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2 years ago

A blue puck has a velocity of 3i –4j m/s. Its mass is 20 kg. What is its momentum?

Physics

2 answers:

damaskus [11]2 years ago

6 0

P = m * v
v = {3i - 4j} = square root (3^2 + 4^2) = 5
P = 20 * 5
P = 100 kg m/s

Tom [10]2 years ago

3 0

Magnitude:

√(3² + 4²) = √(9 + 16) = √25 = 5

Momentum = (mass) x (velocity) = (20kg x 5m/s) = 100 kg-m/s

Direction:

3i - 4j is in the direction tan⁻¹(-4/3) = 53.1 south of east.

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a steel block has a mass of 40g.it is in the form of a cube. each edge is 1.74cm long. calculate the density

Vinil7 [7]

Answer:

d ≈ 7,6 g/cm³

Explanation:

d = m/V = 40g/5,27cm³ ≈ 7,6 g/cm³

V = l³ = (1.74cm)³ ≈ 5,27 cm³

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2 years ago

Which of the following is not a factor in whether a reaction will spontaneously occur? A. Entropy change of the system B. Enthal

Delvig [45]

Answer:

Explanation:

pressure change have nothing to do with the spontaneity.

Entropy change , enthalpy change , temperature have roles in deciding spontaneity.

6 0

2 years ago

An object on a number line moved from x = 15 cm to x = 165 cm and then

olasank [31]

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

$v_{avg}=\frac{x_{all}}{t}$

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm

Then, x_all = 150cm + 140cm = 290cm

The average velocity is, for t = 100s

$v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}$

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s

3 0

2 years ago

Read 2 more answers

A 3kW oven supplied with 9mJ of energy.How many minutes can it run for?

andreev551 [17]

Hello!

We have the following data:

Time (T) = ? (in minutes)
Power (P) = 3 kW → 3000 W
Energy (E) = 9 MJ → 9000000 J or (W/s)

Formula of the consumption of electric energy:

$P = \frac{E}{T}$

Solving:

$P = \frac{E}{T}$

$P = \frac{E}{T} \to T = \frac{E}{P}$

$T = \frac{9000\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W/s}{3\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W}$

$\boxed{T = 3000\:seconds}$

How many minutes can it run for? (Let's convert in minutes)

1 minute --------- 60 seconds
y minute --------- 3000 seconds

$\frac{1}{y} = \frac{60}{3000}$

Product of extremes equals product of means

$60*y = 1*3000$

$60y = 3000$

$y = \frac{3000}{60}$

$\boxed{\boxed{y = 50\:minutes}}\end{array}}\qquad\quad\checkmark$

I hope this helps! =)

7 0

2 years ago

Read 2 more answers

A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th

zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(d) K = 11.8835 J

U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40 ⇒ f = 8.40 / (2π)

⇒ f = 1.3368 Hz

we use the formula

E = m*ω²*A² / 2

⇒ E = (1.15)(8.40)²(0.650)² / 2

⇒ E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²) (the kinetic energy)

and

U = (1/2)*m*ω²*x² (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒ K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒ U = 5.2581 J

4 0

2 years ago