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asambeis [7]
3 years ago
9

4.80-kg bucket of water is accelerated upward by a cord of negligible mass whose breaking strength is 75.0 N. If the bucket star

ts from rest, what is the minimum time required to raise the bucket a vertical distance of 12.0 m without breaking the cord?

Physics
2 answers:
ioda3 years ago
8 0

Answer:

Time,t=2.03seconds

Explanation:

Fg= 4.80×9.80= 47.04N

Maximum accelerating force= 75-47.04=27.96N

F= ma

27.96= 4.80 ×a

27.96/4.80=a

5.825m/s^2 =a

d=1/2at^2

d=12.0m

12=1/2 × 5.825 ×t^2

12 =2.913×t^2

t^2= 12/2.913

t= sqrt(4.120)

t= 2.03seconds

Finger [1]3 years ago
5 0

Explanation:

Below is an attachment containing the solution to the question.

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___________is the tendency of an object to resist change in its state of motion
Mademuasel [1]
Inertia
<span> An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.this tendency to retain its motion is referred to as Inertia </span>
5 0
3 years ago
What is the amount of thermal energy needed to make 5 kg of ice at - 10 °C to
agasfer [191]

Answer:

The amount of thermal energy needed is 15167500 joules.

Explanation:

By First Law of Thermodynamics, we see that amount of thermal energy (Q), in joules, is equal to the change in internal energy. From statement we understand that change in internal energy consisting in two latent components (U_{l,ice}, U_{l,steam}), in joules, and two sensible component (U_{s,w}), in joules, that is:

Q = U_{l,ice} + U_{s, w} + U_{s,ice} + U_{l,steam} (1)

By definitions of Sensible and Latent Heat, we expanded the formula:

Q = m\cdot (h_{f,w}+h_{v,w}+c_{ice}\cdot \Delta T_{ice}+c_{w}\cdot \Delta T_{w}) (2)

Where:

m - Mass, in kilograms.

h_{f,w} - Latent heat of fussion of water, in joules per kilogram.

h_{v,w} - Latent heat of vaporization of water, in joules per kilogram.

c_{ice} - Specific heat of ice, in joules per kilogram per degree Celsius.

c_{w} - Specific heat of water, in joules per kilogram per degree Celsius.

\Delta T_{ice} - Change in temperature of ice, measured in degrees Celsius.

\Delta T_{w} - Change in temperature of water, measured in degrees Celsius.

If we know that m = 5\,kg, h_{f,w} = 3.34\times 10^{5}\,\frac{J}{kg}, h_{v,w} = 2.26\times 10^{6}\,\frac{J}{kg}, c_{ice} = 2.090\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}, c_{w} = 4.186\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}, \Delta T_{ice} = 10\,^{\circ}C and \Delta T_{w} = 100\,^{\circ}C, then the amount of thermal energy is:

Q = 15167500\,J

The amount of thermal energy needed is 15167500 joules.

7 0
2 years ago
Find the kinetic energy of an electron whose de broglie wavelength is 34.0 nm.
dezoksy [38]
The De Broglie wavelength of the electron is
\lambda=34.0 nm=34 \cdot 10^{-9} m
And we can use De Broglie's relationship to find its momentum:
p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{34 \cdot 10^{-9} m}=1.94 \cdot 10^{-26} kg m/s

Given p=mv, with m being the electron mass and v its velocity, we can find the electron's velocity:
v= \frac{p}{m}= \frac{1.94 \cdot 10^{-26} kgm/s}{9.1 \cdot 10^{-31} kg}=  2.13 \cdot 10^4 m/s

This velocity is quite small compared to the speed of light, so the electron is non-relativistic and we can find its kinetic energy by using the non-relativistic formula:
K= \frac{1}{2}mv^2= \frac{1}{2}(9.1 \cdot 10^{-31} kg)(2.13 \cdot 10^4 m/s)^2=2.06 \cdot 10^{-22} J
3 0
3 years ago
A baseball pitcher throws the ball towards the batter at 90 mph. His bat connects with the ball for a line drive, after which th
forsale [732]

Answer:

F=-18412.9N, where the minus indicates the direction is opposite to that of the throw.

Explanation:

a)

Since MKS stands for meter-kilogram-second and we know that:

1\ hour = 3600\ seconds

1\ mile = 1600\ meters

1000g = 1kg

We can write that:

\frac{1\ hour}{3600\ seconds}=1

\frac{1600\ meters}{1\ mile}=1

\frac{1kg}{1000g}=1

These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.

So we have that:

90 mph=90 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=40m/s

110 mph=110 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=48.89m/s

145 g=145 g(\frac{1kg}{1000g})=0.145kg

b)

Newton's 2nd Law tells us that F=ma, and the definition of acceleration is a=\frac{\Delta v}{\Delta t}, so we have:

F=m\frac{\Delta v}{\Delta t}=m\frac{v_f-v_i}{t}

Taking the throw direction as the positive one, for our values we have:

F=m\frac{v_f-v_i}{t}=(0.145kg)\frac{(-48.89m/s)-(+40m/s)}{0.0007s}=-18412.9N

4 0
3 years ago
Check and double-click on the red polygon for the Henry Mountains laccolith complex in the folder labeled Problem 1. This comple
MArishka [77]

Answer: ghg

Explanation:

4 0
3 years ago
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