a spring with a spring constant of 65 n/m is displaced -94cm. what is the magnitude of the force exerted by the spring

Answer:

a

Explanation:

i took the test 100%

Answer:

Explanation:

F = mv²/R

F = 11.0(3.50²)/0.600 = 225 N

Answer:

inertia   .

because yes

Answer:

what will happen if i will answer ur questions?

Explanation:

is there gonna be a bad thing or a good thing

Answer:

Explanation:

F = ma

a = F/m

a = 75/25

a = 3 m/s²

Answer:

[tex]\frac{a_{d}}{a_{i}} = \frac{(1 -mu)}{mu}[/tex]

= (1 - μ)/μ

Explanation:

Always draw a diagram!

Up the incline:

[tex]Fr_{max}[/tex] = maximum friction

[tex]Fr_{max}[/tex] = μk

k = R = mg.cos(45) = mg.sin(45)

Resolution of forces parallel to the slope:

F (Fp in the diagram) = force of propulsion

g = gravity

[tex]F - Fr_{max} = ma_{i}[/tex]

[tex]F -[/tex] μ.mg.cos(45) [tex]= ma_{i}[/tex]

Down the decline:

Resolution of forces:

[tex]mg.sin(45) - Fr_{max} = ma_{d}[/tex]

[tex]mg.sin(45) -[/tex] μ.mg.cos(45) [tex]= ma_{d}[/tex]

Then, find the ratio:

[tex]\frac{ma_{d}}{ma_{i}} = \frac{mg.sin(45) - mu.mg.cos(45)}{-F + mu.mg.cos(45)} \\\\ \frac{a_{d}}{a_{i}} = \frac{k - k.mu}{-F + k.mu} \\\\ = \frac{k(1 -mu)}{-F + k.mu}[/tex]

Potentially, there is no need to consider F in this situation, in which case:

[tex]\frac{a_{d}}{a_{i}} = \frac{k(1 -mu)}{k.mu} \\\\ = \frac{(1 -mu)}{mu}[/tex]

= (1 - μ)/μ

Answer:

Explanation:

vA / vB = √(TA/(m/L)) / √(TB/(m/L))

The lengths are the same, so the L divides out to 1

The material is identical so the mass will be directly proportional to the cross sectional area of the string

vA / vB = √(TA/(πdA²/4)) / √(TB/(πdB²/4))

π/4 is common so divides out to 1

vA / vB = √(TA/dA²) / √(TB/dB²)

vA / vB = √(410/0.489²) / √(809/1.27²)

vA / vB = 41.407 / 23.396

vA / vB = 1.8488961...

vA / vB = 1.85

Answer:

i cant see

Explanation:

but im smart

The muzzle speed of the bullet before the collision is 415.3 m/s.

The given parameters:

Apply the principle of work-energy theorem to determine the final velocity of the block-bullet system;

[tex]F_f \times d = \frac{1}{2} mv^2\\\\\mu_k F_n \times d = \frac{1}{2} mv^2\\\\\mu_ k (m_1 + m_2)g \times d = \frac{1}{2} (m_1 + m_2)v^2\\\\\mu_k g \times d= \frac{1}{2} v^2\\\\2\mu _k gd = v^2\\\\v= \sqrt{2\mu _k gd } \\\\v = \sqrt{2 \times 0.23 \times 9.8 \times 9.5} \\\\v = 6.54 \ m/s[/tex]

Apply the principle of conservation of linear momentum to determine the muzzle speed of the bullet;

[tex]m_1 u_1 \ + \ m_2u_2 = v(m_1 + m_2)\\\\0.024(u_1) \ + \ 1.5(0) = 6.54(0.024 + 1.5)\\\\0.024u_1 = 9.967\\\\u_1 = \frac{9.967}{0.024} \\\\u_1 = 415.3 \ m/s[/tex]

Thus, the muzzle speed of the bullet before the collision is 415.3 m/s.

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Answer:

Explanation:

conservation of momentum

m(u) + M(0) = (m + M)v

u = (m + M)v/m

u = (0.250 + 1.35)(11.9) / 0.250

u = 76.16

u = 76.2 m/s

That's a fairly massive, and slow, bullet.

Answer:

Explanation:

The word "sheet" implies that the copper is quite thin.

Copper is also a very good conductor of heat.

Therefore, with a very short heat flow distance to cover and a high rate of heat transmission, temperature differences on either side of the sheet are almost instantaneously eliminated by heat flow.

Negative

Because the car is moving up and the bug is moving down. but it also depends on the weather so choice between one of those two I think is Negative but I may be wrong.

Answer:

Explanation: 6000z

Answer:

Explanation:

PE = mgh

g = PE/mh

g = 15.0 / (0.200(0.500))

g = 150 m/s²

This is one strong astronaut if he can work in an environment where gravity is more than 15 times stronger than on earth.

Answer:

lithosphere

Explanation:

hope this helps you!!

Answer:

Explanation:

30 m/s • 3 hr •3600 s/hr / 1000 m/km = 324 km west

Answer:

Explanation:

70.0(5.00)(9.81) = 3,433.5 = 3430 N

To solve this we must be knowing each and every concept related to  centripetal force and its calculations. Therefore, the centripetal force acting on her is 3430 N.

The term centripetal relates to a propensity to gravitate toward the center. Centripetal refers to moving in the direction of the center. The force that maintains an item moving in a circular direction and helps it stay on the path is referred to as centripetal force.

