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Lisa [10]
2 years ago
14

One object has twice as much mass as another object. The first object also has twice as much a velocity. b gravitational acceler

ation. c inertia. d all of the above
Physics
1 answer:
Gemiola [76]2 years ago
6 0
It would be c. As it can’t accelerate faster thus not having a faster velocity so it’s inertia
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Compute the density in g/cm^3 of a piece of metal that had mass of 0.485 kg and a volume of 52cm^3
steposvetlana [31]

Answer:

9.3 g/cm³

Explanation:

First, convert kg to g:

0.485 kg × (1000 g / kg) = 485 g

Density is mass divided by volume:

D = (485 g) / (52 cm³)

D = 9.33 g/cm³

Rounding to two significant figures, the density is 9.3 g/cm³.

8 0
3 years ago
A baby carriage is sitting at the top of a hill that is 21m high. The carriage with the baby weighs 12 n. What's the potential e
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A baby carriage is sitting at the top of a hill that is 21m high. The carriage with the baby weighs 12N. The carriage has... energy. Calculate it = <span>252J</span>
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2 years ago
During their physics field trip to the amusement park, Leslie and Maria took a ride on the Whirligig. The Whirligig ride consist
grandymaker [24]

Answer:

a) frequency = 0.1724 Hz

b) Period = 5.8 sec

c) speed = 7.04 m/s

d) acceleration = 7.62 m/s²

Explanation:

Given that;

radius = 6.5m

time period = 5.8 sec every circle

a)  the frequency

frequency is the number of rotation in unit time

frequency = 1 / time period = 1/5.8

frequency = 0.1724 Hz

b)  the period

period is time taken in one rotation

period = total time / rotation = 5.8 / 1

Period = 5.8 sec

c)  the speed

speed = distance/time = circumference/time period = 2πr / t = (2π×6.5) / 5.8

speed = 7.04 m/s

d) acceleration

To find the acceleration we take the linear velocity squared divided by the radius of the circle.

so

acceleration = v² / r = (7.04)² / 6.5 = 49.5616 / 6.5

acceleration = 7.62 m/s²

6 0
3 years ago
Select the correct answer
Rasek [7]

Answer:

I would say the net force acting on the car is in the opposite direction of the car's motion  is correct

5 0
2 years ago
Read 2 more answers
The balmer series is formed by electron transitions in hydrogen that
RSB [31]

Answer:

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The wavelength associated to each spectral line of the Balmer series is given by:

\frac{1}{\lambda}=R_H (\frac{1}{2^2}-\frac{1}{n^2})

where R_H is the Rydberg constant for hydrogen, and where n is the initial level of the electron that jumps to the level n = 2.

The first few spectral lines associated to this series are withing the visible part of the electromagnetic spectrum, and their wavelengths are:

656 nm (red, corresponding to the transition 3 \rightarrow 2)

486 nm (green, 4 \rightarrow 2)

434 nm (blue, 5 \rightarrow 2)

410 nm (violet, 6 \rightarrow 2)

All the following lines lie in the ultraviolet part of the spectrum. The limit of the Balmer series, corresponding to the transition \infty \rightarrow 2, is at 364.6 nm.

4 0
3 years ago
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