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2 years ago

Which of the following is an example of an object that could have a net force greater than zero acting on it?

Physics

2 answers:

mario62 [17]2 years ago

5 0

Net force is basically the force an object has when changing direction, so the answer would be D.

il63 [147K]2 years ago

3 0

An object could have a net force greater than zero acting on it is a toy car moving east at a constant velocity because of the direction it moves.

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A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is 1.90 x 105 P

worty [1.4K]

Answer:

The magnitude of the force is 34.59 N.

Explanation:

Given that,

Inside pressure $P_{in}= 1.90\times10^{5}\ Pa$

Area $A=3.90\times10^{-4}\ m^2$

Outside pressure = 1 atm

We need to calculate the magnitude of the force

Using formula of force

$F_{net}=F_{in}-F_{out}$

$F_{net}=(P_{in}-P_{out})A$

Where, $P_{in}$ =inside Pressure

$P_{out}$ =outside Pressure

A = area

Put the value into the formula

$F_{net}=(1.90\times10^{5}-1.013\times10^{5})3.90\times10^{-4}$

$F_{net}=34.59\ N$

Hence, The magnitude of the force is 34.59 N.

6 0

2 years ago

If the distance of a galaxy is 2,000 Mpc, how many years back into the past are we looking when we observe this galaxy

ruslelena [56]

The age of the galaxy when we look back is 13.97 billion years.

The given parameters:

distance of the galaxy, x = 2,000 Mpc

According Hubble's law the age of the universe is calculated as follows;

v = H₀x

where;

H₀ = 70 km/s/Mpc

$T = \frac{x}{V} \\\\T = \frac{x}{xH_0} \\\\T = \frac{1}{H_0} \\\\T = \frac{1}{70 \ km/s/Mpc} \\\\T = \frac{1 \ sec}{70 \times 3.24 \times 10^{-20} } \\\\T = 4.41 \times 10^{17} \ sec\\\\T = \frac{4.41 \times 10^{17} \ sec\ \times \ years}{3600 \ s \ \times\ 24\ h\ \times \ 365.25 \ days} \\\\T = 1.397 \times 10^{10} \ years\\\\T = 13.97 \ billion \ years$

Thus, the age of the galaxy when we look back is 13.97 billion years.

Learn more about Hubble's law here: brainly.com/question/19819028

8 0

2 years ago

a motorcycle is capable of accelerating at 5.1 m/s starting from rest how far can it travell in 1.5 sec

Katarina [22]

The answer is 21m because the motion is in one dimension with constant acceleration.

The initial velocity is 0, because it started from rest, the acceleration ax is 4.7ms2, and the time t is 3.0s

Plugging in our known values, we have

Δx=(0)(3.0s)+12(4.7ms2)(3.0s)2=21m

5 0

2 years ago

Which two formulas are used to calculate potential and kinetic energy

goblinko [34]

Answer:

gravitational potential energy:

GPE = m g h

kinetic energy:

KE = 1/2 m v^2

6 0

2 years ago

Read 2 more answers

Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is r

Naddik [55]

Explanation:

Fgravity = G*(mass1*mass2)/D²

G is the gravitational constant throughout the universe.

D is the distance between both objects.

D is now reduced by a factor of 5, meaning Dnew = D/5 we get

Fgravitynew = G*(mass1*mass2)/(D/5)² =

= G*(mass1*mass2)/(D²/25) =

= 25* G*(mass1*mass2)/D² = 25* Fgravity

the new force of gravity/attraction is 25×16 = 400 units.

4 0

2 years ago