An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the point of origin of the flight. The plane flies with an airspeed of 120 m/s. If a constant wind blows at 10 m/s toward the west during the flight, what direction must the plane fly relative to the air to arrive at the destination

Answer:

3.86×10⁶ Newton/coulombs

Explaination:

Applying,

E = F/q....................... Equation 1

Where E = Electric Field, F  = Force, q = charge.

From the question,

Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs

Substitute these values into equation 1

E = 5.4×10⁻¹/ -1.4×10⁻⁷

E = -3.86×10⁶ Newtons/coulombs

Hence the magnitude of the electric field created by the

negative test charge is 3.86×10⁶ Newton/coulombs

Answer:

(a) vf = 1.51 m/s

(b) vf = 36.22 m/s

Explanation:

The rate of change of momentum is equal to the force:

[tex]F = \frac{mv_f-mv_i}{t}[/tex]

[tex]Ft = m(v_f-v_i)[/tex]

where,

F = Force = 1035 N

t = time = 0.175 s

vi = initial speed = 0 m /s

vf = final speed = ?

(a)

m = mass of body = 120 kg

Therefore,

[tex](1035\ N)(0.175\ s)=(120\ kg)(v_f - 0\ m/s)\\\\v_f = \frac{181.125\ Ns}{120\ kg} \\\\[/tex]

vf = 1.51 m/s

(b)

m = mass of head = 5 kg

Therefore,

[tex](1035\ N)(0.175\ s)=(5\ kg)(v_f - 0\ m/s)\\\\v_f = \frac{181.125\ Ns}{5\ kg} \\\\[/tex]

vf = 36.22 m/s

Please delete my answer. I made a mistake

Answer:

The speed of the combined mass after the collision is 2.1 m/s.

Explanation:

mass of runner, m = 70 kg

speed  of runner, u = 2.7 m/s

mass of shortstop, m' = 85 kg

speed  of shortstop, u' = 1.6 m/s

Let the velocity of combined system is v.

Use conservation of momentum

Momentum before collision = momentum after collision

m u + m' u' = (m + m') v

70 x 2.7 + 85 x 1.6 = (70 + 85) v

189 + 136 = 155 v

v = 2.1 m/s

Answer:

His height is 6.66 feet or 79.92 inches.

Explanation:

Given that,

An AAMU basketball player is 2.03 meters tall.

Let h is the height.

We know that,

1 m = 3.28 feet

So,

2.03 m = 6.66 feet

Also,

1 m = 39.37 inches

2.03 m = 79.92 inches

Hence, this is the required solution.

Hi there!

[tex]\large\boxed{\text{D. 675000J}}[/tex]

Use the following formula to solve:

KE = 1/2mv², where:

KE = kinetic energy

m = mass (kg)

v = velocity (m/s)

Therefore:

KE = 1/2(1500)(30)²

KE = 1/2(1500)(900)

KE = 675000 J

Answer:

A. the motion of electrically charged particles

Answer:

answer is d

Explanation:

i hope this helps you

Answer:

kinetic

Explanation:

i just remember it from last year

Answer:

kinetic energy

Explanation:

expression for kinetic energy is

kinetic energy = (1/2) × mass × (velocity)^2

so , as velocity increases K.E increases

Answer:

ans: 2.25 meter

explanation

use following equations

F = ma

V = U + aT

S = UT + 1/2 aT^2

Answer:

sports are all forms of physical activity that contribute to physical fitness, mental well-being and social interaction.

hope it is helpful to you

Answer:

Work done by the bullet is 612.26 J.

