What is the net force on a bathroom scale when a 110 pound person stands on it? How do you know?

If we interpret the large, angular rocks to have originated from the outcrop at the top of the hill, we are using __________ rea

IrinaK [193]

The reasoning which is in use when large, angular rocks are interpreted to have originated from the outcrop at the top of the hill is; Fossil succession

<h3>Fossil succession of rocks</h3>

The principle of fossil succession in characterized by the fact that fossil entities succeed one another upward through rock layers in a definite and determinable order.

On this note, any time period can be dated by its fossil content.

Read more on fossil succession;

brainly.com/question/2631497

3 0

2 years ago

Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is descri

Vadim26 [7]

Answer:

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part $kx \pm \omega t$ , the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but here is an explanation of this:

For both cases, + and -, after a certain time $\delta t$ ( $\delta t >0$ ), the displacement y of the wave will be determined by the $kx\pm\omega (t+\delta t)$ term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply $\pm \omega (t+\delta t)$ .

To know which side, right or left of the origin, would go through the origin after a time $\delta t$ (and thus know the direction of propagation) we have to see how we can achieve that same displacement y not by a time variation but by a space variation $\delta x$ (we would be looking where in space is what we would have in the future in time). The term would be then $k(x+\delta x)\pm\omega t$ , which at the origin is $k \delta x \pm \omega t$ . This would mean that, when the original equation has $kx+\omega t$ , we must have that $\delta x>0$ for $k\delta x+\omega t$ to be equal to $kx+\omega\delta t$ , and when the original equation has $kx-\omega t$ , we must have that $\delta x$ for $k\delta x-\omega t$ to be equal to $kx-\omega \delta t$

Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .

In conclusion, when $kx+\omega t$ , the part of the wave on the positive side ( $\delta x>0$ ) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

4 0

3 years ago

Which measure of an earthquake depends on how close you are to the focus?

Vsevolod [243]

The intensity of an earthquake is dependent on one's proximity to the focus of the quake, also called the "epicenter" and is based on observations of the shaking of the ground on humans, structures, and the natural landscape.

8 0

3 years ago

The magnitude of each force is 208 N the force on the right is applied at an angle 36° and the mass of the block is 17 kg the co

djyliett [7]

Answer:

11.06m/s²

Explanation:

According to Newtons second law of motion

$\sm F_x = ma_x\\F_m - F_f = ma_x\\mgsin \theta - \mu R mgcos \theta = ma_x\\$

Given

Mass m = 17kg

Fm = 208N

theta = 36 degrees

g = 9.8m/s²

a is the acceleration

Substitute

208 - 0.148(17)(9.8)cos 36 = 17a

208 - 24.6568cos36 = 17a

208 - 19.9478 = 17a

188.05 = 17a

a = 188.05/17

a = 11.06m/s²

Hence the the magnitude of the resulting acceleration is 11.06m/s²

6 0

2 years ago

Example of centrifugal force (wrong answer will get reported)(right answer will get brainliest)

Lera25 [3.4K]

Vehicle driving round a curve

3 0

3 years ago