An airplane in level flight is acted on by four basic forces. Drag is air resistance, lift is the upward force provided by the wings, thrust is the force provided
by the airplane's engines, and weight is the downward force of gravity acting on the airplane.
Lift
Thrust
Drag
Weight
In level flight at constant speed, which pair of forces must be equal?

The breeze/wind moves from the land to the sea. This statement is NOT correct about sea breezes.

Local wind patterns known as sea breezes flow from the sea to land during the day. When there is no strong large-scale wind system and it is very hot or very cold during the day or at night, sea breezes and land breezes alternate along the coasts of large lakes or oceans.

Because the sea breeze's surface flow ends over land, an area of low-level air convergence is created. Locally, this convergence frequently causes air to rise, which promotes the formation of clouds. Showers over land in the afternoon may result from such clouds.

Hence, the sea  breeze/wind moves from the land to the sea. This statement is NOT correct about sea breezes.

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Answer:

Centripetal acceleration is defined as the property of the motion of an object, traversing a circular path. Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration. ... Centripetal means towards the center.

Answer:

b

Explanation:

Answer:

vavg = 0.37 m/s

Explanation:

       [tex]v_{avg} = \frac{\Delta x}{\Delta t} (1)[/tex]

       [tex]t_{1} =\frac{x_{1}}{v_{1} } = \frac{3.0m}{0.24m/s} = 12.5 s (2)[/tex]

       [tex]\Delta t = t_{1} + \frac{t_{1} }{2} = 12.5 s + 6.25 s = 18.75 s (3)[/tex]

       [tex]v_{avg} = \frac{\Delta x}{\Delta t} = \frac{7.0m}{18.75m} = 0.37 m/s (4)[/tex]

Answer:

τ = RC = 2.17 s

Explanation:

       [tex]V_{C} =V_{o} * e^{-t/RC} (1)[/tex]

       [tex]V_{C} = 1.00 V\\V_{o} = 100V\\t = 10.0 s[/tex]

      [tex]\frac{1.00V}{100 V} = e^{-10s/RC} (2)[/tex]

       [tex]R*C= \frac{-10s}{ln 0.01} = 2.17s (3)[/tex]

Answer:

Answer down below!

Explanation:

Answer:

m

Explanation:

Answer:

A

Explanation:

The angular momentum of a system of particles around a point in a fixed inertial reference frame is conserved if there is no net external torque around that point:

d

→

L

d

t

=

0

or

→

L

=

→

l

1

+

→

l

2

+

⋯

+

→

l

N

=

constant

.

Note that the total angular momentum

→

L

is conserved. Any of the individual angular momenta can change as long as their sum remains constant. This law is analogous to linear momentum being conserved when the external force on a system is zero.

Answer: Because of the amount of energy

Explanation: Heat is an energy, and small amounts of it does not effect large areas, whereas a very large blast of heat energy can greatly effect the air

Answer:

while ice is made by water again it melts and becomes water. water is colourless and odourless and has no taste but ice is only cold and hard. water is used for drinking and other things. but is for freshness and it never flows

Explanation:

so ice reflect more energy compared to water

Answer:

true

Explanation:

i think its true beacuse water can cause electricity

pls let me know if corrct

if yes pls mark brainlest

Answer:

J = 357.5 kg*m/s

Explanation:

       [tex]p_{o} = m*v_{o} = 55 kg * (-3.0 m/s) (1)[/tex]

       [tex]p_{f} = m*v_{f} = 55 kg * (3.5 m/s) (2)[/tex]

      [tex]J = p_{f} - p_{o} = m* (v_{f} -v_{o}) = 55 kg* (3.5m/s- (-3.0m/s)) = 357.5 kg*m/s (3)[/tex]

Explanation:

Assume that mass of Earth is M, radius of earth orbit is R, and rotational period of Earth is T.

