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elena55 [62]
3 years ago
12

a soccer player kicks a ball with a speed of 30 m/s at an angle of 10. how long does the ball stay in the air?

Physics
2 answers:
umka21 [38]3 years ago
6 0
The answer is 3.7 seconds
Brut [27]3 years ago
4 0
3.763 seconds I think
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The volume of a container is reported as 3.49 ft3. what is the volume in units of in3?
Kobotan [32]

The volume in units of 6.03x10^3 in3

1ft = 12 inch

So to convert foot into inches we have to multiply by 12

The volume of container = 3.49ft^3

=3.49×12×12×12

= 6030.72 in^3

<h3>What is the volume of a container?</h3>

The amount of space a container encloses, or how much room is inside of it, is measured by its volume. The volume of a box can be calculated using this straightforward formula: volume V = L × W × H for a box with height H, width W, and length L.

Many different units can be used, however because of the way this formula is written, the volume would have length to the third power dimensions. For instance, if the container's measurements are given in metres, the capacity of the box will be given in metres squared, or m3.

To learn more about volume of a container, visit:

brainly.com/question/9092584

#SPJ4

7 0
2 years ago
A carpenter lifts a 10 kg piece of wood to his shoulder 1.5 m above the ground. What is the wood's potential energy on the carpe
Burka [1]

Answer:

The wood's potential energy on the carpenter's shoulder is 150 J.

Explanation:

Given;

mass of the wood, m = 10 kg

height through which the wood was raised, h = 1.5 m

acceleration due to gravity, g = 10 m/s²

The  wood's potential energy on the carpenter's shoulder is calculated as;

P.E = mgh

P.E = 10 x 10 x 1.5

P.E = 150 J

Therefore, the wood's potential energy on the carpenter's shoulder is 150 J.

8 0
3 years ago
A positively charged particle of mass 7.2 x 10-8 kg is traveling due east with a speed of 88 m/s and enters a 0.6-T uniform magn
Marianna [84]

Answer:

q = 8.57 10⁻⁵ mC

Explanation:

For this exercise let's use Newton's second law

         F = ma

where force is magnetic force

        F = q v x B

the bold are vectors, if we write the module of this expression we have

         F = qv B sin θ

as the particle moves perpendicular to the field, the angle is θ= 90º

        F = q vB

the acceleration of the particle is centripetal

        a = v² / r

we substitute

        qvB = m v² / r

         qBr = m v

          q =\frac{m\  v}{B\  r}

The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use

          v = d / t

the distance is ¼ of the circle,

          d = \frac{1}{4} \  2\pi  r

           d =\frac{\pi }{2r}

we substitute

           v =  \frac{\pi  r}{2t}

           r = \frac{2 \ t  \ v}{\pi }

           

let's calculate

           r =\frac{2 \ 2.2  \ 10^{-3} \ 88}{\pi } 2 2.2 10-3 88 /πpi

           r = 123.25 m

         

let's substitute the values

           q = \frac{ 7.2 \ 10^{-8} \ 88}{ 0.6 \ 123.25}7.2 10-8 88 / 0.6 123.25

            q = 8.57 10⁻⁸ C

Let's reduce to mC

           q = 8.57 10⁻⁸ C (10³ mC / 1C)

           q = 8.57 10⁻⁵ mC

4 0
2 years ago
What do you think we can learn about Earth's history by looking at the Grand Canyon?
viktelen [127]
We can look at all the ages of the earth since it’s a big crack is reveals many layers of the earth and we can know about chemicals and metals that were in earth and diffrent times
5 0
3 years ago
A parallel-plate, air-gap capacitor has a capacitance of 0.14 mu F. The plates are 0.5 mm apart, What is the area of each plate?
Marysya12 [62]

Answer:

7.9060 m²

8.57 Volts

5.142×10⁻⁶ Joule

1.2×10⁻⁶ Coulomb

Explanation:

C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F

d = Distance between plates = 0.5 mm = 0.5×10⁻³ m

Q = Charge = 1.2 μC = 1.2×10⁻⁶ C

ε₀ = Permittivity = 8.854×10⁻¹² F/m

Capacitance

C=\frac{\epsilon_{0}A}{d}\\\Rightarrow A=\frac{Cd}{\epsilon_{0}}\\\Rightarrow A=\frac{0.14\times 10^{-6}\times 0.5\times 10^{-3}}{8.854\times 10^{-12}}\\\Rightarrow A=7.9060\ m^2

∴ Area of each plate is 7.9060 m²

Voltage

V=\frac{Q}{C}\\\Rightarrow V=\frac{1.2\times 10^{-6}}{0.14\times 10^{-6}}\\\Rightarrow V=8.57\ Volts

∴ Potential difference between the plates if the capacitor is charged to 1.2 μC  is 8.57 Volts.

Energy stored

E=0.5CV²

⇒E = 0.5×0.14×10⁻⁶×8.57²

⇒E = 5.142×10⁻⁶ Joule

∴ Stored energy is 5.142×10⁻⁶ Joule

Charge

Q = CV

⇒Q = 0.14×10⁻⁶×8.57

⇒Q = 1.2×10⁻⁶ C

∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb

6 0
3 years ago
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