How do we determine the amount of dark matter in elliptical galaxies?

How could Sam determine the distance he drove and the true displacement between the two locations

irina1246 [14]

Distance is the physical length travelled. While the displacement is the shorter distance between any two given points.

<h3>What is distance?</h3>

Distance is a numerical representation of the distance between two objects or locations.

Distance can refer to a physical length or an estimate based on other factors in physics or common use. |AB| is a symbol for the distance between two points A and B.

Displacement is defined as the shortest distance between the two points.

Sam determines the distance by doing the calculation as;

distance = speed × time

Hence, the Sam determines the distance and displacement by the respective formulas.

To learn more about the distance, refer to the link;

brainly.com/question/26711747

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5 0

2 years ago

Which statement is true of a concave lens?

natka813 [3]

It produces only virtual images is the answer

5 0

3 years ago

Read 2 more answers

a body initially at rest, starts moving with a constant acceleration of 2ms-2 .calculate the velocity acquired and the distance

Marta_Voda [28]

a) 10 m/s

b) 25 m

Explanation:

a)

The body is moving with a constant acceleration, therefore we can solve the problem by using the following suvat equation:

$v=u+at$

where

u is the initial velocity

v is the final velocity

a is the acceleration

t is the time

For the body in this problem:

u = 0 (the body starts from rest)

$a=2 m/s^2$ is the acceleration

t = 5 s is the time

So, the final velocity is

$v=0+(2)(5)=10 m/s$

b)

In this second part, we want to calculate the distance travelled by the body.

We can do it by using another suvat equation:

$v^2-u^2=2as$

where

u is the initial velocity

v is the final velocity

a is the acceleration

s is the distance travelled

Here we have

u = 0 (the body starts from rest)

$a=2 m/s^2$ is the acceleration

v = 10 m/s is the final velocity

Solving for s,

$s=\frac{10^2-0^2}{2(2)}=25 m$

3 0

3 years ago

The work function of a metal surface is 4.80 × 10-19 J. The maximum speed of the electrons emitted from the surface is vA = 7.7

poizon [28]

Answer:

$\lambda_A=2.65177\times 10^{-7}\ m$

$\lambda_B=3.32344\times 10^{-7}\ m$

Explanation:

h = Planck's constant = $6.63\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

$W_0$ = Work function = $4.8\times 10^{-19}\ J$

$v_A$ = Velocity of A particle = $7.7\times 10^5\ m/s$

$v_B$ = Velocity of B particle = $5.1\times 10^5\ m/s$

The wavelength is given by

$\lambda=\frac{hc}{\frac{1}{2}mv^2+W_0}$

$\lambda_A=\frac{6.63\times 10^{-34}\times 3\times 10^8}{\frac{1}{2}9.11\times 10^{-31}(7.7\times 10^5)^2+4.8\times 10^{-19}}\\\Rightarrow \lambda_A=2.65177\times 10^{-7}\ m$

The wavelength $\lambda_A=2.65177\times 10^{-7}\ m$

$\lambda_B=\frac{6.63\times 10^{-34}\times 3\times 10^8}{\frac{1}{2}9.11\times 10^{-31}(5.1\times 10^5)^2+4.8\times 10^{-19}}\\\Rightarrow \lambda_B=3.32344\times 10^{-7}\ m$

The wavelength $\lambda_B=3.32344\times 10^{-7}\ m$

5 0

4 years ago

A driver slows down her car from 32.7 km/hr at a constant rate of 0.63 m/s2 just by taking her foot of the accelerator (the gas

prohojiy [21]

Answer: She moves 5.616 meters in that second.

Explanation:

If we define t = 0s as the moment when she starts decelerating we can write the function of acceleration as:

a(t) = -(0.63 m/s^2)

where the negative sign is because she is slowing down.

The velocity equation can be found if we integrate over time:

v(t) = -(0.63m/s^2)*t + v0

Where v0 is the constant of integration, that represents the initial velocity, in this case is:

v0 = 32.7 km/h

Now, because the acceleration is in m/s^2, we should write this velocity in m/s.

in one km we have 1000 meters, and in one hour we have 3600 seconds, then we have that:

32.7 km/h = 32.7 *(1000/3600) m/s = 9.08 m/s

Then the velocity equation becomes:

v(t) = -(0.63m/s^2)*t + 9.08 m/s

And for the position equation, we should integrate again to get:

p(t) = -(1/2)*(0.63m/s^2)*t^2 + (9.08m/s)*t + p0

Where p0 is the initial position.

For this problem, we want to find the distance that she moved between t = 5s and t = 6s, and that can be calculated as:

D = p(6s) - p(5s)

D = -(1/2)*(0.63m/s^2)*(6s)^2 + (9.08m/s)*6s + p0 +(1/2)*(0.63m/s^2)*(5s)^2 - (9.08m/s)*(5s) - p0

D = -(1/2)*(0.63m/s^2)*((6s)^2 - (5s)^2) + (9.08m/s)*(6s - 5s)

D = 5.615 m

She moves 5.616 meters in that second.

7 0

3 years ago