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3 years ago

Find the frequency of a spring block system if it is doing 4 oscillation in 100s

Physics

1 answer:

Paha777 [63]3 years ago

3 0

Answer:

.004 Hz

Explanation:

Frequeny is cycles per second. An oscillation is 1 cycle

F=cycles/sec

4/100

=0.004Hz (Hz- Hertz=1 cycle/sec)

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A go-kart accelerates at a rate

stira [4]

Answer:

if it is at a speed of 12.5 m / s it would take 2 minutes

Explanation:

4 0

4 years ago

We wish to determine the electric field at a point near a positively charged metal sphere (a good conductor). We do so by bringi

quester [9]

Answer:

Less than.

Explanation:

We have the positive charged metal sphere and we have to determine the electric field at a point near to it. In order to find that if we bring the positive test charge at that point then as we know that "like charges repel" so their electric field lines will repel each other resulting in a weaker electric field.

However if we bring the negative test charge at that point then of course there will be attraction and also the the electric field lines will direct from the positive to negative resulting in a stronger electric field between them. So there will be larger electric field then before.

"In this case, It can be concluded that electric field will be less than it was at this point before the test charge was present."

5 0

3 years ago

An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should happen to break

mamaluj [8]

Answer:

Value that the spring constant k = 12Mg / h

Explanation:

According to 2nd law of Newton:

upward force of the spring= F

The weight of the elevator W = mg

F = Mg = M(5g)

==> F =6Mg.

As the spring is compressed to its maximum distance ie s,the maximum upward acceleration comes just , Hence

F =ks = 6Mg

==> s = 6Mg/k

We have gravitational potential energy turning into elastic potential of the spring as the elevator starts at the top some distance h from the spring, and undergoes a total change in height equal to h + s, so:

Mg(h+s) = 1/2ks2

And plugging in our expression for s:

Mg(h+6Mg/k)= 1/2k(6Mg / k)2

gh + 6M2g2/k = 1/2k(36M2g2 /k2)

Mgh +6M2g2/k = 1/2k(36M2g2 /k2)

gh + 6Mg2/k = 18Mg2 / k

gh = 12Mg2 / k

h = 12Mg / k

k = 12Mg / h

7 0

4 years ago

Read 2 more answers

Cats are groomed often by their owners to remove hair. What logical inference can be made based on this statement?

Tpy6a [65]

Answer:

its keeps the fur straight.....im kinda confused with the question

Explanation:

8 0

3 years ago

Find the electric field at a point midway between two charges of +40.0 x 10^-9 c and.+60.0 x 10^-9 c

PIT_PIT [208]

Missing part in the text: "...the charges are <span>separated by a distance of 30.0 cm."
</span>
Solution:
The point midway between the two charges is located 15.0 cm from one charge and 15.0 from the other charge. The electric field generated by each of the charges is
$E=k_e \frac{q}{r^2}$
where
ke is the Coulomb's constant
Q is the value of the charge
r is the distance of the point at which we calculate the field from the charge (so, in this problem, r=15.0 cm=0.15 m).

Let's calculate the electric field generated by the first charge:
$E_1 = (8.99 \cdot 10^9 Nm^2 C^{-2} ) \frac{+40.0 \cdot 10^{-9} C}{(0.15 m)^2}=1.6 \cdot 10^4 N/C$

While the electric field generated by the second charge is
$E_2 = (8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{+60.0 \cdot 10^{-9} C}{(0.15 m)^2}=2.4 \cdot 10^4 N/C$

Both charges are positive, this means that both electric fields are directed toward the charge. Therefore, at the point midway between the two charges the two electric fields have opposite direction, so the total electric field at that point is given by the difference between the two fields:
$E=E_2 - E_1 = 2.4 \cdot 10^4 N/C - 1.6 \cdot 10^4 N/C = 8000 N/C$

4 0

3 years ago