gravity are at 90º, the work is zero

          W = 0

Answer:

because thermometric liquid readily expands on heating or contracts on cooling even for a small difference in the temperature of the body.

Answer:

v_f = 10.38 m / s

Explanation:

For this exercise we can use the relationship between work and kinetic energy

          W = ΔK

note that the two quantities are scalars

Work is defined by the relation

          W = F. Δx

the bold are vectors.  The displacement is

          Δx = r_f -r₀

          Δx = (11.6 i - 2j) - (4.4 i + 5j)

          Δx = (7.2 i - 7 j) m

          W = (4 i - 9j). (7.2 i - 7 j)

remember that the dot product

           i.i = j.j = 1

           i.j = 0

           W = 4  7.2 + 9  7

           W = 91.8 J

the initial kinetic energy is

           Ko = ½ m vo²

           Ko = ½ 2.0 4.0²

           Ko = 16 J

we substitute in the initial equation

          W = K_f - K₀

          K_f = W + K₀

          ½ m v_f² = W + K₀

          v_f² = 2 / m (W + K₀)

          v_f² = 2/2 (91.8 + 16)

           v_f = √107.8

          v_f = 10.38 m / s

16 because complementary angles equal up to 90. 2x+58 = 90 x= 16

Answer:

- the effective lift area for the aircraft is 8.30 m²

- the required engine thrust is 1275 N

- required power is 79.7 kW

Explanation:

Given the data in the question;

Speed V = 225 km/hr = 62.5 m/s

The lift coefficient CL = 0.45

drag coefficient CD = 0.065

mass = 900 kg

g = 9.81 m/s²

a)  the effective lift area for the aircraft

we know that for a steady level flight, weight = lift and thrust = drag

Using the equation for the lift force

F[tex]_L[/tex] = C[tex]_L[/tex][tex]\frac{1}{2}[/tex]ρV²A = W

we substitute

0.45 × [tex]\frac{1}{2}[/tex] × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )

1081.05 × A = 8829

A = 8829 / 1081.05

A = 8.30 m²

Therefore, the effective lift area for the aircraft is 8.30 m²

b) the required engine thrust and power to maintain level flight.

we use the expression for drag force

F[tex]_D[/tex] = T = C[tex]_D[/tex][tex]\frac{1}{2}[/tex]ρV²A

we substitute

= 0.065 × [tex]\frac{1}{2}[/tex] × 1.21 × ( 62.5 )² × 8.30

T = 1275 N

Since drag and thrust force are the same,

Therefore, the required engine thrust is 1275 N

Power required;

P = TV

p = 1275 × 62.5

p = 79687.5 W

p = ( 79687.5 / 1000 )kW

p = 79.7 kW

Therefore, required power is 79.7 kW

Answer:

Explanation:

The mass of the object can be found by using the formula

[tex]m = \frac{f}{a} \\ [/tex]

f is the force

a is the acceleration

From the question we have

[tex]m = \frac{6}{3} \\ = 2[/tex]

We have the final answer as

Hope this helps you

Answer:

V = 346.15 Volts

Explanation:

Given that,

Diameter of the sphere, d = 0.390 cm

Radius, r = 0.195 cm

Charge, [tex]q=75\ pC =75\times 10^{12}\ C[/tex]

The electric potential near its surface is given by :

[tex]V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 75\times 10^{-12}}{0.195\times 10^{-2}}\\\\V=346.15\ V[/tex]

So, the potential near its surface is equal to 346.15 V.

Answer:

C_d = 0.942

Explanation:

Let's first calculate the angle of inclination.

Formula is;

tan θ = (%slope)

% Slope is given as 12%

Thus;

θ = tan^(-1) (12/100)

θ = 6.843°

Let's now calculate the force due to the weight of the rider and bike combined from;

F = mg sin θ

We are given; m = 100 kg.