Explanation:

mass of bullet, m = 0.5 kg

initial velocity of bullet, u = 50 m/s

coefficient of friction = 0.2

mass of block, M = 3 kg

let the final speed of the bullet block system is v.

use conservation of momentum

Momentum of bullet + momentum of block = momentum of bullet block system

0.5 x 50 + 3 x 0 = (3 + 0.5) v

v = 7.14 m/s

let the stopping distance is

The work done is given by change in kinetic energy of bullet

initial kinetic energy of bullet, K =  0.5 x 0.5 x 50 x 50 = 625 J

Final kinetic energy of bullet, K' = 0.5 x 0.5 x 7.14 x 7.14 = 12.74 J

So, the work done by the bullet

W = 625 - 12.74 = 612.26 J  

Let g be the acceleration due to gravity on the surface of the star. By Newton's second law, the gravitational force felt by the object has a magnitude of

F = GMm/r ² = mg

where

• G = 6.67 × 10⁻¹¹ Nm²/kg² is the gravitational constant,

• M = 2.08 × 10³⁰ kg is the mass of the star,

• m is the unknown mass of the object, and

• r = 6.73 × 10³ m is the radius of the star

Solving for g gives

g = GM/r ²

g = (6.67 × 10⁻¹¹ Nm²/kg²) (2.08 × 10³⁰ kg) / (6.73 × 10³ m)²

g ≈ 3.06 × 10¹² m/s²

The object is in free fall with uniform acceleration and starting from rest, so its speed after falling 0.0093 m is v such that

v ² = 2g (0.0093 m)

v = √(2g (0.0093 m))

v ≈ 240,000 m/s ≈ 240 km/s

I assume you're talking about a pilot. If the ejection seat has an acceleration of 8g, then it would exert a normal force of 8g (70 kg) ≈ 5600 N.

(This is assuming the pilot is flying horizontally at a constant speed, and the seat is ejected vertically upward.)

To reiterate, this is *only* the force exerted by the seat on the pilot. Contrast this with the net force on the pilot, which would be the normal force minus the pilot's weight, 5600 N - (70 kg)g ≈ 4900 N.

If instead the seat ejects the pilot directly downward, the force exerted by the seat would have the same magnitude of 5600 N, but its direction would be reversed to point downward, making it negative. But the net force would change to -5600 N - (70 kg)g ≈ -6300 N

Answer:

4.02×10⁻⁴ K⁻¹

Explanation:

Applying,

γ = (v₂-v₁)/(v₁Δt)................. Equation 1

Where γ  = coefficient of volume expansion, v₂ = Final volume, v₁ = initial volume, Δt = change in temperature.

From the question,

Given: v₂ = 850 m³, v₁ = 830 m³, Δt = (90-30) = 60°C

Substitute these values into equation 1

γ  = (850-830)/(830×60)

γ  = 20/(830×60)

γ  = 4.02×10⁻⁴ K⁻¹

Answer:

Alleles AA and aa

Explanation:

For all the offsprings of the butterflies to have the same heterozygous alleles Aa, it means the two parents have different homogyzous alleles. That is, one of the parents had the alleles AA while the other had the alleles aa. Thus, a combination of the two pairs of alleles will produce 100% Aa alleles in the offspring as seen in the image attached.

When we exhale, 90% waste material is Carbon Dioxide ( CO2 ) , so, it gets exhaled out in the form of CO2 rich air and it gets removed from the body, therefore our internal body becomes more pure and helps in making our internal temperature constant at a suitable level.

Built by 1800, it was the home of inventor Jethro Wood (1774-1834), whose 1819 invention of an iron moldboard plow revolutionized American agriculture.

1) (40 N) cos(30°) ≈ 34.6 N

2) (40 N) sin(30°) = 20 N

3) (65 kg) g = (65 kg) (9.80 m/s²) = 585 N

4) The net force on the sled acting in the vertical direction is made up of

• the sled's weight, 585 N, pointing downward

• the vertical component of the applied force, 20 N, pointing upward

• the normal force, with magnitude n, also pointing upward

The sled does not move up or down, so by Newton's second law,

∑ F = n + 20 N - 585 N = 0   ==>   n = 565 N

5) The net force in the horizontal direction consists of

• the horizontal component of the applied force, 34.6 N, acting in the direction the sled's movement (call this the positive direction)

• kinetic friction, with magnitude f, pointing in the opposite and negative direction

By Newton's second law,

∑ F = 34.6 N - f = 0   ==>   f ≈ 34.6 N

Now if µ is the coefficient of kinetic friction, then

f = µn   ==>   µ = f/n = (34.6 N) / (565 N) ≈ 0.0613

The component of the force is the effective part of that force in that direction.

