Answer:

**Answer:**

3

**Explanation:**

10⁰ = 1 because anything to the power of 0 is 1.

3×1= 3

Suppose that an object undergoes simple harmonic motion, and its displacement has an amplitude A = 15.0 cm and a frequency f = 11.0 cycles/s (Hz). What is the maximum speed ( v ) of the object?A. 165 m/sB. 1.65 m/sC. 10.4 m/sD. 1040 m/s

A thin aluminum meter stick hangs from a string attached to the 50.0 cm mark of the stick. From the 0.00 cm mark on the meter stick hangs a 5.202 kg concrete block. From the 75.0 cm mark on the meter stick hangs a 7.99 kg steel ball. At what mark on the meter stick must a 2.46 kg wooden block be attached so that the meter stick balances horizontally when lowered into fresh water? Assume the densities of concrete, steel, and wood are 2500.0 kg/m3, 8000.0 kg/m3, and 500 kg/m3 respectively. A. 57.6 cm B. 57.4 cm C. 57.2 cm D. 57.8 cm

If you want to play a tune on wine glasses, you’ll need to adjust the oscillation frequencies by adding water to the glasses. This changes the mass that oscillates (more water means more mass) but not the restoring force, which is determined by the stiffness of the glass itself. If you need to raise the frequency of a par- ticular glass, should you add water or remove water?

A CO2 gun shoots a 0.2 gram round pellet (bb) at 2800 ft/sec, and as the bb leaves the gun it gets charged by friction . If Earths magnetic field points South to North at an intensity of 20 uT, and the bb is shot W->E. Find the charge the bb would need to stay level by balancing out the force of gravity.

Which sling can the crane use to lift the 1000kg pipe?A.800kg rated slingB. 1000kg rated slingC. 2000kg rated slingD. Band C

A thin aluminum meter stick hangs from a string attached to the 50.0 cm mark of the stick. From the 0.00 cm mark on the meter stick hangs a 5.202 kg concrete block. From the 75.0 cm mark on the meter stick hangs a 7.99 kg steel ball. At what mark on the meter stick must a 2.46 kg wooden block be attached so that the meter stick balances horizontally when lowered into fresh water? Assume the densities of concrete, steel, and wood are 2500.0 kg/m3, 8000.0 kg/m3, and 500 kg/m3 respectively. A. 57.6 cm B. 57.4 cm C. 57.2 cm D. 57.8 cm

If you want to play a tune on wine glasses, you’ll need to adjust the oscillation frequencies by adding water to the glasses. This changes the mass that oscillates (more water means more mass) but not the restoring force, which is determined by the stiffness of the glass itself. If you need to raise the frequency of a par- ticular glass, should you add water or remove water?

A CO2 gun shoots a 0.2 gram round pellet (bb) at 2800 ft/sec, and as the bb leaves the gun it gets charged by friction . If Earths magnetic field points South to North at an intensity of 20 uT, and the bb is shot W->E. Find the charge the bb would need to stay level by balancing out the force of gravity.

Which sling can the crane use to lift the 1000kg pipe?A.800kg rated slingB. 1000kg rated slingC. 2000kg rated slingD. Band C

**Answer:**

- 93.6 N in the 41° cable
- 155.6 N in the 63° cable
- 200 N in the vertical cable

**Explanation:**

Let T and U represent the tensions in the 41° and 63° cables, respectively. In order for the system to be stationary, the horizontal components of these tensions must balance, and the vertical components of these tensions must total 200 N.

Tcos(41°) =Ucos(63°) . . . . . balance of horizontal components

U = Tcos(41°)/cos(63°) . . . . write an expression for U

__

The vertical components must total 200 N, so we have ....

Tsin(41°) +Usin(63°) = 200

Tsin(41°) +Tcos(41°)sin(63°)/cos(63°) = 200

T(sin(41°)cos(63°) +cos(41°)sin(63°))/cos(63°) = 200

T = 200cos(63°)/sin(41° +63°) ≈ 93.6 . . . newtons

U = 200cos(41°)/sin(41° +63°) ≈ 155.6 . . . newtons

__

The vertical cable must have sufficient tension to balance the weight of the traffic light, so its tension is 200 N.

**Then the tensions in the 3 cables are ...**

** 41°: 93.6 N**

** 63°: 155.6 N**

** 90°: 200 N**

The **tension** in each of the three cables are **94.29**, **155.56** and **200 Newton** respectively.

__Given the following data:__

- Force = 200 Newton.

- Angle 1 = 41°

- Angle 2 = 63°

First of all, we would determine the **third tension** force based on the **vertical component **as follows:

Next, we would apply **Lami's theorem** to resolve the **forces** acting on the **traffic light** at equilibrium:

__For the __**horizontal component**__:__

....equation 1.

