What is the displacement of a person that ran 15 minutes if he or she runs at an average velocity of 16 km/hr west

A tourist drops (from rest) a ping pong ball from the top of the tower, which has a height of 324 meters. Assuming no air resist

Elanso [62]

Answer:

8.13secs

Explanation:

From the question weal are given

Height H =324m

Required

time it takes to drop t

Using the equation of motion

H = ut + 1/2gt²

Substitute the given values

324 = 0(t)+1/2(9.8)t²

324 = 1/2(9.8)t²

324 = 4.9t²

t² =324/4.9

t² = 66.12

t = √66.12

t = 8.13secs

Hence the time taken to drop is 8.13secs

4 0

2 years ago

What were four of Galileo’s discoveries that were important to astronomy?

Scorpion4ik [409]

Explanation:

1. Phases of Venus: Galileo was the first astronomer to use a telescope to observe the celestial objects. Through a telescope he observed that Venus shows the phases just like the Moon. This proved the Heliocentric theory correct against the then prevalent Geocentric theory.

2. Law of Falling bodies: The acceleration due to gravity is independent of weight of the objects that means two bodies of different mass will hit the ground at the same time if dropped from the same height.

3. The uneven surface of the Moon: He observed that the surface of the Moon is uneven and rough.

4. Discovery of the 4 Moons of Jupiter

7 0

3 years ago

In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr

spin [16.1K]

Answer:

a) $v_p=9.35m/s$

Explanation:

From the question we are told that:

Open cart of mass $M_o=50.0 kg$

Speed of cart $V=5.00m/s$

Mass of package $M_p=15.0kg$

Speed of package at end of chute $V_c=3.00m/s$

Angle of inclination $\angle =37$

Distance of chute from bottom of cart $d_x=4.00m$

a)

Generally the equation for work energy theory is mathematically given by

$\frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2$

Therefore

$\frac{1}{2}u^2+gh=\frac{1}{2}v_p^2$

$v_p=\sqrt{2(\frac{1}{2}u^2+gh)}$

$v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}$

$v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}$

$v_p=9.35m/s$

4 0

2 years ago

While walking past a construction site, a person notices a pipe sticking out of a second floor window with water pouring out. As

tia_tia [17]

Answer:

Its diameter increases as it flows down from the pipe. Assuming laminar flow for the water, then Bernoulli's equation can be applied.

P1-P2 + (rho)g(h1 - h2) + 1/2(rho)(v1² - v2²) = 0

Explanation:

P1 = P2 = atmospheric pressure so, P1 - P2 = 0

h1 is greater than h2 so h1-h2 is positive. Rearranging the equation above 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho = v2²

From the continuity equation for fluids

A1v1 = A2v2

v2 = A1v1/A2

Substituting into the equation above

(A1v1/A2)² = 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho

Making A2² the subject of the formula,

A2² = (A1v1)²× rho/(2{ (rho)g(h1-h2) + 1/2(rho)v1²}

The denominator will be greater than the numerator and as a result the diameter of the flowing stream decreases.

Thank you for reading.

4 0

3 years ago

A bobsled has a momentum of 3000 kg•m/s to the east. After the rider gives it a push, its momentum increases to 5500 kg•m/s to t

Archy [21]

Impulse, denoted as J, is defined by the change in momentum. Since we have our initial and our final, we can solve for the change in momentum.

$J = p_{f} - p_{0} = 5500 - 3000 = 2500

J = 2500 kg*m/s$

7 0

3 years ago

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