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uranmaximum [27]

2 years ago

7

You are driving your car and the traffic light ahead turns red.You apply the breaks for 3s and the velocity of the car decreases

to 4.5 m/s . If the cars deceleration has a magnitude of 2.7m/s^2,what is the car's displacement during this time?

1 answer:

hammer [34]2 years ago

7 0

Answer:

The car's displacement during this time is 25.65 meters.

Explanation:

Given that,

Final velocity of the car, v = 4.5 m/s

Deceleration of the car, $a=-2.7\ m/s^2$

Let u is the initial speed of the car. It is given by :

$v=u+at$

$u=v-at$

$u=4.5-(-2.7)\times 3$

u = 12.6 m/s

Let d is the car's displacement during this time. It can be calculated using second equation of motion as :

$d=ut+\dfrac{1}{2}at^2$

$d=12.6\times 3+\dfrac{1}{2}\times (-2.7)\times 3^2$

d = 25.65 meters

So, the car's displacement during this time is 25.65 meters. Hence, this is the required solution.

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When you jump from an elevated position you usually bend your knees upon reaching the ground. By doing this, you make the time o

dsp73

Answer:

c. about 1/10 as great.

Explanation:

While jumping form a certain height when we bend our knees upon reaching the ground such that the time taken to come to complete rest is increased by 10 times then the impact force gets reduced to one-tenth of the initial value when we would not do so.

This is in accordance with the Newton's second law of motion which states that the rate of change in velocity is directly proportional to the force applied on the body.

Mathematically:

$F\propto\frac{d}{dt} (p)$

$\Rightarrow F=\frac{d}{dt} (m.v)$

since mass is constant

$F=m\frac{d}{dt}v$

when $dt=10t$

then,

$F'=m.\frac{v}{10\times t}$

$F'=\frac{1}{10} \times \frac{m.v}{t}$

$F'=\frac{F}{10}$ the body will experience the tenth part of the maximum force.

where:

$\frac{d}{dt} =$ represents the rate of change in dependent quantity with respect to time

$p=$ momentum

$m=$ mass of the person jumping

$v=$ velocity of the body while hitting the ground.

7 0

2 years ago

A coin is placed on a vinyl stereo record that is making 33 1/3 revolutions per minute on a turntable. (a) In what direction is

mamaluj [8]

Answer: a) The acceletarion is directed to the center on the turntable. b) 5 cm; ac= 0.59 m/s^2; 10 cm, ac=1.20 m/s^2; 14 cm, ac=1.66 m/s^2

Explanation: In order to explain this problem we have to consider teh expression of the centripetal accelartion for a circular movement, which is given by:

ac=ω^2*r where ω and r are the angular speed and teh radios of the circular movement.

w=2*π*f

We know that the turntable is set to 33 1/3 rev/m so

the frequency 33.33/60=0.55 Hz

then w=2*π*0.55=3.45 rad/s

Finally the centripetal acceleration at differents radii results equal:

r= 0.05 m ac=3.45^2*0.05=0.50 m/s^2

r=0.1 ac=3.45^2*0.1=1.20 m/s^2

r=0.14 ac=3.45^2*0.14=1.66 m/s^2

4 0

3 years ago

At what position or positions on the x-axis is the electric field zero?

ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

$E=\dfrac{kq}{r^2}$

The electric field vector due to charge one

$\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})$

The electric field vector due to charge second

$\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})$

We need to calculate the electric field

Using formula of net electric field

$\vec{E}=\vec{E_{1}}+\vec{E_{2}}$

$\vec{E_{1}}+\vec{E_{2}}=0$

Put the value into the formula

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0$

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})$

$(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}$

$\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}$

Put the value into the formula

$\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}$

$2.0+x=x$

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

4 0

3 years ago

What is the speed of a wave that has a frequency of 200 Hz and a

Luba_88 [7]

The wave speed to this question is 400 meters

8 0

3 years ago

An object moves from the position +16m to the position +43m in 12s. What us the total displacement

NeX [460]

First method

initial distance = 16m

final distance= 43 m

total distance covered= final -initial

=43m -16m

=27m

Second method

Si= 16m

Sf =43 m

t= 12 s

first we will find V

V = (Sf-Si)/ t

V =( 43- 16)/ 12

V = 27/12 ⇒ V= 9/4

V= distance / time

distance= V×time

distance = (9/4) ×12

distance =27

3 0

3 years ago