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nordsb [41]
3 years ago
11

intial velocity is 45km/h, final velocity is 0, time taken is 3 seconds,calculate the retradation of the vehicle and distance tr

avelled​
Physics
2 answers:
V125BC [204]3 years ago
4 0

Answer:

v= u+at

0=45×1000×60×60+3a

_3a=16200045

a=5400015m per sec

Usimov [2.4K]3 years ago
4 0

Answer:

Acceleration=12.5/3m/s^2=4.17m/s^2

Explanation:

Initial velocity u = 45 km/h = 12.5 m/s; Final velocity v = 0 m/s; time t = 3 s, acceleration a.

[a=v-u/t] : a=0-12.5/3=-12/3m/s^2

(Retaradation=-(-12.5/3)=12.5/3.

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PtichkaEL [24]

During that final period of time,
his acceleration is
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8 0
3 years ago
A baseball is batted from a height of 1.09 m with a speed of
kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0
1 year ago
Sound waves with a constant frequency of 250 hertz are traveling through air at stp. what is the wavelength of the sound waves
Deffense [45]

Answer:

wave length is 1.2m

Explanation:

since formula of wave length is v/f

v(speed of sound in air at stp is 300ms^-1)

f(frequency 250hertz)

then wave length is 300÷250 which give 1.2m

5 0
1 year ago
Nick parks his car at Target and walks 150 ft. north to get to the store. On his way back to the car, he walks 50 ft. before pau
mr Goodwill [35]
75 steps is what he’s missing
7 0
2 years ago
Hello, I wanted an answer from a mathematician. The number 1.04 is closer to the number 1, 2, 1.25 or 1.5.
babymother [125]

Answer:

Explanation:

Of the 4 numbers given, the answer is 1 or A

If you take the absolute value of abs(1 - 1.04) you get 0.04.

(2 - 1.04) = 0.96

1.25 - 1.04 = .21

1.5 - 1.04 = 0.46

The last three are all larger than 0.04

Note: absolute value means the positive difference between 2 numbers (even  though it is negative). If it is negative, absolute value makes it positive.

3 0
2 years ago
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