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Anvisha [2.4K]
3 years ago
11

A body has translatory motion if it moves along a

Physics
1 answer:
crimeas [40]3 years ago
7 0
Straight line, hope this helps
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Disadvantage of intrapreneurship​
adell [148]

Answer:

Another danger that comes with intrapreneurs is the lack of patience from other people in the office. To people who have worked their whole lives in one company, they might expect to see instant results from somebody who has waltzed in and promised big things.

8 0
3 years ago
Assume that, as the battery wears out, the voltage decreases at 0.03 volts per second and, as the resistor heats up, the resista
liraira [26]

Answer:

\frac{dI}{dt} =-3*10^-^4amps/sec

Explanation:

From the question we  are told that

Voltage decreases at   \frac{dv}{dt} =-0.03volts/sec

Resistance increase at \frac{dR}{dt}=0.02ohms /sec

Resistance at R=100ohms

Current at I=0.02amps

Generally the equation for ohms law is mathematically represented as

       V=IR

Therefore

       \frac{dV}{dt} =R\frac{dI}{dt} +I\frac{dR}{dt}

Generally making \frac{dI}{dt} subject of the formula in the above equation mathematically gives

       \frac{dV}{dt} =R\frac{dI}{dt} +I\frac{dR}{dt}

      R\frac{dI}{dt} = \frac{dV}{dt} -I\frac{dR}{dt}

       \frac{dI}{dt} =\frac{1}{R}  (\frac{dV}{dt} -I\frac{dR}{dt})

Therefore

       \frac{dI}{dt} =\frac{1}{100}((-0.03) -(0.02)*(0.02))

Generally it is given that the change in current is

       \frac{dI}{dt} =-3*10^-^4amps/sec

8 0
3 years ago
Water flows at 0.65 m/s through a 3.0 cm diameter hose that terminates in a 0.3 cm diameter nozzle. (Recall, the density of wate
xeze [42]

Answer:

v2 = 65 m/s

the speed of the water leaving the nozzle is 65 m/s

Explanation:

Given;

Water flows at 0.65 m/s through a 3.0 cm diameter hose that terminates in a 0.3 cm diameter nozzle

Initial speed v1 = 0.65 m/s

diameter d1 = 3.0 cm

diameter (nozzle) d2 = 0.3 cm

The volumetric flow rates in both the hose and the nozzle are the same.

V1 = V2 ........1

Volumetric flow rate V = cross sectional area × speed of flow

V = Av

Area = (πd^2)/4

V = v(πd^2)/4 ....2

Substituting equation 2 to 1;

v1(πd1^2)/4 = v2(πd2^2)/4

v1d1^2 = v2d2^2

v2 = (v1d1^2)/d2^2

Substituting the given values;

v2 = (0.65 × 3^2)/0.3^2

v2 = 65 m/s

the speed of the water leaving the nozzle is 65 m/s

6 0
4 years ago
The area around a charged object that can exert a force on other charged objects is an electric ___
Ymorist [56]
Answer is:

Electric field.
8 0
4 years ago
In a standing wave, areas of destructive are the?
Alborosie

antinodes

Hope this helps :)

4 0
3 years ago
Read 2 more answers
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