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valentina_108 [34]
3 years ago
5

The end of a horizontal rope is attatched to a prong of an electricity driven tuning fork that vibrates at 100hz. The other end

passes over a pulley and supports a 2-kg mass. The linear mass of the rope is 0.75kg/m. What is the wavelength of a transverse wave on the rope?
A. 0.25m
B. 0.16m
C. 0.12m
D. 1.2m
Physics
1 answer:
Darina [25.2K]3 years ago
5 0

here since string is attached with a mass of 2 kg

so here tension force in the rope is given as

T = mg

here we will have

T = 2(9.8) = 19.6 N

now we will have speed of wave given as

v = \sqrt{\frac{T}{\mu}}

here we will have

v = \sqrt{\frac{19.6}{0.75\times 10^{-2}}}

v = 16.33 m/s

now we know that frequency is given as

F = 100 Hz

now wavelength is given as

\lambda = \frac{v}{F}

\lambda = \frac{16.33}{100} = 0.16 m

so wavelength will be 0.16 m

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Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00
Elena L [17]

Answer:

2.62898\times 10^{-6}\ C/m^3

1979.99974\ N/C

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C

Volume charge density is given by

\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3

The volume charge density for the sphere is 2.62898\times 10^{-6}\ C/m^3

E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C

The magnitude of the electric field is 1979.99974\ N/C

8 0
3 years ago
A football punker attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in the a
Flauer [41]

To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to

v_x = \frac{x}{t}

v_x = \frac{67}{4.5}

v_x = 14.89m/s

The vertical component of velocity is

-h = v_y t -\frac{1}{2} gt^2

Here,

h = Height

g = Gravitational acceleration

t = Time

v_y = Vertical component of velocity

-1.23 = v_y(4.5)-\frac{1}{2} (9.8)(4.5)^2

-1.23= 4.5v_y - 99.225

v_y = 21.77m/s

The direction of the velocity will be given by the tangent of the components, then

tan\theta = \frac{v_y}{v_x}

\theta = tan^{-1} (\frac{21.77}{14.89})

\theta = 55.59\°

The magnitude is given vectorially as,

|V| = \sqrt{v_x^2+v_y^2}

|V| = \sqrt{14.89^2 +21.77^2}

|V| = 26.37m/s

Therefore the angle is 55.59° and the velocity is 26.37m/s

6 0
3 years ago
If you are given distance and a period of time what can you calculate
Digiron [165]

If you are given distance and a period of time, you can calculate the speed. The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of its speed and direction of motion (e.g. 60 km/h to the north).

6 0
3 years ago
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The 88-lb force P is applied to the 210-lb crate, which is stationary before the force is applied. Determine the magnitude and d
Marina86 [1]

Answer:

F=-88Ib

Explanation:

From the question we are told that:

Force P=88Ib

Mass of crate M_c=210Ib

Generally the equation for Frictional force F is mathematically given by

Friction\ force (f) = friction\ coefficient\ (u) * Normal\ reaction (N)

F=u*N

with \mu =0.47

F=98.7Ib

Therefore since Static Friction supersedes applied force body remains at rest.

Frictional force =88Ib (negative)

F=-88Ib

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A constant force is exerted on a cart that is initially at rest on a frictionless air track. The force acts for a short time int
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<em>v</em> = <em>v</em>₀ + <em>a</em> <em>t</em>

<em>v</em> = (<em>F</em> / <em>m</em>) <em>t</em>

<em />

If we force is halved, so is the accleration:

<em>a</em> = <em>F</em> / <em>m</em>   →   <em>a</em>/2 = <em>F</em> / (2<em>m</em>)

So, in order to get the cart up to the same speed <em>v</em> as before, you need to double the time interval <em>t</em> to 2<em>t</em>, since that would give

(<em>F</em> / (2<em>m</em>)) (2<em>t</em>) = (<em>F</em> / <em>m</em>) <em>t</em> = <em>v</em>

3 0
3 years ago
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