How much power can I generate while lifting a mass for 30 seconds?

A small car of mass m and a large car of mass 4m drive along a highway at constant speeds VS and VL. They approach a curve of ra

Arte-miy333 [17]

Answer:

$v_S=\sqrt{2}v_L$

Explanation:

The acceleration experimented while taking a curve is the centripetal acceleration $a=\frac{v^2}{r}$ . Since $a_S=2a_L$ , we have that: $\frac{v_S^2}{r_S}=\frac{2v_L^2}{r_L}$

They take the same curve, so we have: $r_S=r_L=R$

Which means: $v_S^2=2v_L^2$

And finally we obtain: $v_S=\sqrt{2}v_L$

4 0

3 years ago

How do you break apart a

Tcecarenko [31]

Answer:

B

Explanation:

Depends Mostly on bonds electrolysis can be used, chemical bonding like additional of water or by heating back to their elements.

3 0

2 years ago

Read 2 more answers

A pendulum consists of a large balanced mass hanging on the end of a long wire. At the point where a 28-kg pendulum has the grea

Ray Of Light [21]

Answer:

The length of the wire is approximately 67.1 m

Explanation:

The parameters of the pendulum are;

The mass of the pendulum, m = 28 kg

The angle between the pendulum weight and the wire, θ = 89°

The magnitude of the torque exerted by the pendulum's weight, τ = 1.84 × 10⁴ N·m

We have;

Torque, τ = F·L·sinθ = m·g·l·sinθ

Where;

F = The applies force = The weight of the pendulum = m·g

g = The acceleration due to gravity ≈ 9.8 m/s²

l = The length of the wire

Plugging in the values of the variables gives;

1.84 × 10⁴ N·m = 28 kg × 9.8 m/s² × l × sin(89°)

Therefore;

l = 1.84 × 10⁴ N·m/(28 kg × 9.8 m/s² × sin(89°)) = 67.0656080029 m ≈ 67.1 m

The length of the wire, l ≈ 67.1 m

6 0

3 years ago

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction is 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$

The force needed to stop the wheel is 104.00902 N

5 0

3 years ago

I believe Newton's 1st law is

Bogdan [553]

Answer:

The first law states that if the net force is zero, then the velocity of the object is constant.

5 0

3 years ago