Thus;

F = 100 × 9.81 × sin 6.843

F = 116.885 N

The drag force will also be the same as the force due to the weight of the body. Thus;

Drag force; F = C_d(½ρu²A)

Where;

C_d is drag coefficient

ρ is density

U is terminal speed

A is area

We are given;

A = 0.9 m²

U = 15 m/s

From online tables, density of air is approximately 1.225 kg/m³

Thus;

116.885 = C_d(½ × 1.225 × 15² × 0.9)

116.885 = 124.03125C_d

C_d = 116.885/124.03125

C_d = 0.942

Answer:

carbon d

Explanation:

Answer:

673km

Explanation:

Answer:

[tex]x_2=1.60m[/tex]

Explanation:

From the Question We are told that

Initial Force [tex]F_1=5800N[/tex]

Final Force [tex]F_2=6500N[/tex]

Distance between the front and rear wheels \triangle x=3.20 m

Since

 [tex]\triangle x=3.20 m[/tex]

Therefore

 [tex]x_1+x_2=3.20[/tex]

 [tex]x_1=3.20-x_2[/tex]

Generally the equation for The center of mass is at x_2 is mathematically

given by

 [tex]x_2 =\frac{(F_1x_1+F_2x_2)}{(F_1+F_2)}[/tex]

 [tex]x_2=3.20F_1-\frac{x_2F_1+F_2x_2}{(F_1+F_2)}[/tex]

 [tex]2*F_1*x_2 =3.20F_1[/tex]

 [tex]x_2=1.60m[/tex]

Center of gravity of a body is the  sum of its moments divided by the overall weight of the object. The distance between the front wheels and the truck's center of gravity is 1.6 meters.

Given-

Scale reading value when the front wheels drive over the scale [tex]m_{1}[/tex] is 5800 N.

Scale reading value when the rear wheels drive over the scale [tex]m_{2}[/tex] is 6500 N

Distance between the front and rear wheel [tex]\bigtriangleup x[/tex] is 3.20 meters.

Let, the distance between the front wheels and the truck's center of gravity is [tex]x_{2}[/tex].

Since sum of the distance between front wheel to truck's center of gravity [tex]x_{1}[/tex], and rear wheel to truck's center of gravity [tex]x_{2}[/tex], is equal to the distance between the front and rear wheel [tex]\bigtriangleup x[/tex]. Therefore,

[tex]\bigtriangleup x=x_{1} +x_{2}[/tex]

[tex]3.20=x_{1} +x_{2}[/tex]

[tex]x_{1} =3.20-x_{2}[/tex]

For the distance between the front wheels and the truck's center of gravity is the formula of center of gravity can be written as,

[tex]x_{2} =\dfrac{m_{1}x_{1}+m_{2} x_{2} }{m_{1} +m_{2} }[/tex]

[tex]x_{2} =\dfrac{5800\times (3.20- x_{2})+6500\times x_{2} }{5800 +6500 }[/tex]

[tex]1230 x_{2} ={18560-5800 x_{2}+6500 x_{2} }[/tex]

[tex]x_{2}= 1.6[/tex]

Hence, the distance between the front wheels and the truck's center of gravity is 1.6 meters.

For more about the center of gravity, follow the link below-

https://brainly.com/question/20662119

Answer: D. movement of material along a coast by waves that approach at an angle to the shore.​

Explanation:

Longshore drift is also referred to as the littoral druft and it means the sediment that is moved by the longshore current.

Longshore drift is the movement of material along a coast by waves which approach at an angle to the shore but then recede down the beach.

Therefore, the correct option is D.

Answer:

At the earth's surface  g = G M / R^2        where G is the gravitational constant

at H       g1 = G M / (R + H)^2         using Gauss' theorem for enclosed mass

g1 = G M / (R^2 + 2 R H)    ignoring H^2 as it is small compared to R^2

g / g1 = (R^2 + 2 R H) / R^2 = 1 + 2 R H

g = g1 + 2 R H g1

g1 - g = - 2 R H     or H = (g1 - g) / 2 R

Answer:

Yes

Explanation:

-A solid has a definite shape and volume.

-Solids in general have higher density.

-In solids, intermolecular forces are strong.

la respuesta correcta es la "E"

Answer:

The option (B) is correct.

