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ch4aika [34]
3 years ago
15

1. "Milli" means "thousandth" so a millimeter (symbol: mL) is one thousandth of a liter. Therefore it takes ____(how many?) mL t

o make one L
Physics
1 answer:
koban [17]3 years ago
4 0
One thousand mL to make one L
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A particle travels along a straight line with a velocity v = (12 - 3t) m/s, where t is in seconds. When t = 1 s, the particle is
Simora [160]

Answer:

(1).The acceleration is -24 m/s²

The total distance is 912 m.

(2). The distance traveled before it stop is 78 m.

Explanation:

Given that,

Velocity = (12-3t^2) m/s

When t = 1 s, the particle is located 10 m to the left of the origin.

We need to calculate the acceleration at t = 4 sec

Using formula of acceleration

a=\dfrac{dv}{dt}

Put the value of v

a=\dfrac{d}{dt}(12-3t^2)

a=-6t

Put the value of t

a=-6\times4

a=-24\ m/s^2

The displacement from t = 0,t = 10 s, and the distance the particle travels during this time period.

We need to calculate the distance

Using formula of distance

ds=v\ dt

\int_{-10}^{s}=\int_{1}^{t}{v}dt

Put the value of v

\int_{-10}^{s}=\int_{1}^{t}{ (12-3t^2)}dt

s+10=12t-t^3-11

s=12t-t^3-21

At t = 0,

s_{t=0}=-21

At t = 10,

s_{t=10}=12\times10-10^3-21

s_{t=10}=-901

The displacement is

\Delta s=-901-(-21)

\Delta s=-880\ m

The distance at t= 2 sec

s_{t=2}=12\times2-2^3-21

s_{t=2}=-5

The total distance will be,

s_{T}=(21-5)+(901-5)

s_{T}=912\ m

(2). We need to calculate the distance at 2 sec

Using equation of motion

s=ut-\dfrac{1}{2}at^2

Put the value into the formula

s=27\times 2+\dfrac{1}{2}\times6\times(2)^3

s=78\ m

Hence, (1).The acceleration is -24 m/s²

The total distance is 912 m.

(2). The distance traveled before it stop is 78 m.

3 0
3 years ago
An electric vehicle starts from rest and accelerates at a rate a1 in a straight line until it reaches a speed of v. The vehicle
levacccp [35]

(a) t=\frac{v}{a_1}+\frac{v}{a_2}

In the first part of the motion, the car accelerates at rate a_1, so the final velocity after a time t is:

v = u +a_1t

Since it starts from rest,

u = 0

So the previous equation is

v= a_1 t

So the time taken for this part of the motion is

t_1=\frac{v}{a_1} (1)

In the second part of the motion, the car decelerates at rate a_2, until it reaches a final velocity of v2 = 0. The equation for the velocity is now

v_2 = v - a_2 t

where v is the final velocity of the first part of the motion.

Re-arranging the equation,

t_2=\frac{v}{a_2} (2)

So the total time taken for the trip is

t=\frac{v}{a_1}+\frac{v}{a_2}

(b) d=\frac{v^2}{2a_1}+\frac{v^2}{2a_2}

In the first part of the motion, the distance travelled by the car is

d_1 = u t_1 + \frac{1}{2}a_1 t_1^2

Substituting u = 0 and t_1=\frac{v}{a_1} (1), we find

d_1 = \frac{1}{2}a_1 \frac{v^2}{a_1^2} = \frac{v^2}{2a_1}

In the second part of the motion, the distance travelled is

d_2 = v t_2 - \frac{1}{2}a_2 t_2^2

Substituting t_2=\frac{v}{a_2} (2), we find

d_1 = \frac{v^2}{a_2} - \frac{1}{2} \frac{v^2}{a_2} = \frac{v^2}{2a_2}

So the total distance travelled is

d= d_1 +d_2 = \frac{v^2}{2a_1}+\frac{v^2}{2a_2}

7 0
3 years ago
What is the entire range of em waves called
Olin [163]

The entire range of anything is called its spectrum.

For em wavelengths, it's the electromagnetic spectrum.


7 0
4 years ago
A rock is dropped from a vertical cliff. The rock takes 3.00 s to reach the ground below the cliff. A second rock is thrown vert
Phantasy [73]

Answer:

Velocity v= 12.25 \frac{m}{s}

Explanation:

The first rock dropped give the distance Y in meters

Y_{f}=Y_{o}+v_{o}*t +\frac{1}{2}*a*t^{2}\\   Y_{f}=\frac{1}{2}* 9.8 \frac{m}{s^{2} }* 3^{2}   \\Y_{f}=44.1 m

Now the motion of the second rock the time change so to know the velocity

Y_{f}= Y_{o} +v_{o}*t +\frac{1}{2}*a *t^{2} \\v_{o}*t= -Y_{f} +\frac{1}{2} *a*t^{2} \\v_{o} =\frac{-44.1 +0.5 * 9.8* s^{2} }{2} \\v_{o}=  12.25 \frac{m}{s}

7 0
3 years ago
a 50kg skater on level ice, has built up her speed to 30km/h. how far will she coast before sliding friction dissipates her ener
azamat

Answer:

belpw

Explanation:

7 0
3 years ago
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