All categories

3 years ago

The answer and why it's that one

Physics

1 answer:

scoundrel [369]3 years ago

3 0

The force between two magnets, the gravitational forces between two objects,
and the forces between two charged particles DON't require that the objects
be touching each other.

The forces of friction between an object and air, or the forces of friction
between two objects that are sliding, DO only happen when the two are
in contact with each other.

You might be interested in

Two rods are identical, except that one is brass (Y = 9.0 × 1010 N/m2) and one is tungsten (Y = 3.6 × 1011 N/m2). A force causes

PIT_PIT [208]

To solve this problem it is necessary to apply the concepts related to Young's Module, and find the radius that gives the ratio between the two given materials. Young's module can be defined as,

$Y=\frac{FL}{A \Delta L}$

Where,

F= Force

L = Initial Length

A = Cross-sectional Area

$\Delta L =$ Change in Length

Re-arrange the equation to find the change in Length we have,

$\Delta L = \frac{FL}{AY}$

If both the Force, as the Area and the initial length are considered constant, we can realize directly that the change in length is inversely proportional to Young's Module, therefore

$\Delta L \propto \frac{1}{Y}$

Applying this concept to that of the two materials (Brass and Tungsten),

$\frac{\Delta L_T}{\Delta L_B} = \frac{Y_B}{Y_T}$

$\frac{\Delta L_T}{\Delta L_B} = \frac{9*10^{10}}{3.6*10^{11}}$

$\frac{\Delta L_T}{\Delta L_B} = 0.25$

If the force caused $3 * 10^ {- 6}m$ to be stretched, the tungsten will stretch 0.25 of that ratio

$L_T = 3*10^{-6}*0.25$

$L_T = 7.5*10^{-7}m$

Therefore the amount of stretch of Tungsten is 7.5*10^{-7}m

8 0

3 years ago

Using information about natural laws, explain why some car crashes produce minor injuries and others produce catastrophic injuri

Nataly_w [17]

Don't they have to pay insurance company if something happens to there car.

5 0

3 years ago

A skier with a 65kg mass skies down a 30 degree incline hill. The coefficient of friction is 0.1. a. Draw a free body diagram. b

djyliett [7]

B is the answer! Because no one would be pushing the rock all the way down the hill it would just go down by itself by rolling!!

Hope i helped plz mark as brainlist and 5 star'!

6 0

4 years ago

Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init

Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ= $(m+\frac{1}{2})\frac{lamda}{n}$ ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= $\frac{3*lamda}{2}$

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= $\sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }$

δ= $\sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}$

δ= $\sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }$

δ= $\sqrt{4900+43^2}-\sqrt{4900+23^2}$

δ= $\sqrt{4900+1849}-\sqrt{4900+529}$

δ= $\sqrt{6749}-\sqrt{5429}$

δ= 82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= $\frac{3*lamda}{2}$

δ= 8.47

8.47 = $\frac{3*lamda}{2}$

λ = $\frac{8.47*2}{3}$

λ = 5.65m

3 0

3 years ago

A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes

OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

$y(t) = h +u_y t - \frac{1}{2}gt^2$

where

h = 2.5 km = 2500 m is the initial height

$u_y = 0$ is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

$y(t) = 0$

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

$0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s$

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

$v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s$

So the horizontal distance travelled is

$d=v_x t$

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

$d=(333.3)(22.6)=7533 m = 7.5 km$

7 0

3 years ago