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3 years ago

A basketball rolls down a hill until it comes to a stop in a ditch

Physics

2 answers:

tester [92]3 years ago

8 0

Answer:

1st law.

Explanation:

dmitriy555 [2]3 years ago

3 0

A :-) the first law of motion -
The first law of motion states that a object will be at motion or rest until a unbalanced force is applied on it.

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WHat is the correct definnition for recovery heart rate ?

Ket [755]

Answer:

when your heart rate is not high than normal or lower than normally

Explanation:

Recovery heart rate is when your heart recovers from a certain heart rate eg high heart rate or even low heart rate this means you heart is now functioning normally .

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3 years ago

What rock makes up the majority of the mares?

zzz [600]

Igneous rock makes up the majority of the mares. Because of volcanic eruption.

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4 years ago

Read 2 more answers

A spider spins a web with silk threads of density 1300 kg/m3 and diameter 3.0 μm . a typical tension in the radial threads of su

Tema [17]

Answer:

Explanation:

The velocity of a wave in a string is equal to:

v = √(T / (m/L))

where T is the tension and m/L is the mass per length.

To find the mass per length, we need to find the cross-sectional area of the thread.

A = πr² = π/4 d²

A = π (3.0×10⁻⁶ m)²

A = 2.83×10⁻¹¹ m²

So the mass per length is:

m/L = ρA

m/L = (1300 kg/m³) (2.83×10⁻¹¹ m²)

m/L = 3.68×10⁻⁸ kg/m

So the wave velocity is:

v = √(T / (m/L))

v = √(7.0×10⁻³ N / (3.68×10⁻⁸ kg/m))

v ≈ 440 m/s

The speed of sound in air at sea level is around 340 m/s. So the spider will feel the vibration in the thread before it hears the sound.

5 0

3 years ago

The water stream strikes the inclined surface of the cart. Determine the power produced by the stream if, due to rolling frictio

levacccp [35]

Answer

given,

constant speed of cart on right side = 2 m/s

diameter of nozzle = 50 mm = 0.05 m

discharge flow through nozzle = 0.04 m³

One-fourth of the discharge flows down the incline

three-fourths flows up the incline

Power = ?

Normal force i.e. Fn acting on the cart

$F_n = \rho A (v - u)^2 sin \theta$

v is the velocity of jet

Q = A V

$v = \dfrac{0.04}{\dfrac{\pi}{4}d^2}$

$v= \dfrac{0.04}{\dfrac{\pi}{4}\times 0.05^2}$

v = 20.37 m/s

u be the speed of cart assuming it to be u = 2 m/s

angle angle of inclination be 60°

now,

$F_n = 1000 \times \dfrac{\pi}{4}\times 0.05^2\times (20.37 - 2)^2 sin 60^0$

F n = 2295 N

now force along x direction

$F_x = F_n sin 60^0$

$F_x = 2295 \times sin 60^0$

$F_x = 1987.52\ N$

Power of the cart

P = F x v

P = 1987.52 x 20.37

P = 40485 watt

P = 40.5 kW

3 0

4 years ago

The ball will oscillate along the z axis between z=dz=d and z=−dz=−d in simple harmonic motion. What will be the angular frequen

Eddi Din [679]

Answer:

$\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }$

Explanation:

Additional information:

The ball has charge $-q_0$ , and the ring has positive charge $+Q$ distributed uniformly along its circumference.

The electric field at distance $z$ along the z-axis due to the charged ring is

$E_z= \dfrac{kQz}{(z^2+a^2)^{3/2}}.$

Therefore, the force on the ball with charge $-q_0$ is

$F=-q_oE_z$

$F=- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}$

and according to Newton's second law

$F=ma=m\dfrac{d^2z}{dz^2}$

substituting $F$ we get:

$- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=m\dfrac{d^2z}{dz^2}$

rearranging we get:

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=0$

Now we use the approximation that

$z^2+a^2\approx a^2$ (we use this approximation instead of the original $d^2+a^2\approx a^2$ since $z$ , our assumption still holds )

and get

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(a^2)^{3/2}}=0$

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{a^{3}}=0$

Now the last equation looks like a Simple Harmonic Equation

$m\dfrac{d^2z}{dz^2}+kz=0$

where

$\omega=\sqrt{ \dfrac{k}{m} }$

is the frequency of oscillation. Applying this to our equation we get:

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Q}{a^{3}}z=0\\\\m=m\\\\k= \dfrac{kq_0Q}{a^{3}}$

$\boxed{\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }}$

5 0

4 years ago