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2 years ago

What is the potential energy of a 0.5 kg object sitting on a shelf that is 1.5 meters high?

Physics

2 answers:

snow_lady [41]2 years ago

8 0

Answer:

what is the potential energy of a 5 kg apple that is sitting on a 1.5 m high tree branch

Explanation:

Archy [21]2 years ago

5 0

Gravitational potential energy=mass x height X gravitational field strength
=0.5 x1.5 x9.7(gravity)=7.275J

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Any four function of health post

lesantik [10]

Answer:

1. be the first line of treatment for minor health conditions

2.be the first line of assessment and decision making for further diagnosis and/or treatment and for referral to a higher level facility

3. be a center for all public health activities, such as outreach ...

4. provide basic health services to people who live in rural areas.

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2 years ago

Can someone please answer this, ill give you brainliest and your getting 100 points.

Nastasia [14]

Explanation:

Science says that youngest rock can be found in between ocean .

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Option C is correct

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2 years ago

Read 2 more answers

Chi walked 100 meters in 5 minutes. She took another 15 minutes to walk another 400 meters to reach her house. What is her avera

Law Incorporation [45]

Average speed is the ratio of total distance moved by Chi in total time interval

So here we will have

Total distance = 100 m + 400 m

$d = 500 m = 0.5 km$

Total time taken = 5 min + 15 min = 20 min

$T = \frac{20}{60} = \frac{1}{3} hour$

now by the formula of average speed we know that

$v_{avg} = \frac{d}{T}$

$v_{avg} = \frac{0.5 km}{1/3 h}$

$v_{avg} = 1.5 km/h$

so average speed will be 1.5 km/h

4 0

3 years ago

A neutron consists of one "up" quark of charge +2e/3 and two "down" quarks each having charge -e/3. If we assume that the down q

Anon25 [30]

Answer:

The magnitude of the electrostatic force is 120.85 N

Explanation:

We can use Coulomb's law to find the electrostatic force between the down quarks.

In scalar form, Coulomb's law states that for charges $q_1$ and $q_2$ separated by a distance d, the magnitude of the electrostatic force F between them is:

$F = k \frac{|q_1q_2|}{d^2}$

where $k$ is Coulomb's constant.

Taking the values:

$d = 4.6 \ 10^{-15} m$

$q_1 = q_2 = - \frac{e}{3} = - \frac{1.6 \ 10^{-19} \ C}{3}$

and knowing the value of the Coulomb's constant:

$k = 8.99 \ 10 ^{9} \frac{N m^2}{C^2}$

Taking all this in consideration:

$F = 8.99 \ 10 ^{9} \frac{N m^2}{C^2} \frac{ (- \frac{1.6 \ 10^{-19} \ C}{3} ) ^2}{(4.6 \ 10^{-15} m)^2}$

$F = 120.85 \ N$

8 0

3 years ago

A rifle bullet with mass 8.00g strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless, horizontal s

mafiozo [28]

Answer:

block velocity v = 0.09186 = 9.18 10⁻² m/s and speed bollet v₀ = 11.5 m / s

Explanation:

We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.

Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)

Before the crash

p₀ = m v₀ + 0

After the crash

$p_{f}$ = (m + M) v

p₀ = $p_{f}$

m v₀ = (m + M) v (1)

Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring

Initial

Em₀ = K = ½ m v2

Final

E $m_{f}$ = Ke = ½ k x2

Emo = E $m_{f}$

½ m v² = ½ k x²

v² = k/m x²

Let's look for the spring constant (k), with Hook's law

F = -k x

k = -F / x

k = - 0.75 / -0.25

k = 3 N / m

Let's calculate the speed

v = √(k/m) x

v = √ (3/8.00) 0.15

v = 0.09186 = 9.18 10⁻² m/s

This is the spped of the block plus bullet rsystem right after the crash

We substitute calculate in equation (1)

m v₀ = (m + M) v

v₀ = v (m + M) / m

v₀ = 0.09186 (0.008 + 0.992) /0.008

v₀ = 11.5 m / s

6 0

2 years ago