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VladimirAG [237]
2 years ago
6

A car travels three-quarters of the way around a circle of radius 20.0 m in a time of 3.0 s at a constant speed. the initial vel

ocity is west and the final velocity is south. (a) find its average velocity for this trip. (b) what is the car's average acceleration during these 3.0 s? (c) explain how a car moving at constant speed has a nonzero average acceleration.
Physics
1 answer:
schepotkina [342]2 years ago
3 0
20.3 divided by 3.0 will get u velocity and v times 3.0s 
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Maverick and goose are flying a training mission in their F-14. They are flying at an altitude of 1500 m and are traveling at 68
den301095 [7]

Answer:

The bomb will remain in air for <u>17.5 s</u> before hitting the ground.

Explanation:

Given:

Initial vertical height is, y_0=1500\ m

Initial horizontal velocity is, u_x=688\ m/s

Initial vertical velocity is, u_y=0(\textrm{Horizontal velocity only initially)}

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is S=y-y_0=0-1500=-1500\ m

Acceleration in the vertical direction is due to gravity, g=-9.8\ m/s^2

Therefore, the displacement of the bomb is given as:

S=u_yt+\frac{1}{2}gt^2\\-1500=0-\frac{1}{2}(9.8)(t^2)\\1500=4.9t^2\\t^2=\frac{1500}{4.9}\\t=\sqrt{\frac{1500}{4.9}}=17.5\ s

So, the bomb will remain in air for 17.5 s before hitting the ground.

6 0
2 years ago
A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.12
den301095 [7]

Question

Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s.  What is its angular momentum? Express your answer in kilogram meters squared per second.

Answer:

0.0192 kgm^{2}/s

Explanation:

The angular momentum L of the baton moving about an axis perpendicular to it, passing through the center of the baton is,

L = \frac{1}{{12}}m{l^2}\omega

Here, l is the length of the baton.

Substitute 0.120 kg for m, 3 rads/s for \omega[\tex] and 0.8 m for l [tex]\begin{array}{c}\\L = \frac{1}{{12}}m{l^2}\omega \\\\ = \frac{1}{{12}}\left( {0.120{\rm{ kg}}} \right){\left( {{\rm{80}}{\rm{.0 cm}}} \right)^2}{\left( {\frac{{1 \times {{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)^2}\left( {{\rm{3}}{\rm{.00 rad/s}}} \right)\\\\ = 0.0192{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\\\end{array}

5 0
2 years ago
Read 2 more answers
A machine produces photo detectors in pairs. Tests show that the first photo detector is acceptable with probability 3/5. When t
klasskru [66]

Answer:

a.a. \ \frac{7}{25}

b.\ \ \ P(D_1D_2)=\frac{6}{25}

Explanation:

a. Find the probability that exactly one photo detector of a pair is acceptable:

Let A_i=i^{th} photo is accepted and the probability D_i=i^{th} is defected.

Therefore:

P(A_i)=3/5,\ P(A_2|A_1)=4/5,\ \ P(A_2|D_1)=2/5\\\\\\=P(A_1D_2)+P(D_1A_2)\\\\=\frac{3}{5}\times\frac{1}{5}+\frac{2}{5}\times\frac{2}{5}\\\\=\frac{7}{25}

#The probability of exactly one photo detector of a pair is accepted is 7/25

b.Find the probability that both photo detectors in a pair are defective,P(D1D2):

P(D_1D_2)=\frac{2}{5}\times \frac{3}{5}\\\\=\frac{6}{25}

Hence, from out tree diagram,the probability that both photo detectors in a pair are defective is 6/25

4 0
2 years ago
A developer wants to build a sprawling two-story office complex. The developer's and architect's vision is something low, modern
Lera25 [3.4K]

Answer:

a. directive zoning

Explanation:

Directive zoning is an instrument used in master plans, whereby the city is divided into areas on which differentiated land use and land use guidelines apply, especially urban indexes. Directive zoning acts primarily by controlling two main elements: the use and size (or size) of lots and buildings. It is therefore assumed that the end result achieved through individual actions is in line with the municipality's objectives, which include proportionality between occupation and infrastructure, the need to protect fragile areas and / or cultural interest, the harmony from the volumetric point of view, etc.

4 0
2 years ago
Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous
Orlov [11]

Answer:

Explanation:

The speed of the water in the large section of the pipe is not stated

so i will assume 36m/s

(if its not the said speed, input the figure of your speed and you get it right)

Continuity equation is applicable for ideal, incompressible liquids

Q the flux of water that is  Av with A the cross section area and v the velocity,

so,

A_1V_1=A_2V_2

A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}

the diameter decreases 86% so

d_2 = 0.86d_1

v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}

Thus, speed in smaller section is 48.6 m/s

3 0
2 years ago
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