PHYSICS COLLEGE

A student lifts their 75 N backpack 0.50 m onto their chair. How much work is done?

Answers

Answer 1

Answer:

Answer:

37.5 J

Explanation:

With work done equation: W=Fs

W=75*0.50=37.5 J

or use mgh=(75)(0.5) which is the same

Related Questions

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For a certain transverse wave, the distance between two successive crests is 1.20 m, and eight crests pass a given point along the direction of travel every 12.0 s. calculate the wave speed.
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A periodic wave travels from one medium to another. Which pair of variables are likely to change in the process? A. velocity and wavelength B. velocity and frequency C. frequency and wavelength D. frequency and phase E. wavelength and phase

The left ventricle of a resting adult's heart pumps blood at a flow rate of 85.0 cm3/s, increasing its pressure by 110 mm Hg, its velocity from zero to 25.0 cm/s, and its height by 5.00 cm. (All numbers are averaged over the entire heartbeat.) Calculate the total power output (in W) of the left ventricle. Note that most of the power is used to increase blood pressure.

Answers

Answer:

P = 1.29625 W

Explanation:

Given

Q = 85.0 cm ³ / s, p₁ = 110 mmHg, u₁ = 25.0 cm / s, h = 5.0 cm

Also knowing the density of the blood is

ρₐ = 1.05 x 10 ³ kg / m³

Δp₁ = 110 mmHg * 133.322 Pa / 1 mmHg

Q = 85.0 cm³ / s = 85.0 x 10 ⁻⁶ m³ / s

To calculated the power

P = H * Q

H = Δp₁ + ¹/₂ * ρₐ * ( u₁² - v₂²) + ρₐ * g *Δh

H = 14.666 x 10 ³ Pa + 0.5 * 1.05 x 10 ³ kg / m³ * ( 25 x 10 ² m /s )² + 1.05 x 10 ³ kg / m³ * 9.8 m /s² * 0.05 m

H = 15.25 x 10 ³ Pa

P = 15.25 x 10 ³ Pa * 85.0 x 10 ⁻⁶ m³ / s

P = 1.29625 W

wo parallel plates of area 100cm2are given charges of equal magnitudes 8.9 ×10−7C but opposite signs. The electric field within the dielectric material filling the space between the plates is 1.4 ×106V/m. (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.

Answers

(a) 7.18

The electric field within a parallel plate capacitor with dielectric is given by:

$E=(\sigma)/(k \epsilon_0)$ (1)

where

$\sigma$ is the surface charge density

k is the dielectric constant

$\epsilon_0$ is the vacuum permittivity

The area of the plates in this capacitor is

$A=100 cm^2 = 100\cdot 10^(-4) m^2$

while the charge is

$Q=8.9\cdot 10^(-7)C$

So the surface charge density is

$\sigma = (Q)/(A)=(8.9\cdot 10^(-7) C)/(100\cdot 10^(-4) m^2)=8.9\cdot 10^(-5) C/m^2$

The electric field is

$E=1.4\cdot 10^6 V/m$

So we can re-arrange eq.(1) to find k:

$k=(\sigma)/(E \epsilon_0)=(8.9\cdot 10^(-5) C/m^2)/((1.4\cdot 10^6 V/m)(8.85\cdot 10^(-12) F/m))=7.18$

(b) $7.66\cdot 10^(-7)C$

The surface charge density induced on each dielectric surface is given by

$\sigma' = \sigma (1-(1)/(k))$

where

$\sigma=8.9\cdot 10^(-5) C/m^2$ is the initial charge density

k = 7.18 is the dielectric constant

Substituting,

$\sigma' = (8.9\cdot 10^(-5) C/m^2) (1-(1)/(7.18))=7.66\cdot 10^(5) C/m^2$

And by multiplying by the area, we find the charge induced on each surface:

$Q' = \sigma' A = (7.66\cdot 10^(-5) C/m^2)(100 \cdot 10^(-4)m^2)=7.66\cdot 10^(-7)C$

Nasa's skylab, the largest spacecraft ever to fall back to the earth, reentered the earth's atmosphere on july 11, 1979, and broke into a myriad of pieces. one of the largest fragments was a 1770-kg, lead-lined film vault, which landed with an estimated speed of 120 m/s.

Answers

All of that is fascinating information. Thank you for sharing.

A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air?
(b) What must have been the initial horizontal component of the velocity?
(c) What is the vertical component of the velocity just before the ball hits the ground?
(d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

Answers

Answer:

Explanation:

Given

height of building (h)=60 m

Range of ball=100 m

(a)time travel to cover a vertical distance of 60 m

$h=ut+(at^2)/(2)$

$60=0+(9.8* t^2)/(2)$

$t^2=12.24$

t=3.49 s

(b)To cover a range of 100 m

R=ut

$100=v_x* 3.49$

$v_x=28.57 m/s$

(c)vertical component of velocity just before it hits the ground

$v_y=u+at$

$v_y=0+9.81* 3.49=34.202 m/s$

(d) $v_(net)=√(v_y^2+v_x^2)$

$v_(net)=√(1986.02)=44.56 m/s$

Final answer:

The ball is in the air for about 3.5 seconds. The initial horizontal velocity would have been approximately 28.6 m/s. The vertical component of the velocity just before the ball hits the ground is nearly 34.3 m/s. The overall velocity of the ball just prior to impact is roughly 44.6 m/s.

Explanation:

The problem given is about projectile motion which can be approached by splitting the motion into the horizontal and vertical components. We can work out the durations for each.

For the time the ball is in the air, we know that it falls vertically under gravity. Use the equation of motion, h = 0.5gt^2 where h is the height (60.0 m) and g is the acceleration due to gravity (approx 9.81 m/s^2). Solving for t takes approximately 3.5 seconds.
The initial horizontal component of the velocity can be calculated by the distance it traveled horizontally divided by the time it spent in the air. So, 100 m / 3.5 s = 28.6 m/s.
The vertical component of velocity just before the ball hits the ground can be calculated using v = gt where g is the acceleration due to gravity and t is the time. Solving gives around 34.3 m/s.
The overall velocity can then be calculated using the Pythagorean theorem, combining the horizontal and vertical components of the velocity - sqrt((28.6 m/s)^2 + (34.3 m/s)^2). This will approximately equal 44.6 m/s.

Learn more about Projectile Motion here:

brainly.com/question/29545516

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Find the force on a 5 pС charge in a place where the electric field is 400 N/C.

Answers

Answer:

Electric force, $F=2* 10^(-9)\ N$

Explanation:

It is given that,

Charge on the particle, $q=5\ pC=5* 10^(-12)\ C$

Electric field, $E=400\ N/C$

Let F is the electric force acting on the charged particle. The electric force per unit electric charge is called electric field. Mathematically, it is given by :

$F=qE$

$F=5* 10^(-12)\ C* 400\ N/C$

$F=2* 10^(-9)\ N$

So, the force acting on the charged particle is $2* 10^(-9)\ N$ . Hence, this is the required solution.

On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s meets the runner in a head-on collision. If the two players stick together, a) what is their velocity immediately after collision? b) What is the kinetic energy of the system just before the collision and a moment after the collision?