All categories

3 years ago

Is Mercury a heavier element than tin

Physics

1 answer:

Kryger [21]3 years ago

4 0

Yes because mercury has more protons and electrons that tin. (30 more)

You might be interested in

A satellite is to be launched into an orbit of radius,r Show that v = 2gr, where V is the

Snezhnost [94]

Explanation:

We start by using the conservation law of energy:

$\Delta{K} + \Delta{U} = 0$

$\dfrac{1}{2}mv^2 - G\dfrac{mM}{r} = 0$

Simplifying the above equation, we get

$v^2 = 2G\dfrac{M}{r}$

We can rewrite this as

$v^2 = 2\left(G\dfrac{M}{r^2}\right)r$

Note that the expression inside the parenthesis is simply the acceleration due to gravity $g$ so we can write

$v^2 = 2gr$

where $v$ is the launch velocity.

6 0

3 years ago

What are two differences between gravity and air resistance

Natalka [10]

1. Air resistance force usually upwards, but gravity doenwards.

2. Objects are affected by gravity the same, but air resistance can affect the speed of an object's descent.

Sorry if im wrong

4 0

3 years ago

Find the frequency of a tuning fork that takes2.50×10−3s to complete one oscillation.

maksim [4K]

The frequency of a tuning fork that takes2.50×10⁻³ s to complete one oscillation would be 400 Hz.

<h3>What is the frequency?</h3>

It can be defined as the number of cycles completed per second. It is represented in hertz and inversely proportional to the wavelength.

As given in the problem a tuning fork takes 2.50×10⁻³ s to complete one oscillation.

Frequency = 1/time period of the oscillation

Frequency of the tuning fork = 1/time period

=1/2.50×10⁻³

=400 Hz

Thus the frequency of a tuning fork comes out to be 400 Hz.

Learn more about frequency here, refer to the link;

brainly.com/question/14316711

#SPJ1

6 0

2 years ago

If a spring constant of 128N/m is compressed by 0.18 m , how much potential energy is the spring?

Mademuasel [1]

Answer:

2.1J

Explanation:

Given parameters:

Spring constant = 128N/m

Compression = 0.18m

Unknown:

Potential energy of the spring = ?

Solution

The potential energy of the spring is the elastic potential energy within the spring.

To solve this;

Elastic potential energy = $\frac{1}{2}$ k e²

k is the spring constant

e is the compression

Now;

Elastic potential energy = $\frac{1}{2}$ x 128 x 0.18² = 2.1J

7 0

3 years ago

Chris and Jamie are carrying Wayne on a horizontal stretcher. The uniform stretcher is 2.00 m long and weighs 100 N. Wayne weigh

PIT_PIT [208]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The value is $F_j = 550\ N$

Explanation:

From the question we are told that

The length of the stretcher is $d = 2.0 \ m$

The weight of the stretcher is $W = 100 \ N$

The weight for Wayne is $W_w = 800 \ N$

The distance of center of gravity for Wayne from Chris is $c_w = 75 cm = 0.75 \ m$

Generally taking moment about the first end where Chris is

$F_j * d$ => upward moment

Here $F_j$ is the force applied by Jamie

Generally taking moment about the second end where Jamie is

$W * ( \frac{d}{2} ) + W_w * (d - c_w)$ => downward moment

Generally at equilibrium , the upward moment is equal to the downward moment

$F_j * d = W * ( \frac{d}{2} ) + W_w * (d - c_w)$

=> $F_j * 2 = 100 * ( \frac{ 2}{2} ) + 800 * (2 - 0.75)$

=> $F_j = 550\ N$

3 0

3 years ago