Furthermore, centrifugal force is indeed the tendency of things to deviate from a circular route and fly in a straight line. People frequently confuse centripetal force with centrifugal force.

Mathematically,

F = m a

  = 70 acceleration

  = 70 × 5 × 9.81

 = 3430 N

Therefore, the centripetal force acting on her is 3430 N.

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Answer:

Heat flow

Explanation:

Answer:

The first one is the only one that is true all the time

Explanation:

The second one may be true if friction is high enough.

The other three are false all the time

Answer: the current increases with each 3 volt and 9 volt. The relationship between resistance and current in a circuit is that the greater the resistance the less the current and the greater the current the less the resistance is. yayayay I could answer this I big brain :)

Answer:

*1) 100 Joule energy

*2) 101.2 m/s

Explanation:

*1) 1J = 1w

100J = 100w

*2) A 70-kg person will have to run at a speed of 101.2 m/s to have that amount of kinetic energy.

Crazy Wally Ok Ok ok hhahahaha

Answer:

The removal of heat energy slows the speed of particles

Explanation:

When you add heat to a substance, the heat energy gets transferred to kinetic energy, and the molecules began to move a greater distance at a greater speed. When you remove heat, the opposite happens.

Hi there!

a.

We can use the initial conditions to solve for w₀.

It is given that:

[tex]\frac{d\theta}{dt} = w_0e^{-\sigma t}[/tex]

We are given that at t = 0, ω =  3.7 rad/sec. We can plug this into the equation:

[tex]\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}[/tex]

Now, we can solve for sigma using the other given condition:

[tex]2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}[/tex]

b.

The angular acceleration is the DERIVATIVE of the angular velocity function, so:

[tex]\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}[/tex]

c.

The angular displacement is the INTEGRAL of the angular velocity function.

[tex]\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]

[tex]\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.[/tex]

[tex]\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}[/tex]

[tex]\theta = 8.471 rad[/tex]

Convert this to rev:

[tex]8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}[/tex]

d.

We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.

[tex]0 = 3.7e^{-0.0714t}\\\\t = \infty[/tex]

Evaluate the improper integral:

[tex]\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]

[tex]\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.[/tex]

[tex]\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad[/tex]

Convert to rev:

[tex]51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}[/tex]

Answer:

Adding salt to the water increases the density of the solution because the salt increases the mass without changing the volume very much.

Explanation: the explanation is in a file

Answer:

1.108 ×  [tex]10^{-3}[/tex]J

Explanation:

v=21.0v

Q=52.8×  [tex]10^{-6}[/tex]

E=?

V=E/Q

E=v ×Q

 =21 ×52.8 ×[tex]10^{-6}[/tex]

 =1108.8   ×[tex]10^{-6}[/tex]

E= 1.108 ×  [tex]10^{-3}[/tex]J

Answer:

n

Explanation:

TWO outcomes can be predicted the students will find:

When a constant mass of gas is cooled, its volume falls, and when the temperature is raised, its volume grows. The volume of the gas rises by 1/273 of its initial volume at 0 °C for every degree of temperature rise.

In layman's words, the volume of a fixed mass of gas is exactly proportional to temperature at constant pressure.

The teacher took two latex balloons and blew them up with helium gas to the same size. As she placed Balloon A in a -15° C freezer, its temperature decreases and that's why, the size of balloon A becomes smaller. Again she  placed Balloon B in the sun on 30° C day, its temperature increases and that's why, the size of balloon B becomes larger.

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The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

[tex]\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}[/tex]

[tex]\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}[/tex]

[tex]\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 + \omega^2}[/tex]

[tex]\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2} }[/tex]

[tex]\mathbf{\omega^2=\dfrac{39.24 }{2}}[/tex]

[tex]\mathbf{\omega=\sqrt{19.62 } \ rad/sec}[/tex]

[tex]\mathbf{\omega=4.429 \ rad/sec}[/tex]

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

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The angular velocity of the wheel depends on the mass, radius and the

mode of rotation of the wheel (with or without slipping).

Reasons:

The given parameters are;

Radius of the wheel, r = 2.0 m

Height of the incline, h = 8.0 m

Required:

Angular velocity of the wheel at the bottom of the incline.

Solution:

The potential energy of the wheel at the top of the hill, P.E. = m·g·h

[tex]Sum \ of \ the \ kinetic \ energy \ of \ the \ wheel, \ K.E. = \mathbf{\displaystyle \frac{1}{2} \cdot m \cdot v^2 + \frac{1}{2} \cdot I \cdot \omega ^2}[/tex]

Where;

v = The translational velocity of the wheel = ω·r

I = The moment of inertia of the wheel = m·r²

Therefore'

[tex]Sum \ of \ K.E. = \displaystyle \frac{1}{2} \cdot m \cdot (\omega \cdot r)^2 + \frac{1}{2} \cdot m \cdot r^2 \cdot \omega ^2 = \mathbf{m \cdot r^2 \cdot \omega^2}[/tex]

At the bottom of the hill, the potential energy is converted to kinetic energy

Therefore;

P.E. = Sum of K.E.

m·g·h = m·r²·ω²

g·h = r²·ω²

[tex]\displaystyle \omega = \sqrt{ \frac{g \cdot h}{r^2} } = \mathbf{ \frac{\sqrt{g \cdot h} }{r}}[/tex]

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Therefore;

[tex]\displaystyle \omega = \frac{\sqrt{9.81 \times 8} }{2} \approx \mathbf{ 4.43}[/tex]

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