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2 years ago

Um, I need help finding objects- Name some for me, please :(

Physics

2 answers:

Kryger [21]2 years ago

8 0

Answer:

Every thing around you right now? is an object, your computer is an object also

Explanation:

avanturin [10]2 years ago

4 0

Answer:

table, book, water bottle, bed, vase, statue, etc.

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Starting at 9a.m., you ride your hover board for 3hrs at an average speed of 6 mph. Out of breath, you stop for tea from noon un

Elena-2011 [213]

3 times 6= 18. The average speed is 19 mph.

hope this helps!

6 0

2 years ago

A flying saucer lifts the Physical Science building 10,000 ft into the air before discovering it is useless and discards the rem

scZoUnD [109]

Answer:

500000000 lbft/s

Explanation:

F = Force or weight = 1000000 lbf

s = Displacement = 10000 ft

t = Time taken = 20 seconds

Work done is given by

$W=Fs\\\Rightarrow W=1000000\times 10000\\\Rightarrow W=10000000000\ lb-ft$

Power is given by

$P=\dfrac{W}{t}\\\Rightarrow P=\dfrac{10000000000}{20}\\\Rightarrow P=500000000\ lbft/s$

One Saucer power is 500000000 lbft/s

7 0

3 years ago

Which of the following is NOT an example of a correct interaction between the

makkiz [27]

Answer:

WE NEED ANSWER CHOICES

Explanation:

3 0

3 years ago

A passenger in a helicopter traveling upwards at 15 m/s accidentally drops a package out the window. If it takes 15 seconds to r

Alexeev081 [22]

Answer:

The helicopter was 1103.63 meters high when the package was dropped.

Explanation:

We consider positive speed as a downward movement

y: height (m)

t: time (s)

v₀: initial speed (m/s)

Δy = v₀t + $\frac{1}{2}$ gt²

Δy= 15 $\frac{m}{s}$ ×15 s + $\frac{1}{2}$ ×9.81 $\frac{m}{s^{2} }$ ×(15 s)²

Δy= 1103.63 m

8 0

3 years ago

a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit

ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

$W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)$

We have

$\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}$

$\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}$

Then the total work is

$W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right) \approx \boxed{36.8\,\rm J}$

5 0

2 years ago