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3 years ago

11

If a plane is traveling at a velocity of 100 km/hr East, and if the wind velocity is 25 km/hr to the East, then what is the velo

city of the plane relative to an
observer on the ground below?
100 km/hr East
75 km/hr West
75 km/hr East
125 km/hr East

1 answer:

kkurt [141]3 years ago

5 0

Answer:

75 west

Explanation:

I think not sure lucky guess

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A4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects encounter the same const

Evgesh-ka [11]

Answer:

Both objects travel the same distance.

(c) is correct option

Explanation:

Given that,

Mass of first object = 4.0 kg

Speed of first object = 2.0 m/s

Mass of second object = 1.0 kg

Speed of second object = 4.0 m/s

We need to calculate the stopping distance

For first particle

Using equation of motion

$v^2=u^2+2as$

Where, v = final velocity

u = initial velocity

s = distance

Put the value in the equation

$0= u^2-2as_{1}$

$s_{1}=\dfrac{u^2}{2a}$ ....(I)

Using newton law

$a=\dfrac{F}{m}$

Now, put the value of a in equation (I)

$s_{1}=\dfrac{8}{F}$

Now, For second object

Using equation of motion

$v^2=u^2+2as$

Put the value in the equation

$0= u^2-2as_{2}$

$s_{2}=\dfrac{u^2}{2a}$ ....(I)

Using newton law

$F = ma$

$a=\dfrac{F}{m}$

Now, put the value of a in equation (I)

$s_{2}=\dfrac{8}{F}$

Hence, Both objects travel the same distance.

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2 years ago

PLEASE HELP Which landforms can be associated with the three types of plate boundaries?

Margarita [4]

Convergent, divergent, and transform boundaries

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3 years ago

What electric field strength would store 12.5 JJ of energy in every 6.00 mm3mm3 of space?

Tcecarenko [31]

To develop this problem we will apply the concepts related to the potential energy per unit volume for which we will obtain an energy density relationship that can be related to the electric field. From this formula it will be possible to find the electric field required in the problem. Our values are given as

The potential energy, $U = 13.0 J$

The volume, $V = 6.00 mm^3 = 6.00*10^{-9}m^3$

The potential energy per unit volume is defined as the energy density.

$u = \frac{U}{V}$

$u= \frac{(13.0 J)}{(6.00*10^{-9} m^3)}$

$u= 2.167109 J/m^3$

The energy density related with electric field is given by

$u = \frac{1}{2} \epsilon_0 E^2$

Here, the permitivity of the free space is

$\epsilon_0 = 8.85*10^-{12} C^2/N \cdot m^2$

Therefore, rerranging to find the electric field strength we have,

$E = \sqrt{\frac{2u}{\epsilon_0}}$

$E = \sqrt{\frac{2(2.167109)}{8.85*10^{-12}}}$

$E = 2.211010 V/m$

Therefore the electric field is 2.21V/m

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2 years ago

Two students, standing on skateboards, are initially at rest, when they give each other a shove! After the shove, one student (7

Elden [556K]

Answer:

The other student (59kg) moves right at 7.44 m/s

Explanation:

Given;

mass of the first student, m₁ = 77kg

mass of the second student, m₂ = 59kg

initial velocity of the first student, u₁ = 0

initial velocity of the second student, u₂ = 0

final velocity of the first student, v₁ = 5.7 m/s left

final velocity of the second student, v₂ = ? right

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(77 x 0) + (59 x 0) = (-77 x 5.7) + (59 x v₂)

0 = - 438.9 + 59v₂

59v₂ = 438.9

v₂ = 438.9 / 59

v₂ = 7.44 m/s to the right

Therefore, the other student (59kg) moves right at 7.44 m/s

6 0

3 years ago

In an automobile collision, a 44-kilogram passenger moving at 15 meters per second is brought to rest by an air bag during a 0.1

const2013 [10]

Answer:

6,600N

Explanation:

According to second law of motion, Force = mass × acceleration

If acceleration = change in velocity/time = 15/0.10

Acceleration = 150m/s²

Given mass = 44kg

Force = 44× 150

Force = 6,600N

Magnitude of the average force exerted on the passenger during this time is 6,600N

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3 years ago