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4 years ago

6

An object is in circular motion how will the object behave if the centripetal force is removed

1 answer:

Monica [59]4 years ago

8 0

If the force is removed, then there are no other forces acting on the object, the object will continue to move at constant velocity, meaning that it would move in a line tangent to the circular path.

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A 4000-? resistor is connected across a 220-V power source. What current will flow through the resistor?

Varvara68 [4.7K]

Answer:

55 mA

Explanation:

Ohm's law states:

V = IR

where V is voltage, I is current, and R is resistance.

220 V = I (4000 Ω)

I = 0.055 A

I = 55 mA

7 0

3 years ago

Given (kg/s) (m/s) , what is the resulting unit?

postnew [5]

I think you mean for those fractions to be multiplied.
In that case, you get

kg - m / sec² = Newton .

7 0

3 years ago

Instead of clear-cutting land and moving on, a local lumber business decides to start farming trees: cutting some of the trees a

Nuetrik [128]

I think your answer would be C. because they are replacing the trees they are cutting down.

5 0

3 years ago

Read 2 more answers

The specific heat of soil is 0.20 kcal/kg*C and the specific heat of water is 1.00 kcal/kg*C. This means that if 1 kg of soil an

stiv31 [10]

Answer: The soil will be $4\°C$ warmer than the water.

Explanation:

The heat (thermal energy) absorbed can be found using the following equation:

$Q=m.C.\Delta T$

Where:

$Q$ is the heat

$m$ is the mass of the element

$C$ is the specific heat capacity of the material.

$\Delta T$ is the variation in temperature

<u>In the case of soil we have:</u>

$Q_{soil}=m_{soil}.C_{soil}.\Delta T_{soil}$ (1)

Where:

$Q_{soil}=1 kcal$

$m_{soil}=1 kg$

$C_{soil}=0.2 kcal/kg \°C$

$\Delta T_{soil}$

<u>In the case of water we have:</u>

$Q_{water}=m_{water}.C_{water}.\Delta T_{water}$ (2)

Where:

$Q_{water}=1 kcal$

$m_{water}=1 kg$

$C_{water}=1 kcal/kg \°C$

$\Delta T_{water}$

Isolating $\Delta T$ from both equations:

$\Delta T_{soil}=\frac{Q_{soil}}{m_{soil}.C_{soil}}$ (3)

$\Delta T_{soil}=\frac{1 kcal}{1 kg(0.2 kcal/kg \°C)}$

$\Delta T_{soil}=5\°C$ (4)

$\Delta T_{water}=\frac{Q_{water}}{m_{water}.C_{water}}$ (5)

$\Delta T_{water}=\frac{1 kcal}{1 kg(1 kcal/kg \°C)}$

$\Delta T_{water}=1\°C$ (6)

Comparing (4) and (6) we can find the soil will be $4\°C$ warmer than the water.

8 0

4 years ago

A current loop ABCDA consists of a metal rod. a resistor R, and a pair of conducting rails separated by a distance d. The rod ha

Lilit [14]

Answer:

a) $\vec{d} =B-A$

b) $I=\frac{B*v*d*\sin 90 \textdegree}{R}$

c) $v \approx 0.6m/s$

Explanation:

From the question we are told that

Magnetic field strength $B=3.6T$

Distance traveled $d=7m$

Mass $m=4.7 kg$

Resistance $r=8.2 ohm$

Gravitational acceleration $g=9.8m/s^2$

$\theta =90 \textdegree$ Because of perpendicularity

a)

Generally the direction of the current will be given as

$\vec{d} =B-A$

Because it opposes increases of magnetic flux

b)

Generally the equation for induced EMF $E$ is mathematically given as

$E=B*v*d*\sin \theta$

$E=B*v*d*\sin 90 \textdegree$

Generally the equation for induced current $I$ is mathematically given as

$I=E/R$

$I=\frac{B*v*d*\sin 90 \textdegree}{R}$

c)

Generally the the equation for force F at terminal speed is mathematically given as

$F=mg$

$mg=B*I*d*\sin 90 \textdegree$

$mg=B*(\frac{B * v * d }{R}) *d$

$v=\frac{m*g*R}{B^2*D^2}$

$v=\frac{4.7*9.8*8.2}{3.6^2*7^2}$

$v=0.59475m/s$

$v \approx 0.6m/s$

7 0

3 years ago