According to Newton, which of these will determine if and how an object will move?

one mole of water is equivalent to 18 grams of water. a glass of water has a mass of 200 g. how many moles of water is in this?

ehidna [41]

1 mole = 18 g
200 g = glass of water
200 ÷ 18 = 11.1
11.1 moles of water in 200 g (glass of water)

3 0

3 years ago

Assume that a gravitational anomaly in the solar system has shifted a field of asteroids into Earth’s orbit, and the field is no

Mandarinka [93]

Answer:

An asteroid is a minor planet of the inner Solar System. Historically, these terms have been applied to any astronomical object orbiting the Sun.

7 0

2 years ago

Unlike a roller coaster, the seats in a Ferris wheel swivel so that the rider is always seated upright. An 80-ft-diameter Ferris

telo118 [61]

Answer:

a

$F_A =425.42 \ N$

b

$F_A_H = 358.58 \ N$

Explanation:

From the question we are told that

The diameter of the Ferris wheel is $d = 80 \ ft = \frac{80}{3.281} = 24.383$

The period of the Ferris wheel is $T = 24 \ s$

The mass of the passenger is $m_g = 40 \ kg$

The apparent weight of the passenger at the lowest point is mathematically represented as

$F_A_L = F_c + W$

Where $F_c$ is the centripetal force on the passenger, which is mathematically represented as

$F_c =m * r * w^2$

Where $w$ is the angular velocity which is mathematically represented as

$w = \frac{2* \pi }{T}$

substituting values

$w = \frac{2* 3.142 }{24}$

$w = 0.2618 \ rad/s$

and r is the radius which is evaluated as $r = \frac{d}{2}$

substituting values

$r = \frac{24.383}{2}$

$r = 12.19 \ ft$

So

$F_c = 40 * 12.19* (0.2618)^2$

$F_c = 33.42 \ N$

W is the weight which is mathematically represented as

$W = 40 * 9.8$

$W = 392 \ N$

So

$F_A = 33.42 + 392$

$F_A =425.42 \ N$

The apparent weight of the passenger at the highest point is mathematically represented as

$F_A_H = W- F_c$

substituting values

$F_A_H = 392 - 33.42$

$F_A_H = 358.58 \ N$

4 0

3 years ago

The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 1.10 ✕ 105 kg train mov

Mrac [35]

Answer:

(a) k = 1684.38 N/m = 1.684 KN/m

(b) Vi = 0.105 m/s

(c) F = 1010.62 N = 1.01 KN

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

s = distance covered by train before stopping = 2 m

Therefore,

a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg)(0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

where,

k = Force Constant = ?

Δx = Compression = 2 m

Therefore.

3368.75 N = k(2 m)

k = (3368.75 N)/(2 m)

k = 1684.38 N/m = 1.684 KN/m

(c)

Applying Hooke's Law with:

Δx = 0.6 m

F = (1684.38 N/m)(0.6 m)

F = 1010.62 N = 1.01 KN

(b)

Now, the acceleration required for this force is:

F = ma

1010.62 N = (1.1 kg)a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

Now, we find initial velocity of train by using 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = -0.0092 m/s² (negative sign due to deceleration)

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

Therefore,

-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)

Vi = √(0.0092 m/s²)(1.2 m)

Vi = 0.105 m/s

4 0

3 years ago

A small loop of area A = 5.5 mm2 is inside a long solenoid that has n = 919 turns/cm and carries a sinusoidally varying current

Tema [17]

Answer: $410.90\times 10^{-6} V$

Explanation:

Given

$A=5.5 mm^2$

$n=919 turns/cm\approx 85400 turns/m$

$\omega =226 rad/s$

According to the Faraday law of induction, induced emf is given by

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

Magnetic field $\phi _B=B\cdot A$

$B=\mu _0ni=\mu _0ni_osin\left ( \omega t\right )$

$B=4\pi \times 10^{-7}\times 85400\times 3.08\times \sin (226t)$

$B=0.3305\sin (226t)$

$\phi _{B}=0.3305\times \sin (226t)\times 5.5\times 10^{-6}$

$\phi _{B}=1.818\times 10^{-6}\sin (226t)$

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

$E=-1.818\times 10^{-6}\times 226\cos (226t)$

$E=-410.90\times 10^{-6}\cos (226t)$

Amplitude of EMF induced $=410.90\times 10^{-6} V$

8 0

3 years ago