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3 years ago

C.

Physics

1 answer:

Anna [14]3 years ago

7 0

Answer:

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The diagram below shows a closed system of two tanks that each contain water.

lidiya [134]

Lmalemwlsnlenekenfndelenekf

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2 years ago

How many milligrams in 1 gram. ??? HELP !!!

blagie [28]

1,000 milligrams = 1 gram
2,000 milligrams = 2 grams
3,000 milligrams = 3 grams
4,000 milligrams = 4 grams

3 0

3 years ago

Read 2 more answers

A 1250 kg car, driving 7.39 m/s, runs into the back of a stationary 5380 kg truck. After the collision, the truck moves forward

Leno4ka [110]

Explanation:

Given that,

Mass of the car, m₁ = 1250 kg

Initial speed of the car, u₁ = 7.39 m/s

Mass of the truck, m₂ = 5380 kg

It is stationary, u₂ = 0

Final speed of the truck, v₂ = 2.3 m/s

Let v₁ is the final velocity of the car. Using the conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$1250\times 7.39+5380\times 0=1250\times v_1+5380\times 2.3$

$v_1=-2.5\ m/s$

So, the final velocity of the car is 2.5 m/s but in opposite direction. Hence, this is the required solution.

3 0

3 years ago

The equation Bold r (t )equals(8 t plus 9 )Bold i plus (2 t squared minus 8 )Bold j plus (6 t )Bold k is the position of a parti

KATRIN_1 [288]

Explanation:

It is given that, the position of a particle as as function of time t is given by :

$r(t)=(8t+9)i+(2t^2-8)j+6tk$

Let v is the velocity of the particle. Velocity of an object is given by :

$v=\dfrac{dr(t)}{dt}$

$v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}$

$v=(8i+4tj+6k)\ m/s$

So, the above equation is the velocity vector.

Let a is the acceleration of the particle. Acceleration of an object is given by :

$a=\dfrac{dv(t)}{dt}$

$a=\dfrac{d[8i+4tj+6k]}{dt}$

$a=(4j)\ m/s^2$

At t = 0, $v=(8i+0+6k)\ m/s$

$v(t)=\sqrt{8^2+6^2} =10\ m/s$

Hence, this is the required solution.

7 0

3 years ago

A child’s toy that is made to shoot ping pong balls consists of a tube, a spring (k = 18 N/m) and a catch for the spring that ca

UkoKoshka [18]

Answer:

The height is 3.1m

Explanation:

Here we have a conservation of energy problem, we have a conversion form eslastic potencial energy to gravitational potencial energy, so:

$E_e=\frac{1}{2}K*x^2\\E_e=\frac{1}{2}18N/m*(9.5*10^{-2}m)^2\\E_e=0.081J$

then we have only gravitational potencial energy when the ball is at its maximun height.

$E_g=m*g*h$

because all the energy was transformed Eg=Ee

$h=\frac{0.081J}{9.8m/s^2*m}$

searching the web, the mass of a ping pong ball is 2.7 gr in average. so:

$h=\frac{0.081J}{9.8m/s^2*(2.7*10^{-3}kg)}\\h=3.1m$

6 0

3 years ago