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tresset_1 [31]
3 years ago
10

A car decelerates from 36 meters per second to 12 meters per second. The constant rate of deceleration is 12 meters per second p

er second. How long did it take for the car to decelerate?
Physics
1 answer:
Svetach [21]3 years ago
5 0
Answer:
2s
Explanation:
It takes the car 1 second to decelerate 12m and it decelerated 24m (as 36 - 12 = 24)
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A particle passes through the point P=(−3,1,0)P=(−3,1,0) at time t=3t=3, moving with constant velocity v⃗ =⟨5,3,−2⟩v→=⟨5,3,−2⟩.
eimsori [14]

Answer:

The parametric equation for the position of the particle is (-18+5t,-8+3t,6-2t).

Explanation:

Given that,

The point is

P=(-3,1,0)

Time t = 3

Velocity v=(5,3,-2)

We need to calculate the parametric equation for the position of the particle

Using parametric equation for position

r(t)=r_{0}+v(t)t....(I)

at t = 3,

P=r(t)

Put the value into the formula

(-3,1,0)=r_{0}+(5,3,-2)\times3

(-3,1,0)=r_{0}+(15,9,-6)

r_{0}=(-18,-8,6)

Put the value of r₀ in equation (I)

r(t)=(-18,-8,6)+(5,3,-2)t

r(t)=(-18+5t,-8+3t,6-2t)

Hence, The parametric equation for the position of the particle is (-18+5t,-8+3t,6-2t).

3 0
3 years ago
A uniform thin rod of length 0.11 m and mass 4.6 kg can rotate in a horizontal plane about a vertical axis through its center. T
ale4655 [162]

Answer:

Explanation:

This problem is based on conservation of rotational momentum.

Moment of inertia of rod about its center

= 1/12 m l² , m is mass of the rod and l is its length .

= 1 / 12 x 4.6 x .11²

I = .004638 kg m²

The angular momentum of the bullet about the center of rod = mvr

where m is mass , v is perpendicular component of velocity of bullet and r is distance of point of impact of bullet fro center .

5 x 10⁻³ x v sin60 x .11 x .5  where v is velocity of bullet

According to law of conservation of angular momentum

5 x 10⁻³ x v sin60 x .11 x .5 = ( I + mr²)ω , where ω is angular velocity of bullet rod system and  ( I + mr²) is moment of inertia of bullet rod system .

.238 x 10⁻³ v = ( .004638 + 5 x 10⁻³ x .11² x .5² ) x 12

.238 x 10⁻³ v = ( .004638 + .000015125 ) x 12

.238 x 10⁻³ v = 55.8375 x 10⁻³

.238 v = 55.8375

v = 234.6 m /s

4 0
3 years ago
The group one metals are called alkaline metal and group two metals are called alkaline earth metal why?​
Ilya [14]

Answer:

The group one element are called alkali because when they dissolved in water they form alkaline solutions

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2 years ago
A cyclist racing in the keiran is riding at the top of the track at 5m/s. Then he sprints downhill in the sprinting to the finis
kumpel [21]
The cyclist is moving by uniformly accelerated motion, with an initial velocity of v_i=5~m/s and an acceleration of a=9~m/s^2. 
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a= \frac{v_f-v_i}{\Delta t}
where v_f is the final velocity and \Delta t is the time between the end and the beginning of the motion, and in our case it is 1.75 s. Therefore, from this relationship we can find the final velocity:
v_f=v_i + a \Delta t=5~m/s+9~m/s^2 \cdot 1.75~s=20.75~m/s
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