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ivanzaharov [21]
3 years ago
9

What are three physical properties of a bowling ball

Physics
1 answer:
Sidana [21]3 years ago
3 0

1) round

2)colorful

3)big

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What does this picture mean?
ivann1987 [24]
<span>
be aware of large trucks on the roadway
Tip: usually, these signs are hazard signs or warning.
Also, this isnt physics </span><span />
5 0
2 years ago
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If we heated the ball up and kept the ring room temperature, would the ball be able to fit through the ring? (1 point) Why or wh
Fed [463]

Answer:

no:

Explanation:

it would grow and no longer be able to fit through the loop due to the hot air expanding.

8 0
2 years ago
A spacecraft left Earth to collect soil samples from Mars. Which statement is true about the strength of Earth’s gravity on the
ryzh [129]

Answer:

: It Decreases.

As the spacecraft gets farther and farther from Earth, the gravitational  

forces between the spacecraft and the Earth decrease.

Explanation:

5 0
3 years ago
Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6
Nana76 [90]

Answer:

The answer to the question is as follows

The  acceleration due to gravity for low for orbit is  9.231 m/s²

Explanation:

The gravitational force is given as

F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }

Where F_{G} = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹\frac{Nm^{2} }{kg^{2} }

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by  F_{G} and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and  F_{G} = \frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} } = 9.231 N

Therefore the acceleration due to gravity =  F_{G} /mass  

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is  9.231 m/s²

4 0
3 years ago
Read 2 more answers
An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring
ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

F = k x

mg = k x

k = mg \div x

k = 1.25(9.8) \div 0.0275

k = 445 \frac{5}{11} \texttt{ N/m}

\texttt{ }

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

Ep_1 + Ek_1 = Ep_2 + Ek_2

\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek

Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2

Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh

Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)

\boxed {Ek \approx 1.20 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0
3 years ago
Read 2 more answers
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