You hear a thunder 3 seconds after you see lightning. How far away is the storm?

A particle is released as part of an experiment. Its speed t seconds after release is given by v (t )equalsnegative 0.4 t square

torisob [31]

Answer:

a) 2.933 m

b) 4.534 m

Explanation:

We're given the equation

v(t) = -0.4t² + 2t

If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.

See attachment for the calculations

The conclusion of the attachment will be

7.467 - 2.933 and that is 4.534 m

Thus, The distance it travels in the second 2 sec is 4.534 m

6 0

3 years ago

given a circuit powered at 12V with R1, R2, R3 respectively of 10,20,30 Ohm, determine R4 in such a way that the Wheatstone brid

dalvyx [7]

Answer:

The balanced condition for Wheat stones bridge is

Q

P

=

S

R

as is obvious from the given values.

No, current flows through galvanometer is zero.

Now, P and R are in series, so

Resistance,R

1

=P+R

=10+15=25Ω

Similarly, Q and S are in series, so

Resistance R

2

=R+S

=20+30=50Ω

Net resistance of the network as R

1

and R

2

are in parallel

i=

R

V

=

50

6×3

=0.36 A.

Explanation:

8 0

3 years ago

A steel column is 3 m long and 0.4 m diameter. It carries a load of 50 MN so that 5.967 mm is elongates more. Find the modulus o

Alex787 [66]

Answer:

$E=2.0*10^{11}N/m^{2}$

Explanation:

<u>Relation between stress and Force:</u>

$\sigma=\frac{F}{A}=\frac{F}{\pi*d^{2}/4}$

<u>Relation between stress and strain:</u>

Young's modulus is defined by the ratio of longitudinal stress σ , to the longitudinal strain ε:

$E=\frac{\sigma}{\epsilon}$

$\epsilon=\frac{\Delta l}{l}$

So:

$E=\frac{F*l}{\pi*d^{2}/4*\Delta l}=\frac{50*10^{6}*3}{\pi*(0.4^{2}/4)*5.967*10^{-3}}=2*10^{11}N/m^2$

8 0

3 years ago

An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu

ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, $qE \times S + mg \times S = 0.5 \times mv^{2}$

$1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}$

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight and the above equation will be as follows.

$(1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}$