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2 years ago

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Sonrisa owns a 300 W television. If the total energy usage for February is 32.4 kWh, how many hours per week does Sonrisa watch

television? (Assume that the television does not use energy while it is turned off and that there are 28 days in February.)

2 answers:

Leno4ka [110]2 years ago

7 0

Answer:

27 hr per week

Explanation:

Zielflug [23.3K]2 years ago

6 0

the correct answer is 27 hours per week :) hope this helps

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A cannonball and a marble roll smoothly from rest down an incline. Is the cannonball’s(a) time to the bottom and(b) translationa

Salsk061 [2.6K]

Answer:

Explained

Explanation:

a) Cannonball and marble will reach the bottom at the same time, as acceleration due to gravity is independent of the mass of the body.

b)Translational Kinetic energy of the cannonball will be more than that of the marble as the cannonball has more mass than the marble.

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2 years ago

If a body is moving in the horizontal axis with a velocity Vx= 6m/s and in the vertical axis Vy=8m/s What is the angle Theta abo

cluponka [151]

Answer: C

Explanation: It's a lot of math.

7 0

2 years ago

A uniform stationary ladder of length L = 4.5 m and mass M = 11 kg leans against a smooth vertical wall, while its bottom legs r

Ostrovityanka [42]

Answer:

a) N = 9 Mg

, b)N_w = μ 9M

, c)

Explanation:

a) For this part we write the equations of trslacinal equilibrium

Axis y

N - Mg - 8M g = 0

N = 9 Mg

N = 9 11 9.8

N = 970.2 N

b) the force on the horizontal axis (x) som

fr -N_w = 0

fr = N_w

friction force is

fr = μ N

N_w = μ 9M

g

fr = 0.59 970.2

fr = N_w = 572,418 N

c) For this part we must use rotational equilibrium.

Στ = 0

We set a frame of reference at the bottom of the ladder and assume that the counterclockwise acceleration is positive

the weight of it is at its midpoint (L / 2)

- W L /2 cos 54 - 8M d_max cos 54+ NW L sin 54 = 0

8M d_max cos 54 = - W L / 2 cos 54 + NW L sin 54

d_max = L (-Mg 1/2 cos 54 + NW sin 54) / (8M cos 54)

d_max = L (-g / 16 + μ 9Mg / 8M tan 54)

d_max = L ( 9/8 μ g tan 54- g/16)

3 0

2 years ago

Kayla, a fitness trainer, develops an exercise that involves pulling a heavy crate across a rough surface. The exercise involves

Lisa [10]

Answer:

$a = 3.784\,\frac{m}{s^{2}}$

Explanation:

The Free Body Diagram of the system formed by the crate and the rope and the reference axis are presented below as an attached image. The equations of equilibrium are introduced hereafter:

$\Sigma F_{x} = T\cdot \cos \theta - \mu\cdot N = m\cdot a$

$\Sigma F_{y} = T\cdot \sin \theta + N - m\cdot g = 0$

After some algebraic handling, the system of equations is reduce to a sole expression:

$N = m\cdot g - T\cdot \sin \theta$

$T\cdot \cos \theta - \mu \cdot (m\cdot g - T\cdot \sin \theta) =m\cdot a$

$T\cdot (\cos\theta + \mu \cdot \sin \theta) - \mu \cdot m \cdot g = m\cdot a$

The minimum force to accelerate the crate from rest is:

$T_{min} = \frac{m\cdot \mu_{s}\cdot g}{\cos \theta + \mu_{s} \cdot \sin \theta}$

$T_{min} = \frac{(45\,kg)\cdot (0.770)\cdot (9.807\,\frac{m}{s^{2}} )}{\cos 23^{\textdegree}+(0.770)\cdot \sin 23^{\textdegree}}$

$T_{min} = 278.223\,N$

Since $T > T_{min}$ , the crate will experiment an acceleration due to the tension exerted. The acceleration of the crate is:

$a = \frac{T}{m}\cdot (\cos \theta + \mu_{k}\cdot \sin \theta)-\mu_{k}\cdot g$

$a = \frac{325\,N}{45\,kg}\cdot [\cos 23^{\textdegree}+(0.410)\cdot \sin 23^{\textdegree}]-(0.410)\cdot (9.807\,\frac{m}{s^{2}} )$

$a = 3.784\,\frac{m}{s^{2}}$

5 0

2 years ago

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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1 year ago