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cluponka [151]
3 years ago
7

A hurricane is a huge storm that forms over warm ocean waters. Many hurricanes enter the Gulf of Mexico and move in a northweste

rly direction. A hurricane travels a distance of 20 miles in 1 hour. Determine the hurricane's speed. (SPEED = DISTANCE / TIME)
Physics
1 answer:
labwork [276]3 years ago
5 0

Answer:

536.448m/s or 20miles/hr

Explanation:

Speed is defined as the distance travelled by a body or an object in a specified time.

Speed = Distance/Time

If a hurricane travels a distance of 20 miles in 1 hour.

Distance travelled = 20miles

Time taken = 1hour

Speed = 20miles/1hour

Speed = 20miles/hr

Since 1mile = 1609.344m

20miles = 20×1609.344

20miles = 32186.88m

1hour = 60seconds

Expressing the speed in m/s

Speed = 32186.88m/60s

Speed = 536.448m/s

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The charge per unit length on a long, straight filament is -92.0 μC/m. Find the electric field 10.0 cm above the filament.
Pepsi [2]

Answer:

E = 1.655 x 10⁷ N/C towards the filament

Explanation:

Electric field due to a line charge is given by the expression

E = [tex]\frac{\lambda}{2\pi\times\epsilon_0\times r}[/tex]

where λ is linear charge density of line charge , r is distance of given point from line charge and ε₀ is a constant called permittivity and whose value is

8.85 x 10⁻¹².

Putting the given values in the equation given above

E = \frac{92\times10^{-6}}{2\times3.14\times8.85\times10^{-12}\times10^{-1}}

E = 1.655 x 10⁷ N/C

4 0
3 years ago
What is the farthest distance parallaxes can be used to measure star distances from Earth?
JulsSmile [24]
Using current technology, useful parallax measurements can only be found for stars up to about 340 light years (100 parsecs) away.
8 0
3 years ago
considere que o calor específico de um material presente nas cinzas seja c=0,8j/gc. Supondo que esse material entre na turbina a
drek231 [11]

Answer:

3120J

Explanation:

Given parameters:

C  = Specific heat capacity  = 0.8J/g°C

Initial temperature  = 20°C

Mass given   = 5g

Final temperature  = 800°C

Unknown:

Energy given to the mass  = ?

Solution:

To find the energy given to the mass, let us simply use the expression below:

          H   =   m   c   ΔT

H is the unknown, the energy supplied

m is the mass of the substance

c is the specific heat capacity

ΔT is the change in temperature

Input the variables;

            H    = 5  x   0.8    x    (800 - 20)  = 3120J

7 0
3 years ago
The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?
Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight =   60 .3 (9.81) = 591.54 N

Accel of lift changes this to    60.3 ( 9.81 - L)     where L - accel of lift

                                           60.3 ( 9.81 - L ) = 564

                                               solve for L = .457 m/s^2  DOWNWARD

                                                        so L = - .457 m/s^2

4 0
2 years ago
Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f
vampirchik [111]

Answer:

\large \boxed{\text{30 rev/s}}

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

 I₁ = 3.5    kg·m²s⁻¹

ω₁ = 6.0    rev·s⁻¹

 I₂ = 0.70 kg·m²s⁻¹

Calculation:

\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}

8 0
3 years ago
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