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3 years ago

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UTMSUNG

1 answer:

likoan [24]3 years ago

4 0

Answer:

i think answer id leech because when it move it leave the liner

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The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the part

ololo11 [35]

Answer:

distance/ kinetic

Explanation:

According to the work energy theorem, the work done by all forces is equal to the change in kinetic energy of the body.

So, As the force is applied in the same direction of the distance traveled,so only the kinetic energy of the body changes as after application of force, the speed of the body changes.

6 0

3 years ago

One type of cold remedy is an effervescent tablet that breaks down in water. When the tablet is placed in water, it forms gas bu

oksano4ka [1.4K]

Answer:

is a sign of a chemical change

Explanation:

The tablet reacts with the water to form carbon dioxide

7 0

3 years ago

Read 2 more answers

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the dist

Sidana [21]

Complete Question

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.08m. She sets the pendulum swinging, and her collaborators carefully count 101 complete cycles of oscillation during 2.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2

Answer:

The acceleration due to gravity is $g = 167.2 \ m/s^2$

Explanation:

From the question we are told that

The length of the simple pendulum is $L = 1.081.08 \ m$

The number of cycles is $N = 101$

The time take is $t = 2.00 *10^{2 \ }s$

Generally the period of this oscillation is mathematically evaluated as

$T = \frac{N}{t }$

substituting values

$T = \frac{101}{2.0*10^2 }$

$T = 0.505 \ s$

The period of this oscillation is mathematically represented as

$T = 2 \pi \sqrt{\frac{l}{g} }$

making g the subject of the formula we have

$g = \frac{L}{[\frac{T}{2 \pi } ]^2 }$

$g = \frac{4 \pi ^2 L }{T^2 }$

Substituting values

$g = \frac{4 * 3.142 ^2 * 1.08 }{505.505^2 }$

$g = \frac{4 * 3.142 ^2 * 1.08 }{0.505^2 }$

$g = 167.2 \ m/s^2$

7 0

3 years ago

A bicycle has wheels of 0.8 m diameter. The bicyclist accelerates from rest with constant acceleration to 22 km/h in 11.6 s. Wha

larisa [96]

Answer:

α = 1.32 rad/s²

Explanation:

given,

diameter of the bicycle = 0.8 m

radius of the bicycle = 0.4 m

initial speed of the bicyclist,u = 0 m/s

final speed of the bicyclist,v = 22 Km/h = 22 x 0.278

= 6.12 m/s

time,t = 11.6 s

acceleration = $\dfrac{v-u}{t}$

= $\dfrac{6.12-0}{11.6}$

=0.53 m/s²

we know,

a = α r

$\alpha=\dfrac{a}{r}$

$\alpha=\dfrac{0.53}{0.4}$

α = 1.32 rad/s²

the angular acceleration of the wheels is equal to α = 1.32 rad/s²

7 0

3 years ago

If the 20-mm-diameter rod is made of A-36 steel and the stiffness of the spring is k = 55 MN/m , determine the displacement of e

Alinara [238K]

Answer:

σ = 1.09 mm

Explanation:

<u>Step 1:</u> Identify the given parameters

rod diameter = 20 mm

stiffness constant (k) = 55 MN/m = 55X10⁶N/m

applied force (f) = 60 KN = 60 X 10³N

young modulus (E) = 200 Gpa = 200 X 10⁹pa

<u>Step 2:</u> calculate length of the rod, L

$K = \frac{A*E}{L}$

$L = \frac{A*E}{K}$

$A=\frac{\pi d^{2}}{4}$

d = 20-mm = 0.02 m

$A=\frac{\pi (0.02)^{2}}{4}$

A = 0.0003 m²

$L = \frac{A*E}{K}$

$L = \frac{(0.0003142)*(200X10^9)}{55X10^6}$

L = 1.14 m

<u>Step 3:</u> calculate the displacement of the rod, σ

$\sigma = \frac{F*L}{A*E}$

$\sigma = \frac{(60X10^3)*(1.14)}{(0.0003142)*(200X10^9)}$

σ = 0.00109 m

σ = 1.09 mm

Therefore, the displacement at the end of A is 1.09 mm

3 0

3 years ago