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3 years ago

Hey any physicist or engineer around. am giving brainliest to anyone who will answer this question.

Physics

1 answer:

Oduvanchick [21]3 years ago

8 0

Answer:

N = 167 Newtons

R = 727 Newtons

Explanation:

i) For static equilibrium, moments about any convenient point must sum to zero.

A moment is the product of a force and a moment arm length. Only the force acting perpendicular to a moment arm passing through the pivot point makes a moment.

ii) I will <em>ASSUME </em>the two moment arms are 0.05m and 0.15 m

CCW moments about the fulcrum are

190 N(0.2 m) + 280 N(0.05 m) = 52 N•m

CW moments are (N)N(0.15 m + 90 N(0.3 m) = 27 + 0.15N N•m

For static equilibrium, these must be equal

27 + 0.15N = 52

0.15N = 25

N = 166.6666666...

Sum moments about N to zero

(Same as saying CW and CCW moments must balance)

190(0.2 + 0.15) + 280(0.05 + 0.15) - R(0.15) - 90(0.3 - 0.15) = 0

R = 726.6666666...

We could verify this by summing vertical forces to zero.

R - 190 - 280 - 166.666666 - 90 = 0

R = 726.6666666...

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Two students, Jenny and Cho, are investigating motion.

IgorLugansk [536]

Answer:

1: a measuring instruments the students should use for time is a stopwatch

2: a measuring instruments the students should use for distance is a measuring tape

Explanation:

pls mark brainliest

8 0

2 years ago

A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m

ArbitrLikvidat [17]

Answer:

Explanation:

The sum of force along both components x and y is expressed as;

$\sum Fx = ma_x \ and \ \sum Fy = ma_y$

The magnitude of the net force which is also known as the resultant will be expressed as $R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}$

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

$a_x = \frac{d^2 x }{dt^2}$

$a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}$

Similarly,

$a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2$

$\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N$

$R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N$

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

7 0

3 years ago

Question 6 of 10

Anni [7]

The answer is c or b u choose

6 0

3 years ago

The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur

sp2606 [1]

Answer:

a) q = -9.23 cm, b) h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

1 / f = 0.6 / 4 = 0.15

f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

1 / q = 1 / f - 1 / p

1 / q = 1 / 6.67 - 1/24

1 / q = 0.15 - 0.04167 = 0.10833

q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

M = h ’/ h = - q / p

h’= - q / p h

h’= - (-9.23) / 24.0 0.150

h’= 0.05759 cm

h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

6 0

3 years ago

A truck with 28-in.-diameter wheels is traveling at 50 mi/h. Find the angular speed of the wheels in rad/min, *hint convert mile

solong [7]

Answer:

Angular speed ω=3771.4 rad/min

Revolution=5921 rpm

Explanation:

Given data

$d=28in\\r=d/2=28/2=14in\\v=50mi/hr$

To find

Angular speed ω

Revolution per minute N

Solution

First we need to convert the speed of truck to inches per mile

1 mile=63360 inches

1 hour=60 minutes

$v=(50*\frac{63360}{60} )\\v=52800in/min$

Now to solve for angular speed ω by substituting the speed v and radius r in below equation

$w=\frac{v}{r}\\ w=\frac{52800in/min}{14in}\\ w=3771.4rad/min$

To solve for N(revolutions per minute) by substituting the angular speed ω in the following equation

$N=\frac{w}{2\pi }\\ N=\frac{3771.4rad/min}{2\pi }\\ N=5921RPM$

3 0

3 years ago