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2 years ago

A piano wire has a length of 81 cm and a mass of 2.0

1 answer:

4 0

<span>Frequency = 394 Hz
Length of the string L = 81 cm = 0.81 m
Mass of the string = 0.002 kg
Tension T = ?
Wave length of the string is two times the length.
n x lambda = 2L, we also have lambda = vt = v / f, t is time period and given n = 1.
Therefore L = v / 2f => v = 2fL
Deriving form force equation, force here is tension T so
v = squareroot of (TL/m) hence
2fL = squareroot of (TL/m) => 4 x f^2 x L^2 = (T x L) / m => T = 4 x f^2 x L x m
T = 4 x 0.81 x (394)^2 x 0.002 = 4 x 0.81 x 155236 x 0.002
T = 1005.9 N = 1.006 x 10^3 N</span>

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A bobsled is pushed with a force of 190.08 N. The sled has a mass of 28 kg. What is the acceleration of the bobsled? Report to t

Usimov [2.4K]

By definition it is known that force equals mass by acceleration. In other words F = m * a. To find the acceleration, you must clear the formula mentioned. Therefore, for a force of 190.08N and a mass of 28 Kg, we have that the acceleration is a = F / m = (190.08) / (28) = 6.79 m / s ^ 2

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2 years ago

The force sensor measures the force on the sensor due to the bumper, but the cart's momentum change arises from the force on the

Irina-Kira [14]

Answer:

Answered

Explanation:

1 and 3 are necessary

Every bit of force applied to the bumper will be transmitted to the cart EXCEPT for the force needed to accelerate the bumper. This is the net force on the bumper.

If the bumper was heavy then a significant amount of force might be needed to accelerate the bumper so the amount transmitted to the cart would be substantially reduced.

If the net force on the bumper is small then the amount transmitted to the cart is almost the entire force applied.

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2 years ago

Daniel is 50.0 meters away from a building. He observes that his line-of-sight to the tip of the building makes an angle of 63.0

frez [133]

Since his line of sight 63 degrees makes with the tip of the building

Tan63° = height of building / Horizontal distance

tan63° = Height / 50

50tan63° = Height

Height = 50tan63°

Height ≈ 50*1.9626

Height ≈ 98.13 m

Height of the building is ≈ 98.13 m. Mind you in solving for this height we have neglected the height of Daniel.

The height of building actually should be 98.13 m plus the height of Daniel. Since the 63° was measured from his eye level.

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2 years ago

An inline skater skates on a circular track 120.0 m in diameter at a tangential speed of 9.20 m/s. If the skater’s mass is 68.5

jok3333 [9.3K]

Answer:

The centripetal force acting on the skater is <u>48.32 N.</u>

Explanation:

Given:

Radius of circular track is, $R=120.0\ m$

Tangential speed of the skater is, $v=9.20\ m/s$

Mass of the skater is, $m=68.5\ kg$

We are asked to find the centripetal force acting on the skater.

We know that, when an object is under circular motion, the force acting on the object is directly proportional to the mass and square of tangential speed and inversely proportional to the radius of the circular path. This force is called centripetal force.

Centripetal force acting on the skater is given as:

$F_c=\frac{mv^2}{R}$

Now, plug in the given values of the known quantities and solve for centripetal force, $F_c$ . This gives,

$F_c=\frac{68.5\times (9.20)^2}{120.0}\\\\F_c=\frac{68.5\times 84.64}{120}\\\\F_c=\frac{5797.84}{120}\\\\F_c=48.32\ N$

Therefore, the centripetal force acting on the skater is 48.32 N.

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2 years ago

A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe

Nonamiya [84]

Answer:

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

= surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ / 2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ / 2ε₀

= 6.78 X 10³ N/C

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2 years ago