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alina1380 [7]
3 years ago
11

A spring has a spring constant of 53N/m. How much elastic potential energy is stored in the spring in the spring when it is comp

ressed by p.21 m?
Physics
1 answer:
IRISSAK [1]3 years ago
5 0

Answer:11686.5 joules

Explanation:

elastic constant(k)=53N/m

extension(e)=21m

Elastic potential energy=(k x e^2)/2

Elastic potential energy=(53 x (21)^2)/2

Elastic potential energy=(53x21x21)/2

Elastic potential energy=23373/2

Elastic potential energy=11686.5

Elastic potential energy is 11686.5 joules

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What is the rate of acceleration due to gravity on Earth?
Katena32 [7]

Answer:

Negative 9.8 meters per second squared

Explanation:

The negative is for the direction (down, towards the center of the earth). Often this can be estimated as -10 m/s^2 to make calculations easier.

3 0
3 years ago
Inertia is a measure of
saw5 [17]

Answer:

b.. a difficulty of changing an object's motion, that's my ✨ g u e s s ✨

7 0
2 years ago
A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of
sladkih [1.3K]

Answer:

The magnitude of the resultant of the magnetic field is 4.11\times10^{-5}\ T

Explanation:

Given that,

Current = 40 A

Magnetic field B=3.7\times10^{-5}\ T

Distance = 22 cm

We need to calculate the magnetic field

Using formula of magnetic field

B'=\dfrac{\mu_{0}I}{2\pi r}

Where, r = distance

I = current

Put the value into the formula

B'=\dfrac{4\pi\times10^{-7}\times20}{2\pi\times0.22}

B'=1.8\times10^{-5}\ T

We need to calculate the magnitude of the resultant of the magnetic field

Using formula of resultant

B''=\sqrt{B^2+B'^2}

Put the value into the formula

B''=\sqrt{(3.7\times10^{-5})^2+(1.8\times10^{-5})^2}

B''=4.11\times10^{-5}\ T

Hence, The magnitude of the resultant of the magnetic field is 4.11\times10^{-5}\ T

6 0
3 years ago
A hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake.
Nataliya [291]

To solve the exercise it is necessary to take into account the definition of speed as a function of distance and time, and the speed of air in the sound, as well

v=\frac{d}{t}

Where,

V= Velocity

d= distance

t = time

Re-arrange the equation to find the distance we have,

d=vt

Replacing with our values

d= (343)(3.7)

d= 1269.1m

It is understood that the sound comes and goes across the entire lake therefore, the length of the lake is half the distance found, that is

L_{lake} = \frac{d}{2}

L_{lake} = \frac{1269.1}{2}

L_{lake} = 634.55m

Therefore the length of the lake is 634,55m

8 0
3 years ago
an object is producing a sound that has a wavelength in air of 2.69m. If the speed of sound in air is 346m/s, what is the freque
Misha Larkins [42]

Answer:

129.74 Hz

Explanation:

Given:

Wave velocity ( v ) = 346 m / sec

wavelength ( λ ) = 2.69 m

We have to calculate Frequency ( f ) :

We know:

v = λ / t [ f = 1 / t ]

v = λ f

= > f = v / λ

Putting values here we get:

= > f = 346 / 2.69 Hz

= > f = 34600 / 269 Hz

= > f = 129.74 Hz

Hence, frequency of sound is 129.74 Hz.

5 0
2 years ago
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