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3 years ago

3.

Physics

2 answers:

Alborosie3 years ago

4 0

Answer:

Explanation:

Acceleration is defined as a change in velocity over time. This means that as long as the forces are not zero, the velocity should be changing. The can be negative (as this is negative acceleration also known as deceleration) or positive (known as acceleration). It doesn't matter if they are decimals or fractions.

Nookie1986 [14]3 years ago

4 0

In order for acceleration of an object to occur, the sum of the forces must be anything but zero.

Answer: Option C

<u>Explanation: </u>

Based on Newton second laws of motion, the acceleration due to gravity of any object is directly proportional to the net force acting on the object and inversely proportional to the mass of the object. So, the sum of all the forces acting on the object will decide the acceleration attained by the object.

If the total forces are zero then the acceleration of the object will be zero. In case of deceleration, the net force would be having negative sign and thus increase in net force will lead to increase in acceleration. Hence, for the occurrence of acceleration, the total forces sum would be anything but zero.

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A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magni

mina [271]

Answer:

Induced emf, $\epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}$

Explanation:

The varying magnetic field with time t is given by according to equation as :

$B=B_{max}e^{-t/\tau}$

Where

$B_{max}\ and\ t$ are constant

Let $\epsilon$ is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{d(BA)}{dt}$

$\epsilon=-A\dfrac{d(B)}{dt}$

$\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}$

$\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}$

So, the induced emf in the loop as a function of time is $A\dfrac{B_{max}e^{-t/\tau}}{\tau}$ . Hence, this is the required solution.

7 0

2 years ago

Examples of increase in pressure due to increase in applied force<br>

Nina [5.8K]

Answer:

injecting

Explanation:

5 0

2 years ago

A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside

neonofarm [45]

Answer:

$\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$ , level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

$\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}$

$\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt} = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}$

By replacing all known variables:

$\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$