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3 years ago

Name and describe the apparatus used by Cavendish to discover the universal gravitation constant

Physics

1 answer:

Evgen [1.6K]3 years ago

3 0

Its like a suspended wood with a lead sphere attached to each of its ends

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Margaret wants to go for a swim, and decides to jump in using the diving board that measures 3-m long.

otez555 [7]

Answer:

The horizontal component of her velocity is approximately 1.389 m/s

The vertical component of her velocity is approximately 7.878 m/s

Explanation:

The given question parameters are;

The initial velocity with which Margaret leaps, v = 8.0 m/s

The angle to the horizontal with which she jumps, θ = 80° to the horizontal

The horizontal component of her velocity, vₓ = v × cos(θ)

∴ vₓ = 8.0 × cos(80°) ≈ 1.389

The horizontal component of her velocity, vₓ ≈ 1.389 m/s

The vertical component of her velocity, $v_y$ = v × sin(θ)

∴ $v_y$ = 8.0 × sin(80°) ≈ 7.878

The vertical component of her velocity, $v_y$ ≈ 7.878 m/s.

6 0

3 years ago

To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circus performer wh

sp2606 [1]

Answer:

$v = 15.45 m/s$

Explanation:

As per mechanical energy conservation we can say that here since friction is present in the barrel so we will have

Work done by friction force = Loss in mechanical energy

so we will have

$W_f = (U_i + K_i) - (U_f + K_f)$

here we know that

$W_f = F_f . d$

$W_f = 40 \times 4$

$W_f = 160 J$

Initial compression in the spring is given as

$F = kx$

$4400 = 1100 x$

$x = 4 m$

now from above equation

$W_f = (\frac{1}{2}kx^2 + 0) - (mgh + \frac{1}{2}mv^2)$

$160 = (\frac{1}{2}1100(4^2) + 0) - (60 \times 9.8\times 2.50 + \frac{1}{2}(60)v^2)$

$160 = 8800 - 1470 - 30 v^2$

$v = 15.45 m/s$

3 0

3 years ago

The speed of a wave is 2ms, and its wavelength 0.4 meters. What is the period of the wave?

AnnyKZ [126]

Answer:

Explanation:

6 0

2 years ago

Read 2 more answers

If 0.035pC of charge is transferred via the movement of Al3+ ions, how's many of these must be transferred in total? Please add

mr Goodwill [35]

Each $Al^+^3$ ion contains three extra protons. Hence, the extra charge on each $Al^+^3$ = $3 \times 1.6 \times 10^-^1^9$ C

Total charge = 0.035 pC

Total charge (Q) = $0.035 \times 10^-^1^2$ C

Let the number of $Al^+^3$ ions be n.

According to question:

$n \times 3 \times 1.6 \times 10^-^1^9 =0.035 \times 10^-^1^2$

$n = \frac{0.035 \times 10^-^1^2}{3 \times 1.6 \times 10^-^1^9}$

$n = 7.29167 \times 10^4$

n = 72917

Hence, the total number of ions needed to be transferred is 72917

3 0

3 years ago

What are the units, if any, of the particle in a box wavefunction. What does this mean?

SIZIF [17.4K]

Answer:

$[\psi]= [Length^{-3/2}]$
This means that the integral of the square modulus over the space is dimensionless.

Explanation:

We know that the square modulus of the wavefunction integrated over a volume gives us the probability of finding the particle in that volume. So the result of the integral

$\int\limits^{x_f}_{x_0} \int\limits^{yf}_{y_0} \int\limits^{z_f}_{z_0} |\psi|^2 \, dz \, dy \, dx$

must be dimensionless, as represents a probability.

As the differentials has units of length

$[dx]=[dy]=[dz]=[Length]$

for the integral to be dimensionless, the units of the square modulus of the wavefunction has to be:

$[\psi]^2 = [Length^{-3}]$

taking the square root this gives us :

$[\psi] = [Length^{-3/2}]$

5 0

3 years ago