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2 years ago

How do you use vectors daily without realizing it was a vector? I need help ASAP !!!

Physics

1 answer:

jonny [76]2 years ago

6 0

Explanation:

Vectors are quantities that has both magnitude and direction. A vector is described in terms of quantity and the direction it is headed.

There are different forms of vectors that are used in our daily life.

When we describe the motion of a car and the direction it is heading, we are simply talking about its velocity which is a vector.
When we say James treks 2km from his school to the house everyday, we are simply describing his displacement.
When we apply a force to push the table from one corner of the room to another point, is simply a vector
When we change position, it is a vector application.

You might be interested in

Do frictional forces act in the same direction or in the opposite direction to the applied force?

Evgesh-ka [11]

Frictional forces act in the direction opposite to the MOTION. That direction could be the same OR opposite to applied force.

-- If you push a loaded heavy wagon from behind, trying to get it going faster, friction is acting against you, opposite to your force.

-- If you push a loaded rolling heavy wagon from in front, trying to make it slow down, friction is acting with you, in the same direction as your force.

-- Opposite to the motion both times.

3 0

2 years ago

A motorcycle stops at a red light and then gains a speed of 12 m/s in 15s. What's its acceleration?

erastova [34]

Option a) 0.8 m/s² is the correct answer

the acceleration of the motorcycle is 0.8 m/s²

Given the data in the question;

Since the motorcycle stopped then, initial speed u = 0 m/s

final speed v = 12 m/s

time t = 15 s

acceleration = ?

To get our acceleration, we use the first equation of motion.

This equation states that;

$v = u + at$

so, we simply substitute in our given values

$12 m/s = 0 m/s + ( a * 15s )$

$12 m/s = ( a * 15 s )$

so we find a

$a = \frac{12m/s}{15s}$

$a = 0.8 m/s^2$

Therefore, the acceleration of the motorcycle is 0.8 m/s²

Option a) 0.8 m/s² is the correct answer

for more on this topic, visit brainly.com/question/17418101?referrer=searchResults

8 0

2 years ago

What is the name of the part of the wave that is labeled -

tresset_1 [31]

Answer:

amplitude

Explanation:

sound travel and it amplifies

7 0

2 years ago

A stone is thrown in a vertically

Lemur [1.5K]

Answer:

1.25 m

0.5 s

Explanation:

Given:

v₀ = 5 m/s

v = 0 m/s

a = -10 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (5 m/s)² + 2 (-10 m/s²) Δy

Δy = 1.25 m

Find: t

v = at + v₀

(0 m/s) = (-10 m/s²) t + (5 m/s)

t = 0.5 s

6 0

2 years ago

What fraction of 5 MeV α particles will be scattered through angles greater than 8.5° from a gold foil (Z = 79, density = 19.3 g

aalyn [17]

Answer:

$f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)$

And after do all the operations we got:

$f(8) = 1.96 x10^{-4] atoms/m^2$

And that would be the final answer for this case.

Explanation:

For this case we can use the fomrula for the fraction of incident particles scattered by an angle $\theta$ , given by:

$f(\theta) = \pi nt (\frac{Z_1 e Z_2 e}{8 \pi e_o K})^2 cos^2 (\theta/2)$

Where:

$Z_1 e$ represent the charge of the projectile (Z1=2)

$Z_2 e$ is the target charge (z2=79)

$K= 5x10^6 eV$ represent the kinetic energy of incident particle

n represent the density of target particles (we need to find it first)

$t= 10^{-8] m$ represent the thicknss of the foil

The first step would be calculate the density of target particles with the following formula:

$n =\frac{\rho N_A N_M}{M_g}$

Where:

$\rho = 19.3 g/m^3= 19300 Kg/m^3$

$N_A = 6.022 x10^{23} molecules/mol$ the Avogadro's number

$N_M = 1$ represent the atoms per molecule

$M_g = 197 g/mol = 0.197 Kg/mol$ represent the molecular weigth

If we replace we got:

$n = \frac{19300 kg/m^3 *6.022x10^{23} molecu/mol * 1 atom/mole}{0.197 Kg/mol}= 5.90 x106{28] atoms/m^3$

Now we can calculate the fraction of 5 MeV alpha particles that would be scatteres with angle higher than 8 degrees in a piece of thickness $t=10^{-8}m$

And using the first formula we got:

$f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)$

And after do all the operations we got:

$f(8) = 1.96 x10^{-4] atoms/m^2$

And that would be the final answer for this case.

4 0

2 years ago