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Nana76 [90]
3 years ago
11

A shopper pushes a 5.32 kg grocery cart with a 12.7 N force directed at -28.7° below horizontal. A friction force of 8.33 N push

es back against the motion.
What is the acceleration of the cart?
Physics
1 answer:
atroni [7]3 years ago
3 0

Answer:

0.8214 m/s^2

Explanation:

Fnet= Fpushed - Ffriction

Fpushed = 12.7N      Ffriction = 8.33N

Fnet = 12.7N - 8.33N = 4.37N

Fnet= mass(acceleration)

Fnet = 4.37N    mass = 5.32 kg

4.37N = 5.32 kg(acceleration)

acceleration= 0.8214 m/s^2

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On the ____ scale, an increase of ten dicibels means an intensity increases of ten times.
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QuestionDetails:
sleet_krkn [62]

To solve the two parts of this problem, we will begin by considering the expressions given for gravitational potential energy and finally kinetic energy (to find velocity). From the potential energy we will obtain its derivative that is equivalent to the Force of gravitational attraction. We will start considering that all the points on the ring are same distance:

r = \sqrt{x^2+R^2}

Then the potential energy is

U = \frac{-GMm}{\sqrt{x^2+R^2}}

PART A) The force is excepted to be along x-axis.

Therefore we take a derivative of U with respect to x.

F = -\frac{dU}{dx}

F = -\frac{d}{dx}(GMm(\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}}))

F = \frac{GMmx}{(x^2+R^2)^{3/2}}

This expression is the resultant magnitude of the Force F.

PART B) The magnitude of loss in potential energy as the particle falls to the center

U = GMm(\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}})

According to conservation of energy,

\frac{1}{2}mv^2 = GMm (\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}})

\therefore v = \sqrt{2GM(\frac{1}{R}-\frac{1}{x^2+R^2})}

7 0
3 years ago
A simple pendulum has a period of 2.5 s. What is its period if its length is increased by a factor of four?
Svetach [21]

Answer:

Its period if its length is increased by a factor of four is 5 s.

Explanation:

The period of a simple pendulum is given by;

T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2\pi} = \sqrt{\frac{l}{g} } \\\\ \frac{T^2}{4\pi^2} = \frac{l}{g}\\\\\frac{T^2}{l} = \frac{4\pi^2}{g} \\\\let \ \frac{4\pi^2}{g}  \ be \ constant \\\\\frac{T_1^2}{l_1}  = \frac{T_2^2}{l_2} \\\\

Given;

initial period, T₁ = 2.5

initial length, = L₁

new length, L₂ = 4L₁

the new period, T₂ = ?

\frac{T_1^2}{l_1}  = \frac{T_2^2}{l_2} \\\\T_2^2 = \frac{T_1^2 l_2}{l_1} \\\\T_2 = \sqrt{\frac{T_1^2 l_2}{l_1}} \\\\  T_2 = \sqrt{\frac{(2.5)^2 \ \times \ 4l_1}{l_1}}\\\\  T_2 =\sqrt{(2.5)^2 \ \times \ 4}\\\\T_2 = \sqrt{25} \\\\T_2 = 5\ s

Therefore, its period if its length is increased by a factor of four is 5 s.

5 0
2 years ago
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