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3 years ago

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The driver of a car traveling at a speed of 27 m/s slams on the brakes and comes to a stop in 3 s. If we assume that the speed c

hanged at a constant rate (constant net force), what was the average speed during this 3 s interval

1 answer:

NISA [10]3 years ago

7 0

Answer:

Average speed = 13.5 m/s

Explanation:

Since the car is running at a speed of 27 m/s and it stops after 3 seconds by applying the brake. Therefore, the initial speed of the 27 m/s and final speed is 0.

Use below formula to find the average speed :

Average speed = (Initial speed + final speed ) / 2

Average speed = (27 + 0 ) / 2

Average speed = 13.5 m/s

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A skater extends her arms horizontally, holding a 5-kg mass in each hand. She is rotating about a vertical axis with an angular

sp2606 [1]

Answer:

$1.98$ rev/s

Explanation:

$m$ = mass attached to each hand = 5 kg

$r_{i}$ = initial distance of masses in each hand = 1 m

$r_{f}$ = final distance of masses in each hand = 0.1 m

$I$ = moment of inertia of body = 5 kgm²

$I_{i}$ = initial total moment of inertia = $I + 2 mr_{i}^{2}$

$w_{i}$ = initial angular velocity = 1 rev/s

$I_{f}$ = final total moment of inertia = $I + 2 mr_{f}^{2}$

$w_{f}$ = final angular velocity = ?

Using conservation of angular momentum

$I_{i} w_{i} = I_{f} w_{f}$

$(I + 2 mr_{i}^{2}) w_{i} = (I + 2 mr_{f}^{2}) w_{f}$

$(5 + 2 (5) (1)^{2}) (1) = (5 + 2 (5) (0.1)^{2}) w_{f}$

$w_{f} = 2.94$ rev/s

3 0

3 years ago

Read 2 more answers

If a typical smelt weighs 225 g, what is the total mass of pcbs in a smelt in the great lakes? If a typical smelt weighs 225 g,

photoshop1234 [79]

Answer;

= 2.34 x 10^-4 g pcbs

Explanation;

-A typical smelt has 1.04 ppm of PCB is great lakes.

Therefore;

The amount of PCB in a smelt in great leak can be calculated by multiplying the mass of smelt by the pcbs present in a smelt

= 225 g * ( 1.04/100000)

= 2.34 x10^-4 g pcbs

7 0

3 years ago

From a penalty kick, the ball rebounds off the goalkeeper back to the player who took the kick. That player then kicks the ball

Alexxx [7]

The correct restart is for there to be another kick off taken in this type of scenario.

<h3>What is Kick-off in Soccer?</h3>

This is a method of restarting play in which the ball is put on the center circle and passed by a player.

Kick offs occur at the beginning of any halves or when a goal is scored during the match which is why it's the most appropriate choice.

Read more about Soccer here brainly.com/question/12597997

7 0

2 years ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0

3 years ago

Chegg Given that the mean radius of the Moon’s orbit is 3.84 x 108 m and its period is 2.36 x 106 sec, at what altitude above th

alukav5142 [94]

Answer:

The altitude of geostationary satellite is $3.58\times10^{7}\ m$

Explanation:

Given that,

Radius of moon's orbit $r=3.84\times10^{8}\ m$

Time period $T=2.36\times10^{6}\ sec$

We need to calculate the orbital radius of geostationary satellite is

Using formula of time period

$T=\sqrt{\dfrac{4\pi^2}{GM}a^3}$

$a=((\dfrac{GM}{4\pi^2})T^2)^{\dfrac{1}{3}}$

Where, G = gravitational constant

M = Mass of earth

T = time period of geostationary satellite orbit

Put the value in to the formula

$a=((\dfrac{6.67\times10^{-11}\times5.97\times10^{24}}{4\times\pi^2})\times(86160)^2)^{\dfrac{1}{3}}$

$a=4.217\times10^{7}\ m$

We need to calculate the altitude of geostationary satellite

Using formula of altitude

$h = a-R_{e}$

Where, R = radius of earth

a = radius of geostationary satellite

Put the value into the formula

$h =4.217\times10^{7}-6.38\times10^{6}$

$h =35790000\ m$

$h=3.58\times10^{7}\ m$

Hence, The altitude of geostationary satellite is $3.58\times10^{7}\ m$

4 0

3 years ago