QUICK HELP! Is this statement true or false?

PLEASE ANSWER, I HAVE 5 MINUTES!!!!

zmey [24]

Answer:

<em>I HAD THE SAME QUESTION AND SEARCHED IT UP AND GOT THE ANSWERS.</em>

Explanation:

3 0

3 years ago

Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship jour

Elena-2011 [213]

Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = $\frac{t}{\sqrt{1-\frac{v^2}{c^2} } }$

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

$\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }$

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

$\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }$

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

$\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }$

=1.67 x 10

t₁= 16.7 years

D ) Given t = 10 years , t₁ = ? v = .4c

$\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }$

=1.09 x 10

t₁= 10.9 years

E ) Given t = 20 years , t₁ = ? v = .8c

$\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }$

=1.67 x 20

t₁= 33.4 years

7 0

3 years ago

Physics quiz 11 th-12th grade helpp

FinnZ [79.3K]

5. The jogger's velocity is a constant 3.55 m/s between t = 4 s and t = 8 s.

6. Given a linear plot of velocity, the acceleration is determined by the slope of the line. Take any two points on the part of the plot after t = 8 s - for instance, we see it passes through (8 s, 3.5 m/s) and (10 s, 4 m/s) - and compute the slope:

(4 m/s - 3.5 m/s)/(10 s - 8 s) = (0.5 m/s)/(2 s) = 0.25 m/s^2

7. This amounts to finding the area between the velocity function and the time axis and between t = 4 s and t = 8 s. During this time, the velocity is 3.5 m/s. The time interval lasts 4 s. So the distance covered is

(3.5 m/s)*(4 s) = 14 m

8. After 4 seconds, Jimmy's speed decreases from 30.0 m/s to 27.2 m/s, so his acceleration (assuming it was constant) was

a = (27.2 m/s - 30.0 m/s)/(4 s) = -0.200 m/s^2

It's unclear what is meant by "rate of acceleration", since the acceleration is itself a rate. But maybe they just mean to ask for the acceleration, or possibly the magnitude?

4 0

4 years ago

Read 2 more answers

What is the period of a wave that travels through a spring at 2.5 m/s and has a wavelength of 1.3 meters?

ruslelena [56]

Answer:

v=(1/f) ×λ

(1/f)= v/λ

(1/f)= 2.5 ÷ 1.3

(1/f)= 1.92 seconds

T = 1.92 seconds

Explanation:

period= 1÷ frequency

T=1/f

7 0

2 years ago

A rocket sled accelerates from rest for a distance of 645 m at 16.0 m/s2. A parachute is then used to slow it down to a stop. If

inessss [21]

Answer:

the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

Explanation:

This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period

v² = v₀² + 2 a₁ x

indicate that the initial velocity is zero

v² = 2 a₁ x

let's calculate

v = $\sqrt {2 \ 15.0 \ 645}$

v = 143.666 m / s

now for the second interval let's find the distance it takes to stop

v₂² = v² - 2 a₂ x₂

in this part the final velocity is zero (v₂ = 0)

0 = v² - 2 a₂ x₂

x₂ = v² / 2a₂

let's calculate

x₂ = $\frac{ 143.666^2 }{2 \ 18.2}$

x₂ = 573 m

as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

3 0

3 years ago