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4 years ago

13

A flowerpot that has a mass of 1.5 kg is sitting on a windowsill 30 meters from the ground. Is the energy of the flowerpot poten

tial of kinetic?
How much energy does the flowerpot have?

1 answer:

zzz [600]4 years ago

8 0

Answer:

It is Potential and corresponding to

90

J

.

Explanation:

Potential Energy is energy stored in the system and ready (potentially) to be changed into, say, Kinetic Energy through movement.

The general expression for Gravitational Potential Energy is:

U=mgh

where:

mg is the weight; h is the height relative to a certain reference level (the ground in this case) choosen to have zero Potential Energy.

With your data we get:

U=3⋅30=90J

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The bug has the greater acceleration because it has the smaller mass

The bug has a greater change in velocity (acceleration) due to experiencing equal force as a much smaller mass.

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2 years ago

Why are experiments often performed in laboratories?

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3 years ago

A body of mass 100g moving with a velocity of 10.0m/s collides with a wall .if after the collision it moves with a velocity of 2

ch4aika [34]

Answer:

-1.2 kg - m/s

Explanation:

$\pink{\frak{Given}}\begin{cases}\textsf{ A body of mass 100g moving with a velocity of 10.0m/s collides with a wall .}\\\textsf{ After the collision it moves with a velocity of 2.0m/s in the opposite direction.}\end{cases}$

And we need to find out the change in momentum of the body . Here ,

velocity before collision (u) = 10m/s
velocity after collision (v) = 2m/s .

We know that momentum is defined as amount of motion contained in a body . Mathematically ,

$\sf\longrightarrow momentum (p)= mass(m) * velocity(v)$

Therefore change in momentum will be,

$\sf\longrightarrow \triangle p = mv - mu$

Since the direction of velocity changes after the collision , the velocity will be -2m/s .

$\sf\longrightarrow \Delta p = 100g( -2m/s -10m/s) \\$

$\sf\longrightarrow \Delta p =\dfrac{100}{1000}kg ( -12m/s) \\$

$\sf\longrightarrow \Delta p = 0.1 kg * -12m/s \\$

$\sf\longrightarrow \boxed{\bf \Delta p = -1.2 \ kg-m/s} \\$

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2 years ago

Which would ultimately cause the armature in an electric motor to spin?

malfutka [58]

It would be the electromagnet because that is how it spins

4 0

4 years ago

Read 2 more answers

The fastest time ever run at Pikes Peak is 7:57. That’s way too long. If we removed all the curves, making a straight line from

anygoal [31]

Answer: 271.4 s

Explanation:

We are told the top speed (maximum speed) $V_{max}$ the car has is:

$V_{max}=203 mph=90.74 m/s$ taking into account $1 mile=1609.34 m$

And the car's average acceleration $a_{ave}$ is:

$a_{ave}=0.091 g=2.93 ft/s^{2}=0.89 m/s^{2}$

Since:

$a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}$ (1)

Where:

$V_{f}=V_{max}=90.74 m/s$ is the car's final speed (top speed)

$V_{o}=0 m/s$ because it starts from rest

$\Delta t$ is the time it takes to reach the top speed

Finding this time:

$\Delta t=\frac{V_{f}-V_{o}}{a_{ave}}$ (2)

$\Delta t=\frac{90.74 m/s - 0 m/s}{0.89 m/s^{2}}$ (3)

$\Delta t=t_{1}=101.95 s$ (4)

Now we have to find the distance $d$ the car traveled at this maximum speed with the following equation:

$V_{f}^{2}=V_{o}^{2} + 2a_{ave} d$ (5)

Isolating $d$ :

$d=\frac{V_{f}^{2}}{2a_{ave}}$ (6)

$d=\frac{(90.74 m/s)^{2}}{2(0.89 m/s^{2})}$ (7)

$d=4625.70 m$ (8)

On the other hand, we know the total distance $D$ traveled by the car is:

$D=12.42 miles = 19988.052 m$

Hence the remaining distance is:

$d_{remain}=D-d=19988.052 m - 4625.70 m$ (9)

$d_{remain}=15362.35 m$ (10)

So, we can calculate the time $t_{2}$ it took to this car to travel this remaining distance $d_{remain}$ at its top speed $V_{max}$ , with the following equation:

$V_{max}=\frac{d_{remain}}{t_{2}}$ (11)

Isolating $t_{2}$ :

$t_{2}=\frac{d_{remain}}{V_{max}}$ (12)

$t_{2}=\frac{15362.35 m}{90.74 m/s}$ (13)

$t_{2}=169.45 s$ (14)

With this time $t_{2}$ and the value of $t_{1}$ calculated in (4) we can finally calculate the total time $t_{TOTAL}$ :

$t_{TOTAL}=t_{1}+ t_{2}$ (15)

$t_{TOTAL}=101.95 s + 169.45 s$ (16)

$t_{TOTAL}=271.4 s s$

5 0

4 years ago