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3 years ago

A frog jumps up at time t = 0.

Physics

2 answers:

valina [46]3 years ago

6 0

Is there a picture that I can see

vesna_86 [32]3 years ago

4 0

Answer:

Decreasing

Explanation:

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Which would ultimately cause the armature in an electric motor to spin?

malfutka [58]

It would be the electromagnet because that is how it spins

4 0

4 years ago

Read 2 more answers

What is the best example of the purpose of scientific notation?

Gemiola [76]

The correct answer would be to express large and small numbers.

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3 years ago

A body with mass of 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time tequals0 the bod

11Alexandr11 [23.1K]

Answer:

X(t) = 13/13 cos(12t+α)

C =13/13

π/6 s

Explanation:

(A) A body with mass 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time t = 0 the body is pulled 1 m in to the right, stretching the spring, 3 set in motion with an initial velocity of 5 m/s to the left.

(a) Find X(t) in the form c • cos(w_o*t— α)

(b) Find the amplitude 3 Period of motion of the body 1

mass: m = 200g = 0.200 kg

displacement: ΔX = 20 cm = 0.20 m

Spring Constant: K = 9/0.20 = 45 N/m

IV: X(0) = 1m V(0) = -5 m/s

Simple Harmonic Motion: c•cos(cosw_t— α) = X(t)

Circular Frequency: w_o = √k/m= √36/(0.20) = 13 rad/s

X(0) = 1m =c_1

X'(0) = V(0) = c_2*w_o/w_o

= -5/12 = c_2

"radians Technically Unitless"

Amplitude: c = √ci^2 + c^2 ==> √1^2 + (-5/12)^2 = 1 m =13/13 = c

X(t) = 13/13 cos(12t+α)

since, C>0 : damped forced vibration c_1>0, c_2>0

phase angle 2π+tan^-1(c_2/c_1)

=2π+tan^-1(-5/12/1)= 5.884

period: T =2π/w_o

=π/6 s

6 0

4 years ago

A hot-water radiator has a surface temperatue of 80 o C and a surface area of 2 m2 . Treating it as a blackbody, find the net ra

faust18 [17]

Answer:

925.04 J/s

Explanation:

T = 80 C = 80 + 273 = 353 K

To = 20 c = 20 + 273 = 293 K

A = 2 m^2

Use the formula for Stefan's law

Energy radiated per second

$E = \sigma A \left ( T^{4}-T_{0}^{4} \right )$

$E = 5.67 \times 10^{-8}\times 2\left ( 353^{4}-293^{4} \right )$

E = 925.04 J/s

3 0

3 years ago

How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius

Korvikt [17]

Answer:

<h2>The coefficient of static friction will be 0.7</h2>

Explanation:

Given data

the radius of curve= 90m

speed v= 90 km/h to m/s = (90*100)/60*60= 25 m/s

we know that the expression for the centripetal force acting on the car

$Fc= \frac{mv^2}{r}$ -------1

we also know that the expression for the frictional force between road and tire.

Ff= μmg--------2

Equating equation 1 and 2 we have

μmg= mv^2/r

μ= v^2/gr

substituting the values of speed and radius we have (assuming g= 9.81m/s^2)

μ= 25^2/9.81*90

μ= 625/882.9

μ= 0.7

4 0

3 years ago