The component of the force is the effective part of that force in that direction.

1) The horizontal component of a force = 40 N cos 30 degrees = 34.6 N

2) The vertical component of the force = 40 N sin 30 degrees = 20 N

3) The magnitude of the gravitational force = mg cos 30 degrees  = 65 Kg * 9.8 m/s^2 * cos 30 degrees = 551.7 N

4) The normal force = 551.7 N

5) The coefficient of friction = F/R =  40 N /551.7 N = 0.07

Learn more about component of a force:https://brainly.com/question/15529350

#SPJ6

Answer:

B = 4.51×10⁻⁵ T

Explanation:

Given that,

The average magnetic field strength of the Earth is 0.0000451 T.

We need to write the value in the form of scientific notation. Any number in scientific notation is written as follows :

N=a×bⁿ

Where

n is any integer and a is a real no

So,

0.0000451 = 4.51×10⁻⁵ T

So, the required answer is equal to 4.51×10⁻⁵ T.

Answer:pp

Explanation:

ii

Answer:

When the blood passes into the smaller branches, its average velocity reduces by a factor of 0.12

Explanation:

Given;

initial area of the artery, A₁ = 1.3 cm²

Area of each smaller 18 arteries, a₂ = 0.6 cm²

Total area of the smaller 18 arteries, A₂ = 18 x 0.6 cm²

Apply flow rate equation;

Q = AV

where;

Q is the flow rate of the blood

V is the average velocity of the blood

If the flow rate is constant, then;

A₁V₁ = A₂V₂

[tex]V_2 = \frac{A_1V_1}{A_2} = \frac{1.3\times V_1}{18\times 0.6} \\\\V_2 = 0.12 \ V_1[/tex]

When the blood passes into the smaller branches, its average velocity reduces by a factor of 0.12

Answer:

hey. i dont know what you tryna say but if u replying to someone else, you should use the comments section. in that way you won't lose points.

Answer:

1) 0.3675

2) 0.367

3) 72.6186

4) 0.5

5) 0.000671

Answer:

1) 367.5 mg = 0.3675 g

2) 367 mL = 0.367 L

3) 28.59 in = 72.61 cm

4) 8 0z = 0.5 lb

5) 0.671 mm = 0.0000671 m

Answer:

(a) The acceleration is 2.35 m/s^2.

(b) The time is 5.53 s.

Explanation:

coefficient of friction = 0.24

(a) The acceleration of the car is

[tex]a =\mu g\\\\a = 0.24 \times9.8\\\\a = 2.35 m/s^2[/tex]

(b) initial velocity, u =0, final velocity, v = 13 m/s

Let the time is t.

Use first equation of motion

v = u + a t

13 = 0 + 2.35 x t

t = 5.53 seconds

Answer:

B.

Explanation:

sana makatulong sayo

Answer:

sorry dear

Explanation:

sorry dear but ur question is hard to understand can u try to edit it so i can tell u the answer?

I suppose the hill makes an angle of 5.0° with the horizontal.

• F acts parallel to the road and in the direction of the car's motion, so it contributes a positive amount of work, F (290 m).

• Friction does negative work on the car since it opposes the car's motion. As the car moves up the slope, the work done by friction is (-524 N) (290 m) = -151,960 J.

• The car's weight has components that act parallel and perpendicular to the road. The parallel component has a magnitude of W sin(5.0°) and points down the slope, so it contributes negative work of -(1200 kg) g sin(5.0°) ≈ 1,024.95 J. The perpendicular component of W does not do any work.

• The normal force FN also doesn't do any work to move the car up the slope because it points perpendicular to the road, so we can ignore it, too.

The net work done on the car is then

F (290 m) + (-151,960 J) + 1,024.95 J = 150,000 J

==>   F (290 m) ≈ 300,935 J

==>   F ≈ (300,935 J) / (290 m) ≈ 1,037.71 N