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Pachacha [2.7K]

3 years ago

13

They are 4 springs stretched by the same mass. Spring a stretches 25cm, spring B stretches 10cm, spring C stretches 100cm, and s

pring B stretches 1 cm, which spring has the largest spring constant

1 answer:

Wewaii [24]3 years ago

6 0

Answer:

Spring C

Explanation:

Multiply 25 and 4

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What is the % error in using g = 10.0 m/s^2? (Take the value ofg as 9.8 m/s^2)

Alenkasestr [34]

Answer:

So percentage error will be 2 %

Explanation:

We have given initial value of acceleration due to gravity $g=10m/sec^2$

And final value of acceleration due to gravity $g=9.8m/sec^2$

We have to find the percentage error

We know that percentage error is given by $percentage\ error=\frac{initial\ value-final\ value}{initial\ value}\times 100$

So $percentage\ error=\frac{10-9.8}{10}\times 100=2$ %

4 0

4 years ago

The distance that a person can see to the horizon on a clear day from a point above the surface of Earth varies directly as the

lara [203]

Answer:

900 m ( 0.9 km)

Explanation:

from the question we have the following

first person's height (h) = 100 m above the earths surface

distance to the horizon that can be seen (s) = 15 km = 15,000 m

second person's height (h1) = 36 m

distance to the horizon that can be seen (s1) = ?

take note that the distance (s) a person can see = square root of the person's height

s = $\sqrt{h}$ x K

where k is a constant

from the height and the distance the first person can see we can get the value of K

15000 = $\sqrt{ 100}$ x K

15000 = 10 x K

K = 150

now putting the value of K and the height of the second person into the equation we can get the distance into the horizon the person can see

s = $\sqrt{36}$ x 150

s = 900 m

therefore the second person who is 36 m above the surface can see 900 m ( 0.9 km) into the horizon

5 0

3 years ago

A coil has N turns enclosing an area of A. In a physics laboratory experiment, the coil is rotated during the time interval delt

zmey [24]

Answer:

a) Magnetic flux before the plane is rotated = (BA) Wb

b) Magnetic flux after the plane is rotated = 0 Wb

c) Magnitude of the average induced emf =

(NBA)/Δt

Explanation:

The magnetic flux is given as

Φ = BA cos θ

where B = magnetic field strength

A = Cross sectional Area of the loop enclosed

θ = the angle in the equation above is between the line normal to plane (NOT the plane itself!) and the magnetic field.

a) Given that the position of the plane of each turn is perpendicular to Earth's magnetic field before being rotated.

Before the plane is rotated, θ = 0°

Φ = BA cos θ = BA cos 0° = BA

b) Given that the position of the plane of each turn is parallel to Earth's magnetic field after being rotated.

After the plane is rotated, θ = 90°

Φ = BA cos θ = BA cos 90° = 0

c) According to the Faraday's law of electromagnetic induction,

E = - N (ΔΦ/Δt) (minus sign to indicate that the direction of the induced emf is opposite the direction of the change of magnetic flux)

ΔΦ = (final Φ) - (initial Φ) = (0 - BA) = - BA

E = - N (- BA)/Δt

E = (NBA)/Δt

Hope this Helps!!!

8 0

3 years ago

Read 2 more answers

I need help on this question, please.

frez [133]

Answer:

|X| = 72 cm

Explanation:

We need to find the magnitude of vector X.

Vector Y = 21 cm

Vector Z = 75 cm

We know that, the magnitude of resultant vector is given by :

$Z={X^2+Y^2} \\\\X^2=Z^2-Y^2\\\\X^2=(75)^2-(21)^2\\\\X=72\ cm$

So, the value of magnitude of vector X is 72 cm

4 0

3 years ago

A 13.0 g piece of Styrofoam carries a net charge of â5.40 Î¼C and floats above the center of a large horizontal sheet of plastic

kati45 [8]

Answer:

σ =4.180×10^{-9} C/m^2

Explanation:

electric field due to non conducting sheet is

$E=\frac{\sigma}{2\epsilon_0}$

the force acting on the piece of Styrofoam

Eq= mg

⇒E= mg/q

now,

$\frac{mg}{q} = \frac{\sigma}{2\epsilon_0}$

⇒ $\sigma= \frac{2mg\epsilon_0}{q}$

$\sigma= \frac{2\times0.013\times9.81\times8.85\times10^{-12}}{5.40\times10^{-4}}$

charge per unit area (in C/m2) on the plastic sheet σ =4.180×10^{-9} C/m^2

6 0

4 years ago