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3 years ago

A 1.0 kg toy car is released at the top of a frictionless track on the left and rolls off of the track from its right

Physics

2 answers:

lara31 [8.8K]3 years ago

7 0

Answer: -5.4

Explanation:

Khan said so

lapo4ka [179]3 years ago

7 0

Answer:

2.5 J

Explanation:

What is the change in the car gravitational potential energy from A-B

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3. What does the difference in force depend on?

dangina [55]

Answer:

it depends on the relative masses of the objects.

Explanation:

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3 years ago

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shtirl [24]

5.6 g/ml. That is the density.

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Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water

Ksenya-84 [330]

To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

$v_2^2 = v_1^2 + 2gh$

Where,

$V_i =$ Velocity in each state

g= Gravity

h = Height

Our values are given as,

$A_1 = 2.4*10^{-4} m^2$

$v_1 = 0.8 m/s$

$h = 0.11m$

Replacing at the kinetic equation to find $V_2$ we have,

$v_2 = \sqrt{v_1^2 + 2gh}$

$v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}$

$v_2= 1.67 m/s$

Applying the concepts of continuity,

$A_1v_1 = A_2v_2$

We need to find A_2 then,

$A_2= \frac{A_1v_1 }{v_2}$

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

$A_2= \frac{A_1v_1 }{v_2}$

$A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}$

$A_2= 1.14*10^{-4} m2$

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is $1.14*10^{-4} m2$

8 0

3 years ago

A spherical balloon has a radius of 6.55 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of a

Ksju [112]

Answer:

$M_1 = 317.7 kg$

Explanation:

Mass of the helium gas filled inside the volume of balloon is given as

$m = \rho V$

$m = 0.179(\frac{4}{3}\pi R^3)$

$m = 0.179(\frac{4}{3}\pi 6.55^3)$

$m = 210.7 kg$

now total mass of balloon + helium inside balloon is given as

$M = 210.7 + 990$

$M = 1200.7 kg$

now we know that total weight of balloon + cargo = buoyancy force on the balloon

so we will have

$(M + M_1)g = \rho_{air} V g$

$(1200.7 + M_1) = (\frac{4}{3}\pi 6.55^3) (1.29)$

$1200.7 + M_1 = 1518.4$

$M_1 = 317.7 kg$

3 0

3 years ago

Which object has the most gravitational potential energy?

Kipish [7]

Answer: An 8 kg book at a height of 3 m has the most gravitational potential energy.

Explanation:

Gravitational potential energy is the product of mass of object, height of object and gravitational field.

So, formula to calculate gravitational potential energy is as follows.

U = mgh

where,

m = mass of object

g = gravitational field = $9.81 m/s^{2}$

h = height of object

(A) m = 5 kg and h = 2m

Therefore, its gravitational potential energy is calculated as follows.

$U = mgh\\= 5 kg \times 9.81 m/s^{2} \times 2 m\\= 98.1 J (1 J = kg m^{2}/s^{2})$

(B) m = 8 kg and h = 2 m

Therefore, its gravitational potential energy is calculated as follows.

$U = mgh\\= 8 kg \times 9.81 m/s^{2} \times 2 m\\= 156.96 J (1 J = kg m^{2}/s^{2})$

(C) m = 8 kg and h = 3 m

Therefore, its gravitational potential energy is calculated as follows.

$U = mgh\\= 8 kg \times 9.81 m/s^{2} \times 3 m\\= 235.44 J (1 J = kg m^{2}/s^{2})$

(D) m = 5 kg and h = 3 m

Therefore, its gravitational potential energy is calculated as follows.

$U = mgh\\= 5 kg \times 9.81 m/s^{2} \times 3 m\\= 147.15 J (1 J = kg m^{2}/s^{2})$

Thus, we can conclude that an 8 kg book at a height of 3 m has the most gravitational potential energy.

3 0

3 years ago