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fiasKO
1 month ago
9

Theresa is swinging a 4.0 kg ball on a 1.2 m long rope with a tangential speed of 1.8 m/s. What is the centripetal force exerted

on the rope? Round to the tenths place.
Physics
2 answers:
Keith_Richards [3.2K]1 month ago
8 0
Rounding the number gives us a result of 10.8N.
ValentinkaMS [3.4K]1 month ago
5 0

In this scenario, the centripetal force is expressed as:

F = m * (\frac{v^2}{r})

Where,

m: object mass

v: speed along a tangent

r: radius of the rope

By substituting the values into the equation, we find:

F = (4.0) * (\frac{1.8^2}{1.2})

Next, we perform the necessary calculations:

F = 10.8 N

Result:

The force acting on the rope is:

F = 10.8 N

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A beam of electrons moves at right angles to a magnetic field of 4.5 × 10-2 tesla. If the electrons have a velocity of 6.5 × 106
Softa [3030]
Greetings!

Using the formula F = Bqv sin theta, we define F as Force, B as magnetic flux density, q as charge, v as velocity, and theta as the angle created by the moving electrons in relation to the magnetic field.

^^^You can compute the force using that equation^^^

In conclusion, your result would MOST LIKELY be "B".

"<span>-3.9 × 10-14 N"
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6 0
17 days ago
Read 2 more answers
A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re
Softa [3030]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

P_1 = Reservoir pressure = 10 atm

T_1 = Reservoir temperature = 300 K

P_2 = Exit pressure = 1 atm

T_2 = Exit temperature

R_s = Specific gas constant = 287 J/kgK

\gamma = Specific heat ratio = 1.4 for air

Assuming isentropic flow

\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K

Flow temperature at exit is 155.38424 K

Density at exit can be derived using the ideal gas equation

\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3

Flow density at exit measures 2.2721 kg/m³

4 0
1 month ago
If 100 grams of vinegar and 5 grams of baking soda are poured in a container, a small amount of gas will be produced. What will
Maru [3345]
The final mass will be slightly lower due to evaporation. I learned this back in third grade, so it's surprising you're in high school and don't know this.
5 0
1 month ago
When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t
Keith_Richards [3271]

Response:

The intensity of light 18 feet underwater is about 0.02%

Clarification:

Employing Lambert's law

Let dI / dt = kI, where k is a proportionality factor, I represents the intensity of incident light, and t indicates the thickness of the medium

Then dI / I = kdt

Taking logarithms,

ln(I) = kt + ln C

I = Ce^kt

At t=0, I=I(0) implies C=I(0)

I = I(0)e^kt

At t=3 & I=0.25I(0), we find 0.25=e^3k

Solving for k gives k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

The intensity of light 18 feet underwater is about 0.2%

3 0
1 month ago
the minute hand on a clock is 9 cm long and travels through an arc of 252 degrees every 42 minutes. To the nearest tenth of a ce
Ostrovityanka [3204]

Answer:

The distance covered by the minutes hand is 39.60 cm.

Explanation:

Note: A clock has a circular shape, where the minutes hand acts as the radius, and its motion creates an arc.

Length of an arc is calculated as ∅/360(2πr)

L = ∅/360(2πr).................... Equation 1π

Here, L represents the arc’s length, ∅ is the angle made by the arc, and r is the arc’s radius.

Given: ∅ = 252°, r = 9 cm, π = 3.143.

Upon substituting these values into equation 1,

L = 252/360(2×3.143×9)

L = 0.7×2×3.143×9

L = 39.60 cm.

Thus, the distance traversed by the minutes hand is 39.60 cm.

4 0
2 months ago
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