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3 years ago

7

An imaginary dense star z has 1/5 the radius of earth, but 1000 times the earths. how much would a mass weighing 1.0n on earth w

eigh on star z

1 answer:

andrew11 [14]3 years ago

8 0

A 1-newton mass on earth would be 1000 newtons on star Z.

Funny enough, stars like this exist~! They're called "Neutron Stars."

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Two people push on the same door from opposite sides as shown.

tangare [24]

The correct answer is
<span>c. one person exerts more force than the other so that the forces are unbalanced.

In fact, the door is initially at rest. In order to move the door, a net force different from zero should be applied, according to Newton's second law:
</span> $\sum F = ma$
<span>where the term on the left is the resultant of the forces acting on the door, m is the door mass and a its acceleration.

In order to move the door, the acceleration must be different from zero. But this means that the resultant of the forces acting on it must be different from zero: this is possible only if the forces applied by the two persons are unbalanced, i.e. one person exerts more force than the other.</span>

3 0

3 years ago

Read 2 more answers

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at

yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

$\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)$

Therefore, the work done by the worker in lifting the bucket is given as:

$W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)$

Now, plug in the values given and solve for 'W'. This gives,

$W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J$

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

7 0

3 years ago

Yamin is running 50 feet of No. 14 wire (with a cross section of 4,110 cmils) to a load that draws current of 11 amps. What appr

castortr0y [4]

Resistance per 1000 feet for gauge 14 wire is given as

R = 2.525 ohm

now if wire is of length 50 feet only then the resistance is given as

$R = \frac{2.525}{1000}\times 50$

$R = 0.126 ohm$

now if 11 A current flows through the wire then the voltage drop is given by ohm's law

$V = iR$

$V = 11 \times 0.126$

$V = 1.4 Volts$

so most appropriate answer in given options is

A. 1.8 Volts

8 0

3 years ago

A tank contains 350 liters of fluid in which 10 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

aleksandr82 [10.1K]

Answer:

$A(t) = -340e^{-t/70} + 350$

Explanation:

Since fluid is pumping in and out at the same rate (5L/min), the total fluid volume in the tank stays constant at 350L. Only the amount of salt and its concentration changed overtime.

Let A(t) be the amount of salt (g) at time t and C(t) (g/L) be the concentration at time t

A(0) = 10 g

Brine with concentration of 1g/L is pouring in at the rate of 5L/min so the salt income rate is 5 g/min

The well-mixed solution is pouring out at the rate of 5L/min at concentration C(t) so the salt outcome rate is 5C g/min

But the concentration is total amount of salt over 350L constant volume

C = A / 350

Therefore our rate of change for salt A' is

A' = 5 - 5A/350 = 5 - A/70

This is a first-order linear ordinary differential equation and it has the form of y' = a + by. The solution of this is

$y = ce^{bt} + \frac{a}{b}$

So $A = ce^{\frac{-t}{70}} + \frac{5}{1/70} = ce^{-t/70} + 350$

with A(0) = 10

c + 350 = 10

c = 10 - 350 = -340

$A(t) = -340e^{-t/70} + 350$

4 0

3 years ago

Do objects tend to stay moving because of a force called Interia

ahrayia [7]

The first law of Newton’s law state an object in motion will stay in motion and an object at rest will stay at rest unless acted upon with an outside force. Other know as the law of inertia so yes it does

6 0

3 years ago