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3 years ago

As an object is raised to a higher position,Which type of energy increases?

2 answers:

8 0

As the object is raised higher, its gravitational potential energy increases which is denoted in a formulae as gpe = mgh where m is mass, g is gravitational accelaration and h is height.
Therefore height increases proportionally to gpe if mass and acceleration are kept constant!

If you need further help over this topic, I will be more than happy to help you out, just message me!

Hope this helps!

Marizza181 [45]3 years ago

7 0

By raising a position of an object, Its potential energy increases.

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Suppose that you measure a galaxy's redshift, and from the redshift you determine that its recession velocity is 30,000 (3×10^4)

anyanavicka [17]

Answer:

1.4 billion light years away

Explanation:

v = Recessional velocity = 30000 km/s[/tex]

$H_0$ = Hubble constant = $\frac{65}{3.2\times 10^6}\ ly$

D = Distance to the galaxy

According to Hubble's law

$v=H_0D\\\Rightarrow D=\frac{v}{H_0}\\\Rightarrow D=\frac{3\times 10^{4}}{65}\times 3.2\times 10^6\\\Rightarrow D=1476923076.92307\ ly\\\Rightarrow D=1.4\times 10^9\ ly$

The galaxy is 1.4 billion light years away

5 0

3 years ago

Metals are good conductors because

Bingel [31]

I think it’s A I’m not too sure sorry if I’m wrong

6 0

3 years ago

Read 2 more answers

why is it important for digestive system functions that mucous membranes can secrete and absorb susbstances?

MissTica

Answer:

It pushes food through the GI tract. The inside of GI tract is lined with mucous membranes. Mucous membranes are moist tissues that can secrete and absorb substances. The ability to secrete and absorb substances is necessary for the functions of the digestive system.

Explanation:

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6 0

4 years ago

The 3.00 kg cube in fig. 15-47 has edge lengths d 6.00 cm and is mounted on an axle through its center. a spring (k 1200 n/m con

ira [324]

The Period of the resulting shm will be T=39.7

Explanation:

Given data

m=3kg

d=.06m

k=1200 N/m

Θ=3 °

T=?

we have the formulas,

I = (1/6)Md2

F = ma

F = -kx = -(mω2x)

k = mω2 τ = -d(FgsinΘ)

T=2 x 3.14/ √(m/k)

Solution for the given problem would be,

F=-Kx (where x= dsin Θ)

F=-k dsin Θ

F=-(1200)(.06)sin(3 °)

F=-10.16N

By newton's second law.

F = ma

a= F/m

a=(-10.16N)/3

a=3.38

using the k=mω value

k=mω

ω=k/m

ω=1200/3

ω=400

Using F = -kx value

x = F/-k

x=(-10.16)/1200

x=0.00847m

Restoring the torque value

τ = -dmgsinΘ where( τ = Iα so.).. Iα = -dmgsinΘ α = -(.06)(4)α =

α =(.06)(4)(9.81)sin(4°)

α=-1.781

Rotational to linear form

a = αr

r = .1131 m

a=-1.781 x .1131 m

a=-0.2015233664

Time Period

T=2 x 3.14/ √(m/k)

T=6.28/√(3/1200)

T=6.28/0.158

T=39.7

6 0

4 years ago

A train travels 90 kilometers in 4 hours, and then 82 kilometers in 1 hours. What is its average speed?

kvasek [131]

Average speed = total distance / total time.

That's 172 km / 5 hr = 34.4 km//hr

5 0

3 years ago