The angular momentum of Earth is,

 [tex]L_{z} &=M R^{2} \omega \\ &=M R^{2}\left(\frac{2 \pi}{T}\right) \\ &=\frac{2 \pi M R^{2}}{T}[/tex]

The mass of mars is, mass of Earth

=0.11 M

The radius of mars orbit is, of radius of earth

=0.53 R

The rotational period of mars is, of period of Earth

=1.03 T

The angular momentum of mars is,

[tex]L_{m}=\frac{2 \pi(0.11 M)(0.53 R)^{2}}{1.03 T}[/tex]

The angular momentum of mars is,

[tex]L_{m}=\frac{2 \pi(0.11 M)(0.53 R)^{2}}{1.03 T}[/tex]

The ratio of angular momentum of mars to that of earth is,

[tex]\frac{L_{m}}{L_{E}}=\frac{\frac{2 \pi(0.11 M)(0.53 R)^{2}}{1.03 T}}{\frac{2 \pi M R^{2}}{T}} \\ \frac{L_{m}}{L_{E}}=0.03 \\ \frac{L_{m}}{L_{B}}=3.0 \times 10^{-2}[/tex]

Answer:

72 J

Explanation:

Use the Work formula

W= F x d

Given:

F - 48 N

d - 1.5 m

Solution:

W= F x d

W=  48 N x 1.5 m

W= 72 J

Explanation:

Its D. The warm air from the land moves towards the water

Answer:

1) Linear motion

2) Rotational motion

3) Rotational Motion

4) Random Motion ( The ball can be rolling in any direction)

5) Linear motion

The types of motion in each activity include the following:

This involves an object or a body changing position over time. There are

different types of motion with different examples as can be seen above in

this scenario.

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Answer:

0.15 s

Explanation:

From the question given above, the following data were obtained:

Speed of sound (v) = 330 m/s

Distance (x) = 25 m

Time (t) =?

The time taken for the echo of the sound to the bat can be obtained as follow:

v = 2x / t

330 = 2 × 25 / t

330 = 50 / t

Cross multiply

330 × t = 50

Divide both side by 330

t = 50 / 330

t = 0.15 s

Thus, it will take 0.15 s for the echo of the sound to the bat

Answer:

A Data is transmitted in the form of electromagnetic radiation.

B Signals travel faster through space than through fiber-optic cables.

C No ground equipment is needed to access the internet.

D The network is made of many satellites organized in a grid pattern.

Explanation:

THose are the options

Answer:

B

Explanation:

Answer:

B I believe, because the axe is wedged into the log

Answer:

B

Explanation:

I think just rhis is the answer

Answer:

They're going to come home as soon as the movie is over.

Answer:

B: They're going to come home as soon as the movie is over.

Explanation:

One of the major effects of heat transfer is temperature change: heating increases the temperature while cooling decreases it. We assume that there is no phase change and that no work is done on or by the system. Experiments show that the transferred heat depends on three factors—the change in temperature, the mass of the system, and the substance and phase of the substance.

Figure a shows a copper-colored cylinder of mass m and temperature change delta T. The heat Q, shown as a wavy rightward horizontal arrow, is transferred to the cylinder from the left. To the right of this image is a similar image, except that the heat transferred Q prime is twice the heat Q. The temperature change of this second cylinder, which is also labeled m, is two delta T. This cylinder is surrounded by small black wavy lines radiating outward. Figure b shows the same two cylinders as in Figure a. The left cylinder is labeled m and delta T and has a wavy heat arrow pointing at it from the left that is labeled Q. The right cylinder is labeled two m and delta T and has a wavy heat arrow pointing to it from the left labeled Q prime equals two Q. Figure c shows the same copper cylinder of mass m and with temperature change delta T, with heat Q being transferred to it. To the right of this cylinder, Q prime equals ten point eight times Q is being transferred to another cylinder filled with water whose mass and change in temperature are the same as that of the copper cylinder.

Answer:

its b

Explanation:

Amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. It is equal to one-half the length of the vibration path.

Answer:

C1 + C2 = 30     parallel connection

C1 * C2 / (C1 + C2) = 7.2  series connection

C1 * C2 = 7.2 * (C1 + C2) = 216

C2 + 216 / C2 = 30    using first equation

C2^2 + 216 = 30 C2

C2^2 - 30 C2 + 216 = 0

C2 = 12 or 18    solving the quadratic

Then C1 = 18 or 12

Answer:the answer is galaxy

Explanation:

Answer:

(a) Q = 142.67 W

(b) Basal Metabolic Rate = 178.33 W

Explanation:

(a)

We can find the heat radiated by the person by using Stefan-Boltzman's law:

[tex]Q = \sigma A (T^4 - T_{s}^4)\\[/tex]

where,

Q = heat radiated per second = ?

