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3 years ago

13

A proton in the nucleus of an atom has an electrical charge of: neutral - + zero

1 answer:

dlinn [17]3 years ago

6 0

Answer:

proton is positively charged changechar

Explanation:

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two objects are in uniform circular motion at the same speed but at different radii. the ones with the __ radius has the largest

Cerrena [4.2K]

Answer:the one with the smaller radius has the highest centripetal force

Explanation:

5 0

3 years ago

Read 2 more answers

In a hydraulic lift, if the pressure exerted on the liquid by one piston is increased by 100 N/m2 , how much additional weight c

shepuryov [24]

Answer:

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

Explanation:

By means of the Pascal's Principle, the hydraulic lift can be modelled by the following two equations:

Hydraulic Lift - Before change

$P = \frac{F}{A}$

Hydraulic Lift - After change

$P + \Delta P = \frac{F + \Delta F}{A}$

Where:

$P$ - Hydrostatic pressure, measured in pascals.

$\Delta P$ - Change in hydrostatic pressure, measured in pascals.

$A$ - Cross sectional area of the hydraulic lift, measured in square meters.

$F$ - Hydrostatic force, measured in newtons.

$\Delta F$ - Change in hydrostatic force, measured in newtons.

The additional weight is obtained after some algebraic handling and the replacing of all inputs:

$\frac{F}{A} + \Delta P = \frac{F}{A} + \frac{\Delta F}{A}$

$\Delta P = \frac{\Delta F}{A}$

$\Delta F = A\cdot \Delta P$

Given that $\Delta P = 100\,Pa$ and $A = 25\,m^{2}$ , the additional weight is:

$\Delta F = (25\,m^{2})\cdot (100\,Pa)$

$\Delta F = 2500\,N$

The additional mass needed for the additional weight is:

$\Delta m = \frac{\Delta F}{g}$

Where:

$\Delta F$ - Additional weight, measured in newtons.

$\Delta m$ - Additional mass, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

If $\Delta F = 2500\,N$ and $g = 9.807\,\frac{m}{s^{2}}$ , then:

$\Delta m = \frac{2500\,N}{9.807\,\frac{m}{s^{2}} }$

$\Delta m = 254.92\,kg$

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

3 0

3 years ago

An all-electric car (not a hybrid) is designed to run from a bank of 12.0 V batteries with total energy storage of 2.30 ✕ 107 J.

AnnyKZ [126]

Answer:

a) $I=733.33\ A$

b) $d=52272.7273\ m$

c) $d'=51948.0519\ m$

Explanation:

Given:

voltage of the battery, $V=12\ V$

energy storage capacity of the battery, $E=2.3\times 10^7\ J$

speed of the car, $v=20\ m.s^{-1}$

a)

power drawn by the car, $P=8.8\ kW$

<u>Now the Current delivered to the motor:</u>

we the relation between the power and electrical current,

$P=V.I$

$8800=12\times I$

$I=733.33\ A$

b)

<u>Distance travelled before battery is out of juice:</u>

we first find the time before the battery runs out,

$t=\frac{E}{P}$

$t=\frac{2.3\times 10^7}{8800}$

$t=2613.636\ s$

Now the distance:

$d=v.t$

$d=20\times 2613.636$

$d=52272.7273\ m$

c)

When the head light of 55 W power is kept on while moving then the power consumption of the car is:

$P'=P+55$

$P'=8800+55$

$P'=8855\ W$

<u>Now the time of operation of the car:</u>

$t'=\frac{E}{P'}$

$t'=\frac{2.3\times 10^7}{8855}$

$t'=2597.4026\ s$

<u>Now the distance travelled:</u>

$d'=v.t'$

$d'=20\times 2597.4025$

$d'=51948.0519\ m$

5 0

3 years ago

The acceleration of an object is ____________________ related to the net force exerted upon it and _____________________ related

Anettt [7]

The acceleration of an object is directly related to the net force exerted upon it and inversely related to the mass of the object.

8 0

2 years ago

a cannon sits on a stationary railroad flatcar with a total mass of 1000 kg. when a 10 kg cannonball is fired to the left at a s

marishachu [46]

Answer:

0.5 m/s

Explanation:

From Newton's Third law of motion,

Momentum of the cannon = momentum of the flatcar.

mv = MV........................ Equation 1

Where m = mass of the cannon, v = velocity of the cannon, M = mass of the flatcar, V = Velocity of the flat car.

make V the subject of the equation

V = mv/M................... Equation 2

Given: m = 10 kg, v = 50 m/s, M = 1000 kg

Substitute into equation 2

V = 10×50/1000

V = 500/1000

V = 0.5 m/s

Hence the speed of the flatcar = 0.5 m/s

6 0

3 years ago