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2 years ago

According to Newton’s law of universal gravitation, which statements are true?

Physics

2 answers:

kvasek [131]2 years ago

8 0

Answer:

-As we gain mass, the force of gravity on us increases.

-As we move to a higher altitudes, the force of gravity on us decreases.

Explanation:

I just took the test! :)

Arte-miy333 [17]2 years ago

7 0

Answer: The statement first and the fourth statement are true.

Explanation:

According to Newton's gravitational law, every particle in the universe attracts every other particle with the force of attraction between the masses is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

As we move to higher altitude, the force of gravity on use decreases because the force of gravity is inversely proportional to the distance.

If the masses of the two objects are more then there will be greater force of gravity between them.

Therefore, the statement first and the fourth statement are true.

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An 80 kg skateboarder moving at 3 m/s pushes off with her back foot to move faster. If her velocity increases to 5 m/s, what is

Arturiano [62]

1) The kinetic energy of an object is given by:
$K= \frac{1}{2}mv^2$
where m is the object's mass and v its speed.

By using this equation, we find the initial kinetic energy of the skateboarder:
$K_i= \frac{1}{2}(80 kg)(3 m/s)^2=360 J$
and the final kinetic energy as well:
$K_f= \frac{1}{2}(80 kg)(5 m/s)^2=1000 J$

So, her change in kinetic energy is
$\Delta K=K_f-K_i=1000 J-360 J=640 J$

2) The work-energy theorem states that the work done to increase the speed of an object is equal to the variation of kinetic energy of the object:
$W=\Delta K$
Therefore, the work done by the skateboarder is
$W=\Delta K=640 J$

7 0

2 years ago

Read 2 more answers

Steam enters a counterflow heat exchanger operating at steady state at 0.07 MPa with a specific enthalpy of 2431.6 kJ/kg and exi

PIT_PIT [208]

Answer:

Explanation:

Given:

Steam Mass rate, ms = 1.5 kg/min

= 1.5 kg/min × 1 min/60 sec

= 0.025 kg/s

Air Mass rate, ma = 100 kg/min

= 100 kg/min × 1 min/60 sec

= 1.67 kg/s

Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.

xf, quality = 0.9.

Tsat = 89.9°C

hf = 376.57 kJ/kg

hfg = 2283.38 kJ/kg

Using the equation for specific enthalpy,

hi = hf + (hfg × xf)

= 376.57 + (2283.38 × 0.9)

= 2431.552 kJ/kg

The specific enthalpy of the outlet, h2 = hf

= 376.57 kJ/kg

Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy

= ms × (hi - h2)

= 0.025 × (2431.552 - 376.57)

= 0.025 × 2055.042

= 51.37455 kW

= 51.38 kW.

5 0

2 years ago

A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .

nevsk [136]

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

v is image distance

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm$

Magnification,

$m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3$

So, the magnification of the lens is 3.

3 0

2 years ago

A cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position x

mash [69]

Answer:

The acceleration of the cart is 1.0 m\s^2 in the negative direction.

Explanation:

Using the equation of motion:

Vf^2 = Vi^2 + 2*a*x

2*a*x = Vf^2 - Vi^2

a = (Vf^2 - Vi^2)/ 2*x

Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.

Let x = Xf -Xi

Where Xf is the final position of the cart and Xi the initial position of the cart.

x = 12.5 - 0

x = 12.5

The cart comes to a stop before changing direction

Vf = 0 m/s

a = (0^2 - 5^2)/ 2*12.5

a = - 1 m/s^2

The cart is decelerating

Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.

5 0

2 years ago

A 35-g block of ice at -14°C is dropped into a calorimeter (of negligible heat capacity) containing 400 g of water at 0°C. When

kompoz [17]

Answer:

Total mass of ice = 38.06g

Explanation:

Since the heat capacity of calorimeter is negligible.

The water is already at 0°C, so the heat loss can no longer reduce the temperature of the water. It is used for fusion and forming more ice.

The equilibrium temperature will be 0°C, because the heat gain by ice is only enough to bring it down to 0°C.

Heat gained by ice = heat loss by water

Heat gained by ice (from -14°C to 0°C) = heat lost to fusion by water (heat of fusion of some amount of the water present in the calorimeter)

mi Ci ∆Ti = mw . L ......1

Where;

mi = mass of ice = 35g = 0.035 kg

Ci = specific heat capacity of ice = 2090 J/kg ∙ K

∆Ti = change in temperature of ice = 0-(-14) = 14 K

mw = the mass of water that have gained enough heat for fusion ( mass of water converted to ice)

L = latent heat of fusion of water = 33.5 × 10^4 J/kg.

From equation 1;

mw = (mi Ci ∆Ti )/L

mw = (0.035×2090×14)/335000

mw = 0.00306 kg

mw = 3.06 g

Therefore, 3.06 g of water has been converted to ice.

When combined with the initial amount of ice initially in the calorimeter (at 0°C)

Total mass of ice = mi + 3.06g = 35g + 3.06g = 38.06g

Total mass of ice = 38.06g

6 0

2 years ago

Read 2 more answers