What did Edwin Hubble discover that supported the Big Bang Theory?

A gazelle leaps from a cliff 2.5 m high with a speed of 5.6m/s.

trapecia [35]

Initial speed of Gazelle is along x direction and its value will be

$v_x = 5.6 m/s$

also its initial height is given as

$y = 2.5 m$

Part a)

now from kinematics along Y direction

$\Delta y = v_y t + \frac{1}{2} at^2$

as we know that

$\Delta y = 0$

$v_y = 0$

$a = 9.8 m/s^2$

$2.5 = 0 + \frac{1}{2} (9.8) t^2$

$t = 0.714 s$

Part b)

distance moved horizontally

$\Delta x = v_x t$

as we know that

$v_x = 5.6 m/s$

now we will have

$v_x = 5.6 (0.714) = 4m$

so it will lend at distance of 4 m.

Part c)

final velocity in vertical direction

$v_{fy} = v_y + at$

$v_{fy} = 0 + (9.8)(0.714) = 7 m/s$

$v_x = 5.6 m/s$

so net speed will be

$v^2 = v_x^2 + v_y^2$

$v^2 = 7^2 + 5.6^2$

$v = 8.96 m/s$

7 0

4 years ago

How is radiometric dating done? A. The ratio of the isotopes from two different elements is compared over time. B. The ratio of

SpyIntel [72]

The answer should be C.

Ratio it has two radioactive isotopes is compared by overtime.

Hope it helped you.

-Charlie

3 0

3 years ago

An archer shot a 0.06 kg arrow at a target. The arrow accelerated at 5,000 m/s2 to reach a speed of 50.0 m/s as it left the bow.

laila [671]

Answer:

300 N

Explanation:

The net force acting on the arrow is given by Newton's Second Law:

$F=ma$

where

m = 0.06 kg is the mass of the arrow

a = 5,000 m/s^2 is the acceleration of the arrow

Substituting the numbers into the equation, we find

$F=(0.06 kg)(5,000 m/s^2)=300 N$

7 0

4 years ago

_______ are different forms of a single element. A) Atoms B) Elements C) Ions D) Isotopes

ELEN [110]

The answer is D. Isotopes.
Hope that helped.

7 0

4 years ago

Read 2 more answers

A 25,000-kg train car moving at 2.50 m/s collides with and connects to a train car of equal mass moving in the same direction at

Margarita [4]

Answer:

14062.5 J

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

V = (m₁u₁ + m₂u₂)/(m₁+m₂).................1

Where V = common velocity after collision

Given: m₁ = m₂ = 25000 kg, u₁ = 2.5 m/s, u₂ = 1 m/s

Substitute into equation 1

V = [25000(2.5) + 25000(1)]/(25000+25000)

V = (62500+25000)/50000

V = 87500/50000

V = 1.75 m/s.

Note: The collision is an inelastic collision as such there is lost in kinetic energy of the system.

Total Kinetic energy before collision = kinetic energy of the first train car + kinetic energy of the second train car

E₁ = 1/2m₁u₁² + 1/2m₂u₂²........................ Equation 2

Where E₁ = Total kinetic energy of the body before collision, m₁ and m₂ = mass of the first train car and second train car respectively. u₁ and u₂ = initial velocity of the first train car and second train car respectively.

Given: m₁ = m₂ = 25000 kg, u₁ = 2.5 m/s, u₂ = 1 m/s

Substitute into equation 2

E₁ = 1/2(25000)(2.5)² + 1/2(25000)(1.0)²

E₁ = 12500(6.25) + 12500

E₁ = 78125+12500

E₁ = 90625 J.

Also

E₂ = 1/2V²(m₁+m₂)....................... Equation 3

Where E₂ = total kinetic energy of the system after collision, V = common velocity, m₁ and m₂ = mass of the first and second train car respectively.

Given: V = 1.75 m/s, m₁ = m₂ = 25000 kg

Substitute into equation 3

E₂ = 1/2(1.75)²(25000+25000)

E₂ = 1/2(3.0625)(50000)

E₂ = (3.0625)(25000)

E₂ = 76562.5 J.

Lost in kinetic Energy of the system = E₁ - E₂ = 90625 - 76562.5

Lost in kinetic energy of the system = 14062.5 J

5 0

3 years ago