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zloy xaker [14]

2 years ago

15

Can the velocity of a body revese the direction when acceleration is constant?

1 answer:

andriy [413]2 years ago

7 0

Answer:

Yes, the velocity of the object can reverse direction when its acceleration is constant. For example consider that the velocity of any object at any time t is given as: ... At At t = 0 sec, the magnitude of velocity is 2m/s and is moving in the forward direction i.e.v (t) = -2.

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The magnetic field of a long, straight, and closely-wound solenoid, inside the solenoid at a point near the center, is 0.645 T.

Savatey [412]

Answer:

B'=1.935 T

Explanation:

Given that

magnetic field ,B= 0.645 T

We know that magnetic filed in the solenoid is given as

$B=\mu _0 n\ I$

I=Current

n=Number of turn per unit length

μ0 =magnetic permeability

Now when the current increased by 3 factors

I'=3 I

Then the magnetic filed

$B'=\mu _0 n\ I'$

$B'=\mu _0 n\ (3I)$

B'=3 B

That is why

B' = 3 x 0.645 T

B'=1.935 T

Therefore the new magnetic filed will be 1.935 T.

3 0

3 years ago

What is the most likely evidence of an energy transformation taking place when a skater glides over ice?

Triss [41]

Small tracks of water are left on the ice as kinetic energy is transformed into thermal energy

5 0

3 years ago

Read 2 more answers

Determine the capacitive reactance for a 20 uF capacitor that is across a 20 volt, 60 Hz source

aleksandr82 [10.1K]

Answer:

Capacitive reactance is 132.6 Ω.

Explanation:

It is given that,

Capacitance, $C=20\ \mu F=20\times 10^{-6}\ F=2\times 10^{-5}\ F$

Voltage source, V = 20 volt

Frequency of source, f = 60 Hz

We need to find the capacitive reactance. It is defined as the reactance for a capacitor. It is given by :

$X_C=\dfrac{1}{2\pi fC}$

$X_C=\dfrac{1}{2\pi \times 60\ Hz\times 2\times 10^{-5}\ F}$

$X_C=132.6\ \Omega$

So, the capacitive reactance of the capacitor is 132.6 Ω. Hence, this is the required solution.

4 0

3 years ago

Snezhnost [94]

Frozen water has move volume than water in liquid form

8 0

2 years ago

In Trial II, the same spring is used as in Trial I. Let us use this information to find the suspended mass in Trial II. Use 0.51

Masteriza [31]

Answer:

$M_2=0.79kg$

Explanation:

From the question we are told that:

Period $T=0.517s$

Trial 1

Spring constant $\mu=117N/m$

Period $T_1=0.37$

Mass $m=0.400kg$

Trial 2

Period $T_2=0.52$

Generally the equation for Spring Constant is mathematically given by

\mu=\frac{4 \pi^2 M}{T^2}

Since

$\mu _1=\mu_2$

Therefore

$\frac{4 \pi^2 M_1}{T_1^2}=\frac{4 \pi^2 M_2}{T_2^2}$

$M_2=M_1*(\frac{T_2}{T_1})^2$

$M_2=0.400*(\frac{0.52}{0.37}})^2$

$M_2=0.79kg$

5 0

3 years ago