__For the __**vertical component**__:__

...equation 2.

Substituting eqn. 1 into eqn. 2, we have:

__For the __**first tension**__:__

Read more on **tension** here: brainly.com/question/4080400

**Answer:**

Observed time, t = 5.58 s

**Explanation:**

Given that,

Speed of light in a vacuum has the hypothetical value of, c = 18 m/s

Speed of car, v = 14 m/s along a straight road.

A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s.

We need to find the time the driver of the car measure for his trip between the poles. The relation between real and observed time is given by :

t is observed time.

**So, the time observed by the driver of the car measure for his trip between the poles is 5.58 seconds.**

**Answer:**

The value of g is

**Explanation:**

From the question we are told that

The mass of the weight is

The spring constant

The second harmonic frequency is

The number of oscillation is

The time taken is

Generally the frequency is mathematically represented as

At second harmonic frequency the length of the string vibrating is equal to the wavelength of the wave generated

Noe from the question the vibrating string is just half of the length of the main string so

Let assume the length of the main string is

So

The velocity of the vibrating string is mathematically represented as

Where T is the tension on the string which can be mathematically represented as

So

Then

=>

=>

=>

substituting values

Generally the period of oscillation is mathematically represented as

=>

The period can be mathematically evaluated as

substituting values

Therefore

so

substituting for L

=>

with the attached objects will balance there?

**Answer:**

See answer below

**Explanation:**

Hi there,

To get started, recall the Center of Mass formula for **two masses**:

where m is mass and x is displacement *from the center of the shape.*

Since masses at the center of a geometric shape have a displacement (x) value of 0, as the mass is already of the center, and does not affect Xcm. So, we can disregard the central mass, hence we use the above formula for two masses.

We can arbitrarily define left to be a negative (-) displacement, and vice versa for right direction. We proceed with the formula:

Since we defined left (-) and right (+), we notice the center of mass is (+) value. This makes sense, as there is slightly more mass on the right side. Hence, you should place a support 1/6 of the rod's length away from the rod's center.

Study well and persevere.

thanks,

To balance the rod with the attached objects, place a support at a **distance** of L/3 from the left end of the rod.

To balance the rod with the attached objects, you need to place a support at a distance of L/3 from the left end of the rod. This is because the **center of gravity **of the system should be directly above the support.

The center of gravity is given by the equation xcg = (m*0 + m*L/2 + 2m*L)/ (m + m + 2m). Solving this equation, we get xcg = 2L/3. Therefore, the support should be placed at a distance of L/3 from the left end of the rod.

#SPJ3

**Answer:**

** The add mass = 5.465 kg**

**Explanation:**

**Note: Since the spring is the same, the length and Tension are constant.**

f ∝ √(1/m)........................ Equation 1 (length and Tension are constant.)

Where f = frequency, m = mass of the spring.

But f = 1/T ..................... Equation 2

Substituting Equation 2 into equation 1.

1/T ∝ √(1/m)

Therefore,

T ∝ √(m)

Therefore,

T₁/√m₁ = k

where k = Constant of proportionality.

T₁/√m₁ = T₂/√m₂ ........................ Equation 3

making m₂ the subject of the equation

m₂ = T₂²(m₁)/T₁²........................... Equation 4

Where T₁ = initial, m₁ = initial mass, T₂ = final period, m₂ = final mass.

**Given: T₁ = 1.18 s m₁ = 0.50 kg, T₂ = 2.07 s.**

*Substituting into equation 4*

*m₂ = (2.07)²(0.5)/(1.18)²*

*m₂ = 4.285(1.392)*

*m₂ = 5.965 kg.*

Added mass = m₂ - m₁

Added mass = 5.965 - 0.5

Added mass = 5.465 kg.

**Thus the add mass = 5.465 kg**

**Answer:**

**- z direction**

**Explanation:**

To find the direction of the magnetic field, you take into account that the magnetic force over a charge, is given by the following cross product:

(1)

F_B: magnetic force

q: charge of the particle

v: velocity of the charge

B: magnetic field

In this case you have that the electron is moving along x-axis. You can consider this direction as the **^i** direction. The electron experiences a magnetic deflection in the -y direction, that is, in the **-^j **direction.

By the cross products between unit vectors, you have that:

**-^j = ^i X ^k**

That is, the cross product between two vectors, one in the +x direction, and another one in the +z direction, generates a vector in the -y direction. However, it is necessary to take into account that the negative charge of the electron change the sign of the result of the cross product, which demands that the second vector is in the -z direction. That is:

** -^i X -k^ = ^i X ^k = - ^j**

**Hence, the direction of the magnetic field is in the -z direction**