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3 years ago

PLEASE HELP! Explain what maria and craters are and how they formed This is 6th GRADE SCIENCE

Physics

2 answers:

oksian1 [2.3K]3 years ago

8 0

Answer:

anstroded hut them most of the time

Explanation:

patriot [66]3 years ago

7 0

Anstroded hut them most of the time

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A 1,450 kg car drives toward a 60 kg shopping cart that has a velocity of -1.2 m/s toward the car. The two objects collide, givi

Y_Kistochka [10]

Answer:

A) v₁ = 5.66 [m/s]

Explanation:

To solve this problem we must use the definition of linear momentum conservation, which tells us that momentum is conservation before and after a collision.

The linear momentum is equal to the mass by the product of the Velocity.

P = m*v

where:

P = lineal momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

Now, to the right side of the equal sign will take the linear momentum before the collision and to the left side of the equal sign as after the collision.

Pbefore = Pafter

(m₁*v₁) - (m₂*v₂) = (m₁*v₃) + (m₂*v₄)

where:

m₁ = mass of the car = 1450 [kg]

v₁ = velocity of the car before the collision [m/s]

m₂ = mass of the shopping cart = 60 [kg]

v₂ = velocity of the shopping cart before the collision = -1.2 [m/s]

v₃ = velocity of the car after the collision = 5.13 [m/s]

v₄ = velocity of the shopping cart after the collision = 11.75 [m/s]

Now replacing:

(1450*v₁) - (60*1.2) = (1450*5.13) + (60*11.75)

1450*v₁ - 72 = 7438.5 + 705

1450*v₁ = 7438.5 + 705 + 72

1450*v₁ = 8215.5

v₁ = 5.66 [m/s]

4 0

3 years ago

Read 2 more answers

Two objects of different masses accelerate at the same rate. According to Newtonâ€™s second law of motion, the more massive obje

babymother [125]

Newton's Second law of motion is

F = m * a ( force = mass * acceleration).

So if a is constant then a = F 1 / m1 and a = F2 / m2 then

F1 /m1 = F2 / m2 and if m1 > m2 then

F1 must be greater than F2.

Answer is C.

7 0

3 years ago

Please help me with this i dont know it

Inessa05 [86]

Answer:

no it would not. that is an open circuit and it would need to be closed at the switch for current to flow.

5 0

3 years ago

why does a spark jump from your hand to the doorknob as you reach for the door after walking across a carpeted floor?

Anna11 [10]

<span>because electrical charge builds up otherwise know as static electricity</span>

8 0

3 years ago

Read 2 more answers

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

larisa86 [58]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

Mass of the larger disk = $M_2\ =\ 1.60\ kg$
Mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the larger disk = $R_2\ =\ 5.00\ cm\ =\ 0.05\ m$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$
Mass of the block = M = 1.60 kg

Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

$\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

$mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)$

From the pulley,

$\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)$

From the equation (1) and (2),

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2$

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = $R_2\ =\ 0.05\ m$

Let $a_2$ be the acceleration of the block while connecting to the larger pulley. $\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2$

4 0

3 years ago