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marusya05 [52]
3 years ago
11

A flight attendant pushes a beverage car on an aircraft. Her mass is 75 kg and the cart’s mass is 27 kg. Calculate the accelerat

ion produced when the flight attendant exerts a backward force of 155 N on the floor. All forces opposing the motion, such as friction on the cart’s wheels and air resistance, total 18 N.
Physics
1 answer:
tankabanditka [31]3 years ago
6 0

Answer:

a = 1.34 m/s²

Explanation:

given,

mass of the attendant(M) = 75 Kg

mass of cart (m)= 27 Kg

Backward force exerted on floor (F)= 155 N

frictional force on the cart (f)= 18 N

acceleration

a = \dfrac{F_{net}}{M+m}

F_{net} = F - frictional force

F_{net} = 155 N - 18 N

F_{net} = 137 N

now,

a = \dfrac{137}{75 + 27}

a = \dfrac{137}{102}

a = 1.34 m/s²

the acceleration produced when flight attendant is equal to  a = 1.34 m/s²

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A convex lens can produce a real image but not a viral image<br> a. true<br> b. false
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The answer is true a convex lens can produce a real image but not a viral image
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3 years ago
In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that
erik [133]

Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m

This means that the relation between the wavelength and the length of the string is

3\lambda/2 = L

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)

a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0

For this equation to be equal to zero, sin(59.94t) = 0. So,

59.94t = \pi\\t = \pi/59.94 = 0.0524~s

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s

4 0
3 years ago
Boy or girl and age you can still comment if its answered
Ksenya-84 [330]

I am a girl and i have a virgina and im 13 tho so dont touch me lolz mark brainly

6 0
2 years ago
The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh
Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

v^2=v^2_o-2ax

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}

Next, consider that the formula for the force is:

F=ma

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

F=(820kg)(0.66\frac{m}{s^2})=542.20N

Hence, the required force to stop the car is 542.20N

4 0
1 year ago
Help me pls its for my science class i need to show my work
Lostsunrise [7]

Answer:

P = 5880  J

Explanation:

Given that,

The mass of the block, m = 30 kg

The block is sitting at a height of 20 m.

The block will have gravitational potential energy. The formula for gravitational potential energy is given by :

P=mgh\\\\=30\times 9.8\times 20\\\\=5880\ J

So, the required potential energy is equal to 5880  J.

6 0
2 years ago
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