Explanation:

The magnetic force between the two current carrying wires is given by

[tex]F =\frac{\mu o}{4\pi}\times \frac{2 I I'}{r}[/tex]

where,  I and I' be the currents in the wires and r is the distance between the two wires.

Here, we observe that the force between the wires is inversely proportional to the distance between them.

So, when the distance is doubled , let the new force is F'.

[tex]\frac{F'}{F}=\frac{r}{r'}\\\\\frac{F'}{F}=\frac{d}{2d}\\\\F'=\frac{F}{2}[/tex]

So, option (B) is correct.

Answer:

The answer is "0.2711  m/s".

Explanation:

Potential energy = Kinetic energy + Potential energy

[tex]m_1 gh =\frac{1}{2} m_1v^2 +\frac{1}{2} m_2v^2 + \frac{1}{2} I\omega^2 + m_1gh\\\\[/tex]

[tex](m_1- m_2)gh =\frac{1}{2} m_1v^2 +\frac{1}{2} m_2v^2 +\frac{1}{2} I\omega^2\\\\2(m_1 - m_2)gh = m_1v^2 + m_1v^2 + I\omega^2\\\\solid \ disk (I) = \frac{1}{2} \ \ M r^2 \\\\[/tex]

When there is no slipping, \omega  =\frac{ v]{r}\\\\  

[tex]2(m_1 - m_2)gh = m_1v^2 + m_2v^2 + (\frac{1}{2} Mr^2) (\frac{v}{r})^2\\\\2(m_1 -m_2)gh = m_1v^2 + m_2v^2 + \frac{1}{2} Mv^2\\\\4(m_1 -m_2)gh = 2m_1v^2 + 2m_2v^2 + Mv^2\\\\4(m_1 - m_2)gh = (2m_1 + 2m_2 + M) v^2\\\\[/tex]

[tex]v^2 = \frac{4(m_1 - m_2)gh}{(2m_1 + 2m_2 + M)}v[/tex]

[tex]v^2 = \frac{4 (0.25 \ kg - 0.20 \ kg) (9.8 \frac{m}{s^2}) (0.21 m)}{ (2 \times 0.25 kg + 2 \times 0.20 kg + 0.50 kg)}[/tex]

[tex]=\frac{0.1029}{1.4} \ \ \frac{m^2}{s^2}\\\\=0.0735\ \ \frac{m^2}{s^2}\\\\= 0.2711 \ \frac{m}{s}[/tex]

Answer:

The magnitude of the friction force is also F.

Explanation:

By the second Newton's law, we know that:

F = m*a

Net force is equal to mass times acceleration.

Here, we know that the box moves with constant speed, thus, the box has no acceleration, then the net force applied to the box is zero.

Also, remember that the friction force is given by:

[tex]F_f = -\mu*W[/tex]

Where mu is the coefficient of friction, and this force opposes to the direction of motion (that coincides with the direction of our forward force, that is why this has a negative sign)

The net force will be equal to the sum of our two horizontal forces (as the weight is already canceled by the normal force)

[tex]F_{total} = F + F_f[/tex]

And this is equal to zero, because we know that the box is non-accelerated.

Then we must have that:

[tex]F_f = -F[/tex]

Then we can conclude that the magnitude of the friction force is F.

Answer:

Following are the solution to the given question:

Explanation:

Its strength from both charges is equivalent or identical. The power is equal. And it is passed down

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

Therefore, the extent doesn't rely on the fact that charges are the same or different. Newton's third law complies with Electrostatic Charges due to a couple of charges. They are similar in magnitude, and they're in the other way.

[tex]|F_{12}| = |F_{21}|[/tex]

Answer:

v = 0.5 m/s.

Explanation:

Total distance, d = 20 m + 20 m = 40 m

Total time taken, t = 30 s + 50 s = 80 s

The average speed of an object is the total distance divided by time taken. So,

[tex]v=\dfrac{40}{80}\\\\=0.5\ m/s[/tex]

So, the average speed of the object is 0.5 m/s.