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3 years ago

I'm pretty sure it's density but I need another opinion.

Physics

2 answers:

Alecsey [184]3 years ago

8 0

It is density.

The formula of density is p=m/V, where p is the density, m is the mass, and V is the volume.

Therefore, if you're working out mass divided by volume, you're measuring the total amount of mass that can be contained in that volume.

bixtya [17]3 years ago

6 0

I agree with density

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Does the uphill or downhill direction matter to the speed of the marble, or is the height the only contributing variable?

USPshnik [31]

B down hill is the fastest

3 0

2 years ago

Height of cannon 5 m, initial speed of projectile 15m/s, angle of launch 0 degrees. What is the range and time in the air? Pleas

Westkost [7]

Answer:

<em>The range is 15.15 m and the time in the air is 1.01 s</em>

Explanation:

<u>Horizontal Motion</u>

When an object is thrown horizontally (with angle 0°) with a speed v from a height h, it follows a curved path ruled exclusively by gravity until it eventually hits the ground.

The range or maximum horizontal distance traveled by the object can be calculated as follows:

$\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}$

To calculate the time the object takes to hit the ground, we use the equation below:

$\displaystyle t=\sqrt{\frac{2h}{g}}$

The cannon is shot from a height of h=5 m with an initial speed of v=15 m/s. The range is calculated below:

$\displaystyle d=15\cdot\sqrt{\frac {2*5}{9.8}}=15*1.01$

d = 15.15 m

The time in the air is:

$\displaystyle t=\sqrt{\frac{2*5}{9.8}}$

t = 1.01 s

The range is 15.15 m and the time in the air is 1.01 s

8 0

2 years ago

Design a voltage divider to provide the following approximate voltages with respect to ground using a 30 V source: 8.18 V, 14.7

yarga [219]

Answer:

R₁ = 14.7 10³ Ω , R₂ = 8.18 10³ Ω , R₃ = 1.72 10³ Ω , R₄ = 5.4 10³ Ω 1/8 W resistor

Explanation:

For this exercise we must use a series circuit since the sum of the voltage on each resin is equal to the source voltage (V = 30 V)

Therefore we build a circuit with 4 resistors in series, in such a way that

V = i R

let the voltage

1st resistance

V = i R

R₁ = V / i

R₁ = 14.7 / 1 10⁻³

R₁ = 14.7 10³ Ω

power is

P = V i

P = 14.7 1 10⁻³

P = 14.7 10⁻³ W = 0.0147 W

a resistance of ⅛ W is indicated

2nd resistance

R₂ = 8.18 / 1 10⁻³

R₂ = 8.18 10³ Ω

Power

P = 8.18 1 10⁻³

P = 0.00818W

a 1/8 W resistor

3rd resistance

this resistance is calculated in such a way that

V₁ + V₂ + V₃ = 24.6

V₃ = 24.6 - V₁ -V₂

V₃ = 24.6 - 14.7 - 8.18

V₃ = 1.72 V

R₃ = 1.72 / 1 10⁻³

R₃ = 1.72 10³ Ω

power

P = Vi

P = 1.72 10⁻³

P = 0.00172 W

a resistance of ⅛ W

To obtain the voltage of 24.6 we use this three resistors together

4th resistance

The value of this resistance is calculated so that the sum of all the voltages reaches the source voltage

30 = V₁ + V₂ + V₃ + V₄

V₄ = 30 - V₁ -V₂ -V₃

V₄ = 30 -14.7 - 8.18 - 1.72

V₄ = 5.4 V

R₄ = 5.4 / 1 10⁻³

R₄ = 5.4 10³ Ω

Power

P = V i

P = 5.4 10⁻³

P = 0.0054 W

⅛ W resistance

The values of these resistance are commercially

Let's check the consumption of the circuit

R_total = R₁ + R₂ + R₃ + R₄

R_total = (14.7 + 8.18 + 1.72 + 5.4) 10³

R_total = 30 10³

the current circulating in the circuit is

i = V / R_total

i = 30/30 10³

i = 1 10⁻³ A

therefore it is within the order requirement.

for connections see attached diagram

8 0

3 years ago

If light moves from left to right along the x-axis, then the electric field line vibrates along what?? :(

sasho [114]

All you can say about it is that it then vibrates perpendicular to the x-axis. But that could be up and down parallel to the y-axis, in and out parallel to the z-axis, or some of it in every possible direction perpendicular to the x-axis. We "polarize" the light when we want to pick out only one perpendicular direction and stop all the others.

6 0

3 years ago

A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.

alex41 [277]

The magnitude of the unknown height of the projectile is determined as 16.1 m.

<h3>Magnitude of the height</h3>

The magnitude of the height of the projectile is calculated as follows;

H = u²sin²θ/2g

H = (36.6² x (sin 29)²)/(2 x 9.8)

H = 16.1 m

Thus, the magnitude of the unknown height of the projectile is determined as 16.1 m.

Learn more about height here: brainly.com/question/1739912

#SPJ1

3 0

1 year ago