All categories

4 years ago

An Alaskan rescue plane traveling 41 m/s

Physics

1 answer:

densk [106]4 years ago

8 0

Answer:

-56.9 m/s

Explanation:

Given:

Δy = -165 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-165 m)

v = -56.9 m/s

You might be interested in

When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____

valentinak56 [21]

Answer and explanation:

When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.

The total car tire diameter can be calculated as:

$D_{tot orig}=\frac{195 \cdot 75}{2540 \cdot 2}+15''=26.5''$

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:

pr=0.95·75=71.25

The problem is that the profiles are normalized and the nearest profile available is 70.

If we take a theorical tire with a profile of 71.25:

$D_{tot orig}=D_{new}\\\frac{195 \cdot 75}{2540 \cdot 2}+15''=\frac{X \cdot 71.25}{2540 \cdot 2}+15''\\X=205.26$

The theorical tire size should be 205/71 R15.

If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.

The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').

3 0

4 years ago

By how many newtons does the weight of a 85.9-kg person lose when he goes from sea level to an altitude of 6.33 km if we neglect

sergejj [24]

Answer:

$Weight\ loss=1.6321N$

Explanation:

From the question we are told that:

Weight $W=85.9kg$

Altitude $h= 6.33 km$

Let

Radius of Earth $r=6380km$

Gravity $g=9.8m/s^2$

Generally the equation for Gravity at altitude is mathematically given by

$g_s=9.8(\frac{6380}{6380+6.33})^2$

$g_s=9.781m/s^2$

Therefore

Weight at sea level

$W_s=9.8*85.9$

$W_s=841.82N$

Weight at 6.33 altitude

$W_a=9.781*85.9$

$W_a=840.2N$

Therefore

$Weight loss=W_s-W_b$

$Weight loss=841.82-840.2$

$Weight loss=1.6321N$

3 0

3 years ago

Please help! I'd really appreciate it :-)

Alexxx [7]

Answer:

0.31429 Amps

Explanation:

Using the "Ohm's Law" we can look for the current (amps) with a formula of I = V/R

Given:

Resistance (Ohm) = 350

Volts = 110

Solution:

I = 110v/350r

I = 0.31429 amps

6 0

3 years ago

A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05

7nadin3 [17]

Answer:

$W_{drag} = 4.223\,J$

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

$U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}$

The work done on the ball due to drag is:

$W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})$

$W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})$

$W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]$

$W_{drag} = 4.223\,J$

7 0

4 years ago

Read 2 more answers

The following questions present a twist on the scenario above to test your understanding. Suppose another stone is thrown horizo

Ipatiy [6.2K]

The first part of the text is missing, you can find on google:

"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."

Let's now solve the different parts.

(a) 3.03 s

The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by

$y(t) = h -\frac{1}{2}gt^2$

where

h = 45 m is the initial height

g = 9.8 m/s^2 is the acceleration of gravity

When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:

$0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s$

(b) 18.5 m/s

For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.

The horizontal position is given by

$x=v_x t$

where

$v_x$ is the horizontal speed, which is constant

t is the time

At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:

$v_x = \frac{x}{t}=\frac{56}{3.03}=18.5 m/s$