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3 years ago

A spring has a constant of 875 N/m. What hanging mass will cause this spring to stretch 4.5 m?

Physics

2 answers:

Ivenika [448]3 years ago

8 0

Answer:401.79 kg

Explanation:

andrey2020 [161]3 years ago

3 0

Answer:

8) A hanging 7.5 kg object stretches a spring 1.1 m. What is the spring constant? 1 . 1 66. 82 NIMO 9) A spring has a constant of 875 N/m. What hanging mass will cause this spring to stretch 4.5 m? Iouliply then divide 875 X 4. 2- - (4 04. 7919) 10) A spring with a spring constant of 25 N/m is stretched 65 cm. What was the force used? 11) A 25 N force stretches a spring 280 cm. What was the spring constant? 12) A 75 N force stretches a spring 175 cm. What was the proportionality constant? 8) 66.82 N/m 9) 401.79 kg 10) 16.25 N 11) 8.93 N/m 12) 42.86 N/m

Explanation:

You might be interested in

How can shorter string produce more fundamental frequency in two string of different length?

Nana76 [90]

Answer:

Shorter string produces more frequency in two different strings because the equation for frequency is velocity/wavelength , this means that a shorter string creates a shorter wavelength which essentially increases the total frequency produced

5 0

3 years ago

A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium p

saul85 [17]

Answer:

0.67 s

Explanation:

This is a simple harmonic motion (SHM).

The displacement, $x$ , of an SHM is given by

$x = A\cos(\omega t)$

A is the amplitude and $\omega$ is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or $\frac{\pi}{4}$ radian.

From trigonometry, $\sin A =\cos B$ if A and B are complementary.

At $t = 0$ , $x = 3.5$

$3.5 = A\cos(\omega \times0)$

$A =3.5$

$x = 3.5\cos(\omega t)$

At $t = 0.12$ , $x = 1.5$

$1.5 = 3.5\cos(0.12\omega)$

$\cos(0.12\omega)=\dfrac{1.5}{3.5}=0.4286$

$0.12\omega =\cos^{-1}0.4286$

$0.12\omega = 1.13$

$\omega = 9.4$

The period, $T$ , is related to $\omega$ by

$T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67$

5 0

3 years ago

A solenoidal coil with 26 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 20.0 cm long

noname [10]

Answer:

Part a)

$\phi = 2.76 \times 10^{-7} T m^2$

Part B)

$M = 5.52 \times 10^{-5} H$

Part C)

$EMF = 0.1 V/s$

Explanation:

Part a)

Magnetic field due to a long ideal solenoid is given by

$B = \mu_0 n i$

n = number of turns per unit length

$n = \frac{N}{L}$

$n = \frac{350}{0.20}$

$n = 1750 turn/m$

now we know that magnetic field due to solenoid is

$B = (4\pi \times 10^{-7})(1750)(0.100)$

$B = 2.2 \times 10^{-4} T$

Now magnetic flux due to this magnetic field is given by

$\phi = B.A$

$\phi = (2.2 \times 10^{-4})(\pi r^2)$

$\phi = (2.2 \times 10^{-4})(\pi(0.02)^2)$

$\phi = 2.76 \times 10^{-7} T m^2$

Part B)

Now for mutual inductance we know that

$\phi_{total} = M i$

$\phi_{total} = N\phi$

$\phi_{total} = 20(2.76 \times 10^{-4})$

$\phi_{total} = 5.52 \times 10^{-6}$

now we have

$M = \frac{5.52 \times 10^{-6}}{0.100}$

$M = 5.52 \times 10^{-5} H$

Part C)

As we know that induced EMF is given as

$EMF = M \frac{di}{dt}$

$EMF = 5.52 \times 10^{-5} (1800)$

$EMF = 0.1 V/s$

3 0

4 years ago

Please help ill mark youas brainliest !!

USPshnik [31]

Answer:

can you put on a clearer image this one is hard to see

8 0

2 years ago

Monochromatic light is incident on (and perpendicular to) two slits separated by 0.200 mm, which causes an interference pattern

marusya05 [52]

Answer:

I = 0.636*Imax

Explanation:

(a) To find the fraction of the maximum intensity at a distance y from the central maximum you use the following formula:

$I=I_{max}cos^2(\frac{\pi d}{\lambda L}y)$ (1)

I: intensity of light

Imax: maximum intensity of light

d: separation between slits = 0.200mm = 0.200 *10^-3 m

L: distance from the screen = 613cm = 0.613 m

y: distance to the central peak of the interference pattern

λ: wavelength of light = 656.3 nm = 656.3 *10^-9 m

You replace the values of all variables in the equation (1):

$I=I_{max}cos^2(\frac{\pi (0.200*10^{-3}m)}{(656.3*10^{-9}m)(0.613m)}0.600m)\\\\I=I_{max}cos^2(937.06)=0.636I_{max}$

Hence, the fraction of the maximum intensity is I = 0.636*Imax

6 0

3 years ago