Does mercury have more protons and electrons than tin

Calculate the change in length of concrete sidewalk (coefficient of linear expansion for concrete is 12*10^-6/celcius) that is 1

anyanavicka [17]

Answer:

The answer to your question is 5.4 cm

Explanation:

This problem refers to calculate the change in length in one dimension due to a change in temperature.

Data

α = 12 x 10⁻⁶

Lo = 150 meters

ΔT = 30 °C

Formula

ΔL/Lo = αΔT

solve for ΔL

ΔL = αLoΔT

Substitution

ΔL = (12 x 10⁻⁶)(150)(30)

Simplification

ΔL = 0054 m = 5.4 cm

7 0

3 years ago

A power supply is connected to a 59 Ohm resistor and a 53 Ohm resistor in series. The total current is found to be 0.15 A. What

kirill [66]

the answer is 47265.dug

Explanation:

im bot sure

4 0

2 years ago

A child goes down the slide,starting from rest. If the length of the slide is 2m and it takes the child 3 seconds to go down the

lapo4ka [179]

Answer:

$0.44 m/s^2$

Explanation:

We have the following data:

- distance covered by the child: d = 2 m (length of the slide)

- time taken to cover this distance: t = 3 s

- initial velocity of the child: 0 m/s (he starts from rest)

So we can find the acceleration by using the equation:

$d=ut+\frac{1}{2}at^2$

Where a is the acceleration.

Substituting the values and solving for a,

$a=\frac{2d}{t^2}=\frac{2(2)}{3^2}=0.44 m/s^2$

3 0

3 years ago

When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the

Taya2010 [7]

Answer:

The angular velocity is 15.37 rad/s

Solution:

As per the question:

$\theta = 54.6^{\circ}$

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v = $\sqrt{\frac{gx}{sin2\theta}}$

v = $\sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}$

v = $\sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s$

Now,

The angular velocity can be calculated as:

$v = \omega R$

Thus

$\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s$

8 0

3 years ago

An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part

guajiro [1.7K]

Explanation:

The electric field at a distance r from the charged particle is given by :

$E=\dfrac{kq}{r^2}$

k is electrostatic constant

if r = 2 m, electric field is given by :

$E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)$

If r = 1 m, electric field is given by :

$E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)$

Dividing equation (1) and (2) we get :

$\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1$

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

4 0

3 years ago