All categories

1 year ago

PLS HELP. WILL GIVE BRAINLIEST. PLS THIS IS WORTH 35 POINTS ON MY TEST

Physics

2 answers:

Nata [24]1 year ago

5 0

Answer:

18m/s^2

Explanation:

Vf = Vi + at

t = distance/ average velocity

(120 + 0)/2 = 60 (average velocity)

400m/60m/s = 20/3 s

insert into first equation:

120 = 0 + a(20/3)

360 = 20a

18 = a

HOPE THIS HELPS!!!

Greeley [361]1 year ago

3 0

Initial velocity=0=u
Final velocity=v=120m/s
Distance=s=400m
Acceleration=a

Using third equation of kinematics

$\\ \rm\hookrightarrow v^2-u^2=2as$

$\\ \rm\hookrightarrow 120^2=800a$

$\\ \rm\hookrightarrow 14400=800a$

$\\ \rm\hookrightarrow a=16m/s^2$

You might be interested in

The acceleration due to gravity on the surface of Mars is gmars = 3.7 m/s2. How much would an 11 kg bag of potatoes weigh on Mar

allsm [11]

Fnet = (mass) (acceleration)
= 11 kg x 3.7m/s^2
= 41 N

6 0

3 years ago

Which of the following correctly arranges the planets by the size of their diameter in increasing order?

valentinak56 [21]

Increasing Order..
So, it's from the little one into the biggest one..

My Answer is
C: Mercury < Earth < Saturn < Jupiter

What Shape of The Earth ?
Is it Round or Flat ?
Subjective Answer..

7 0

2 years ago

Read 2 more answers

Light of wavelength 580 nm is incident on a slit of width 0.30 mm. An observing screen is placed 2.0 m past the slit. Find the d

uysha [10]

Answer:

Y = 3.87 x 10⁻³ m = 3.87 mm

Explanation:

This problem can be solved by using Young's double-slit experiment formula:

$Y = \frac{\lambda L}{d}$

where,

Y = fringe spacing = ?

L = slit to screen distance = 2 m

λ = wavelength of light = 580 nm = 5.8 x 10⁻⁷ m

d = slit width = 0.3 mm = 3 x 10⁻⁴ m

Therefore,

$Y = \frac{(5.8\ x\ 10^{-7}\ m)(2\ m)}{3\ x\ 10^{-4}\ m}$

3 0

2 years ago

Light in vacuum is incident on the surface of a slab of transparent material. In the vacuum the beam makes an angle of 39.9° wit

wariber [46]

Answer:

n=2.053

Explanation:

We will use Snell's Law defined as:

$n_{1}*Sin\theta_{1}=n_{2}*Sin\theta_{2}$

Where n values are indexes of refraction and $\theta$ values are the angles in each medium. For vacuum, the index of refraction in n=1. With this we have enough information to state: