How much does the Earth weigh? Why does the moon appear in the daytime?

Doss [256]

Answer:

The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.

Explanation:

Considering west direction along negative x-axis and north direction along positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

The car then travels at the same speed in the north direction.

Now, considering the given directions, the velocities are given as:

Velocity in west direction is, $\overrightarrow{v_1}=-120\ \vec{i}$

Velocity in north direction is, $\overrightarrow{v_2}=120\ \vec{j}$

Now, since $v_1\ and\ v_2$ are perpendicular to each other, their resultant magnitude is given as:

$|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}$

Plug in the given values and solve for the magnitude of the resultant.This gives,

$|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h$

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

$x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)$

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

4 0

3 years ago

Based on its location on the periodic table, which element is a metal?

goldfiish [28.3K]

Sodium (Na) is a Alkali metal so that's the answer

6 0

3 years ago

Read 2 more answers

"My distance to the center of the earth is about 4000 miles when I am on the surface. If I go to a height of 8000 miles above th

lord [1]

Given,

Distance from the surface to the center of the earth, d=4000 miles

Distance from the center to you at a height of 8000 miles, a= 8000+4000=12000 miles

The gravitational force acting on a person at the surface is equal to his weight.

From Newton's Universal Law of Gravitation, the gravitational force is

$F=\frac{G\times M\times m}{r^2}$

Where G is the gravitational constant, M is the mass of the earth, m is the mass of the object/person, r is the distance between the center of the earth and the object/person

At the surface, this force is equal to the weight of the person, W=mg

i.e.

$F_s=\frac{G\times M\times m}{d^2}=W$

On substituting the of d,

$W=\frac{\text{GMm}}{4000^2}$

At a height of 8000 miles from the surface, the gravitational force is equal to,

$F_a=\frac{GMm}{12000^2}$

On dividing the above two equations,

$\frac{F_a}{W}=\frac{4000^2^{}}{12000^2}=\frac{1}{9}$

Therefore,

$F_a=\frac{1}{9}W$

Therefore at a height of 8000 miles above the surface of the earth, the force of gravity becomes 1/9 time your weight.

5 0

1 year ago

Three balls are kicked from the ground level at some angles above horizontal with different initial speeds. All three balls reac

Charra [1.4K]

Answer:

Time of flight A is greatest

Explanation:

Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.

So

H = u₁² sin²θ₁ /2g

H = u₂² sin²θ₂ /2g

H = u₃² sin²θ₃ /2g

On the basis of these equation we can write

u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃

For maximum range we can write

D = u₁² sin2θ₁ /g

1.5 D = u₂² sin2θ₂ / g

2 D =u₃² sin2θ₃ / g

1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁

1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )

u₂ cosθ₂ >u₁ cosθ₁

u₂ sinθ₂ < u₁ sinθ₁

2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g

Time of flight B < Time of flight A

Similarly we can prove

Time of flight C < Time of flight B

Hence Time of flight A is greatest .

8 0

3 years ago

When you jump from an elevated position you usually bend your knees upon reaching the ground. By doing this, you make the time o

dsp73

Answer:

c. about 1/10 as great.

Explanation:

While jumping form a certain height when we bend our knees upon reaching the ground such that the time taken to come to complete rest is increased by 10 times then the impact force gets reduced to one-tenth of the initial value when we would not do so.

This is in accordance with the Newton's second law of motion which states that the rate of change in velocity is directly proportional to the force applied on the body.

Mathematically:

$F\propto\frac{d}{dt} (p)$