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3 years ago

4. A spring is stretched 0.5 m from equilibrium. The force constant (k) of the

Physics

2 answers:

makkiz [27]3 years ago

8 0

Answer:

31.25

Explanation:

The formula for the potential energy of a spring is $\frac{1}{2}kx^2$ , where x is the distance in meters the spring is stretched. Plugging in the numbers that you are given, you get $\frac{1}{2}\cdot 250\cdot 0.25=31.25$ . Hope this helps!

hodyreva [135]3 years ago

8 0

31.25 hope this helps:)

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As a 5.0 × 10^2-newton basketball player jumps

AysviL [449]

This player is initially at rest, then the Force Weight (W) and the Normal Force (N) have a same module. When applies a Extra Force (E) on the floor, this reacts with this increase, causing the player to be released upwards.
Using the Newton's Secound Law, we have:

$F_{R}=ma \\ W+E-N=ma \\ N =W+E \\ N=5*10^2+10*10^2 \\ \boxed {N=1.5*10^3N}$

Number 3

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4 0

2 years ago

A ball is batted straight up into the air and reaches a maxium height 65.6 m (a) How long did it take to reach this height? (b)

kondaur [170]

Answer:

a) 3.65 seconds

b) 35.87 m/s

Explanation:

s = Displacement = 65.6 m

u = Initial velocity

v = Final velocity

t = Time taken

a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive and upward is taken as negative)

b) Equation of motion

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2\times -9.81\times 65.6\\\Rightarrow u=\sqrt{2\times 9.81\times 65.6}\\\Rightarrow u=35.87\ m/s$

Initial pop up velocity is 35.87 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-35.87}{-9.81}\\\Rightarrow t=3.65\ s$

It took 3.65 seconds to reach this height

6 0

3 years ago

An electromagnetic flowmeter is useful when it is desirable not to interrupt the system in which the fluid is flowing (e.g. for

AURORKA [14]

Answer:

2 m/s

Explanation:

The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as

$E = Blv$

where $E$ is the induced voltage = 2.88 mV = 2.88 x 10^-3 V

$l$ is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m

$v$ is the velocity of the fluid through the field = ?

$B$ is the magnetic field = 0.120 T

substituting, we have

2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x $v$

2.88 x 10^-3 = 1.44 x 10^-3 x $v$

$v$ = 2.88/1.44 = 2 m/s

8 0

3 years ago

Structures on a bird feather act like a diffraction grating having 7000 7000 lines per centimeter. What is the angle of the firs

Anestetic [448]

Answer:

θ = 28.9°

Explanation:

We are given;

Wavelength; λ = 602nm = 602 x 10^(-9) m

Lines per centimetre = 7000 /cm = 700000 /m

Thus, the distance between 2 adjacent lines is;

d = 1/700000 = 1.43 x 10^(-6) m

The angle at which diffracted light is formed is given by the formula

mλ = d sinθ

Where;

m is the mth order of the diffraction

λ is the wavelength of the incident light

d is the distance separating the centres of 2 adjacent slits

θ is the angle at which diffraction occurs

From the question, m is 1 because it says first order.

Thus, plugging in the relevant values into mλ = d sinθ, we have;

1 x 602 x 10^(-9) = 1.43 x 10^(-6) sinθ

sinθ = 602 x 10^(-9)/(1.43 x 10^(-6))

sinθ = 0.42098

θ = sin^(-1) 0.42098

θ = 28.9°

8 0

2 years ago

A bicycle wheel rotates at a constant 25 rev/min. What is true about its angular acceleration?

Nostrana [21]

Answer:

The angular acceleration is zero

Explanation:

When an object is in rotational motion, it has a certain angular velocity, which is the rate of displacement of its angular position.

This angular velocity can change or remain constant - this is given by the angular acceleration, which is:

$\alpha =\frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time elapsed

Therefore, the angular acceleration is the rate of change of angular velocity.

In this problem, the bicycle rotates at a constant angular velocity of

$\omega=25 rev/min$

This means that the change in angular velocity is zero:

$\Delta \omega=0$

And so, that the angular acceleration is zero:

$\alpha=0$

8 0

3 years ago