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harina [27]
1 year ago
13

a car company wants to ensure its newest model can stop in 50 m when traveling at 30 m/s (which is about 108 km/h). if we assume

constant deceleration, find the value of deceleration that accomplishes this
Physics
1 answer:
Brut [27]1 year ago
8 0

A decrease in velocity is referred to as deceleration. If car is moving at 30 m/s and stop in 50 m .The value of deceleration is 11.56 ms−2.

<h3>How to calculate deceleration ?</h3>

While acceleration is motion in which an object's speed varies every second, deceleration is motion that causes an object to slow down.

We are aware that acceleration refers to an object's rate of increase in speed, and deceleration refers to an object's rate of decrease in speed. For instance, when we apply the brakes while driving, we benefit from the vehicle's ability to decelerate and slow down.

The Deceleration Formula is the final velocity minus the initial velocity, with a negative sign in the result because the velocity is decreasing, if starting velocity, final velocity, and time taken are given.

velocity of car = 30 m/s

car need to stop in 50m

Deceleration a = v^2 –  u^2 / 2s

                          = 0^2 - 50^2 / 2*30

                          = 11.56

Deceleration of the care = 11.56 ms−2

To learn more about deceleration refer :

brainly.com/question/75351

#SPJ4

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a 0.25 kg arrow is moving at 5 m/s and hits a 0.10 kg apple. The apple sticks to the arrow, and both move to the right together.
MatroZZZ [7]

Answer:

The arrow-apple combo will move with a velocity 3.6 m/s to the right.

Explanation:

Use the law of conservation of momentum to solve this problem. In this case the law can be written as follows:

m_{arrow}v_{arrow}+m_{apple}\cdot 0= (m_{arrow}+m_{apple})v_{both}

from which the desired velocity can be isolated:

\frac{m_{arrow}v_{arrow}}{m_{arrow}+m_{apple}}= v_{both}\\v_{both} = \frac{0.25kg\cdot 5\frac{m}{s}}{0.35kg}=3.6\frac{m}{s}

The arrow-apple combo will move with a velocity 3.6 m/s to the right.

3 0
3 years ago
Salmon often jump waterfalls to reach their
PilotLPTM [1.2K]

The minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

The given parameters;

  • height of the waterfall, h = 0.432 m
  • distance of the Salmon from the waterfall, s = 3.17 m
  • angle of projection of the Salmon, = 30.8º

The time of motion to fall from 0.432 m is calculated as;

h = v_0_y + \frac{1}{2} gt^2\\\\0.432 = 0 + (0.5\times 9.8)t^2\\\\0.432 = 4.9t^2\\\\t^2 = \frac{0.432}{4.9} \\\\t^2 = 0.088\\\\t = \sqrt{0.088} \\\\t = 0.3 \ s

The minimum velocity of the Salmon jumping at the given angle is calculated as;

X = v_0_x t\\\\3.17 = (v_0\times cos(30.8)) \times 0.3\\\\10.567 = v_0\times cos(30.8)\\\\v_0 = \frac{10.567}{cos(30.8)} \\\\v_0 = 12.3 \ m/s

Thus, the minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

Learn more here: brainly.com/question/20064545

8 0
2 years ago
A satellite in outer space is moving at a constant velocity of 20.5 m/s in the +y direction when one of its on board thruster tu
Katyanochek1 [597]

Answer:

a)  V_f = 25.514 m/s

b)  Q =53.46 degrees CCW from + x-axis

Explanation:

Given:

- Initial speed V_i = 20.5 j m/s

- Acceleration a = 0.31 i m/s^2

- Time duration for acceleration t = 49.0 s

Find:

(a) What is the magnitude of the satellite's velocity when the thruster turns off?

(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.

Solution:

- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:

                                   V_f = V_i + a*t

                                   V_f = 20.5 j + 0.31 i *49

                                   V_f = 20.5 j + 15.19 i

- The magnitude of the velocity vector is given by:

                                   V_f = sqrt ( 20.5^2 + 15.19^2)

                                   V_f = sqrt(650.9861)

                                  V_f = 25.514 m/s

- The direction of the velocity vector can be computed by using x and y components of velocity found above:

                                 tan(Q) = (V_y / V_x)

                                 Q = arctan (20.5 / 15.19)

                                 Q =53.46 degrees

- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.

4 0
3 years ago
lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

5 0
3 years ago
What is the definition of power.
Mandarinka [93]

Answer:

What is the definition of power.

Explanation:

ability to do or act; capability of doing or accomplishing something.

6 0
2 years ago
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