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3 years ago

5

The strength of the gravitational pull between two objects is determined by what

2 answers:

Dvinal [7]3 years ago

4 0

by mass and distance.

Annette [7]3 years ago

4 0

The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases.

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PEDALING A BIKE : ACCELERATION:: PULLING A DOGS LEASH: _____.

Alex_Xolod [135]

Pulling a dogs leash: inertia

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4 years ago

A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that

Virty [35]

Answer:

$\boxed {\boxed {\sf 29,400 \ Joules}}$

Explanation:

Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.

$E_P= m \times g \times h$

The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.

m= 150 kg
g= 9.8 m/s²
h= 20 m

Substitute the values into the formula.

$E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m$

Multiply the three numbers and their units together.

$E_p=1470 \ kg*m/s^2 \times 20 m$

$E_p=29400 \ kg*m^2/s^2$

Convert the units.

1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.

$E_p= 29,400 \ J$

The crate has <u>29,400 Joules</u> of potential energy.

7 0

3 years ago

Explain the following (using the concepts of moment, center of gravity and stability.) a) Racing cars are low, with wide wheels.

LUCKY_DIMON [66]

Answer:

b

Explanation:

3 0

3 years ago

Heat is extracted from a certain quantity of steam at

vodomira [7]

Answer: $v=2452.91 m/s$

Explanation:

Given

initially steam is at $100^{\circ}C$ and converted to $0^{\circ} C$ ice

Let m be the mass of steam

latent heat of fusion and vaporization for water is

$L_f=3.33\times 10^5 J/kg$

$L_v=2.26\times 10^6 J/kg$

Heat required to convert steam in to water at $100^{\circ}C$

$Q_1=m\times L_v=m\cdot 2.26\times 10^6 J$

Heat required to lower water temperature to $0^{\circ}C$

$Q_2=m\times c\times \Delta T$

$Q_2=m\times 4.184\times (100)$

$Q_2=4.184m\times 10^5 J$

Heat required to convert $0^{\circ}C$ water to ice at $0^{\circ}C$ is

$Q_3=m\times L_f$

$Q_3=m\times 3.33\times 10^5=3.33m\times 10^5 J$

$Q=Q_1+Q_2+Q_3$

$Q=(2.26+0.4184+0.33)m\times 10^6 J$

$Q=3.0084m\times 10^6 J$

So this energy is equal to kinetic energy of bullet of mass m moving with velocity v

$Q=\frac{1}{2}mv^2$

$3.0084m\times 10^6=\frac{1}{2}mv^2$

$v^2=3.0084\times 2\times 10^6$

$v=2.452\times 10^3 m/s$

$v=2452.91 m/s$

5 0

4 years ago

If you walk 3 meters north and then 4 meters west, what is the magnitude of your displacement?

ankoles [38]

S^2=3^2+4^2
S^2=9+16
S^2=25
S=√25
S=5m

7 0

3 years ago