σ = Stefan-Boltzman Constant = 5.6703 x 10⁻⁸ W/m².k⁴

A = Surface Area = 2 m²

T = Temperature of Skin = 30° C + 273 = 303 k

Ts = Temperature of room = 18° C +273 = 291 k

Therefore,

[tex]Q = (5.6703\ x\ 10^{-8}\ W/m^2.k^4)(2\ m^2)[(303\ k)^4-(291\ k)^4][/tex]

Q = 142.67 W

(b)

Since the heat calculated in part (a) is 80 percent of basal metabolic rate. Therefore,

[tex]Q = (0.8)(Basal\ Metabolic\ Rate)\\Basal\ Metabolic\ Rate = \frac{Q}{0.8}\\\\Basal\ Metabolic\ Rate = \frac{142.67\ W}{0.8}[/tex]

Basal Metabolic Rate = 178.33 W

(a)  The  excess heat just through radiation Q = 142.67 W

(b) Basal Metabolic Rate = 178.33 W

Basal metabolic rate is the number of calories your body needs to accomplish its most basic (basal) life-sustaining functions.

(a) We can find the heat radiated by the person by using Stefan-Boltzman's law:

[tex]Q=\sigma A(T^4-T_s^4)[/tex]

where,

Q = heat radiated per second = ?

σ = Stefan-Boltzman Constant = 5.6703 x 10⁻⁸ W/m².k⁴

A = Surface Area = 2 m²

T = Temperature of Skin = 30° C + 273 = 303 k

Ts = Temperature of room = 18° C +273 = 291 k

Therefore,

[tex]Q=(5.6703\times 10^{-8}\times (2)\times[(303)-(291)][/tex]

Q = 142.67 W

(b) Since the heat calculated in part (a) is 80 percent of basal metabolic rate. Therefore,

[tex]\rm Q=0.8\times (Basal\ Metabolic\ Rate)[/tex]

[tex]\rm Basal\ metabolic \ Rate = \dfrac{Q}{0.8}[/tex]

[tex]\rm Basal\ Metabolic\ Rate = \dfrac{142.67}{0.8} =178.330W[/tex]

Basal Metabolic Rate = 178.33 W

Hence the  excess heat just through radiation Q = 142.67 and Basal Metabolic Rate = 178.33 W

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Answer:

speed =distance /time

speed =4.82/0.00360

speed =1338.8m/s

Answer:

his distance is too small (r = 0.01 mm), therefore the cut can be at any distance

Explanation:

For this exercise let's use the ampere law.

Let's use a cylinder as the circulating surface

          ∫ B. ds = μ₀ I

in this case the field is circular and ds is circular therefore the angle between them is zero and cos 0 = 1

          B 2π r =  μ₀ I

          r =  [tex]\frac{\mu_o I}{2\pi B}[/tex]

The field needed to demagnetize the card is B = 1000 gauss = 0.1 T

          r = [tex]\frac{4\pi 10^{-7} 5.0 }{2\pi \ 0.1}[/tex]

           r = 2 10⁻⁷  5.0/0.1

          r = 1  10⁻⁵ m

this distance is too small (r = 0.01 mm), therefore the cut can be at any distance

Answer:

a) 6636 km

b) 0.0154

Explanation:

The height above the earth at its furthest point is 368 km

The height above the earth at its closest point is 164 km

Radius of the Earth is 6370 km

The distance of the satellite from the center of the earth to the furthest point is 6370 + 368 km = 6738 km

The distance of the satellite from the center of the earth to the closest point is 6370 + 164 = 6534 km

If we add together the sum of the distance of the satellite from the furthest and its closest distance, it is equal to the 2 major semi axis.

Basically,

2a = R + r

a = (R + r) / 2

a = (6738 + 6534) / 2

a = 13272 / 2

a = 6636 km

Eccentricity, e = (a - r) / a

Eccentricity, e = (6636 - 6534) / 6636

Eccentricity, e = 102 / 6636

Eccentricity, e = 0.0154