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3 years ago

At 15°C air is transmitted at 340 m/s. Express this speedin Kilometers per hour.

Physics

2 answers:

EleoNora [17]3 years ago

7 0

Answer:

1224km/hr

Explanation:

To convert from m/s to km/hr

1000m = 1km

Divide both sides by 1000

1m = 1/1000 km................. (1)

60×60 seconds = 1 hr

3600s = 1hr

Divide both sides by 3600

1s = 1/3600 .............(2)

Divide (2) by (1)

1m/s = 1/1000 ÷ 1/3600 km/hr

1m/s = 1/1000 × 3600/1 km/hr

1m/s = 3600/1000 km/hr

1m/s = 3.6 km/hr .............(3)

To convert 340m/s to km/hr

Multiply (3) by 340

1× 340m/s = 3.6 × 340 km/hr

340m/s = 1224km/hr

I hope this was helpful, please mark as brainliest

Jet001 [13]3 years ago

7 0

Answer:

1224 km/hr

Explanation:

<h2><u>Method 1</u></h2><h3><u><em>to convert m/s to km/hr</em></u></h3>

recall that :

1km = 1000m

firstly lets convert m to km

340 m to km

340 / 1000

= 0.34 km

recall that :

1hr = 60 min

1 min = 60 seconds

hence 1 hr = 60*60

1 hr = 3600 seconds

<h3><u>now to convert seconds to hr :</u></h3>

1 second / 3600

=1/3600 hours

hence to get the answer you say :

$\frac{0.34}{\frac{1}{3600} }$

<h2><u>Method 2</u></h2>

recall : 1km = 1000m

recall: 1hr = 60 min

1 min = 60 seconds

hence 1 hr = 60*60

1 hr = 3600 seconds

now to get this answer you have to divide m by 1000 to get km

and you have to divide seconds by 3600 to get hours

= $\frac{340 / 1000}{1/3600}$

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Answer:

$A = 0.2875 m^2$

Explanation:

As we know that

$Q = 4.6 \mu C$

E = 1.8 kV/mm

now we know that electric field between the plated of capacitor is given as

$E = \frac{\sigma}{\epsilon_0}$

now we will have

$1.8 \times 10^6 = \frac{\sigma}{\epsilon_0}$

$\sigma = (1.8 \times 10^6)(8.85 \times 10^{-12})$

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now we have

$\sigma = \frac{Q}{A}$

now we have area of the plates of capacitor

$A = \frac{Q}{\sigma}$

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3 years ago

Quentin is playing baseball and is batting. He swings and hits the ball sending it flying to the outfield. Which is greater- the

Bess [88]

Answer:

The force of the ball on the bat is same as the force of the bat on the ball.

Explanation:

A bat hits the ball and the ball moves to the out filed.

According to the Newton's third law, for every action there is an equal and opposite reaction.

The action and the reaction forces acts on the two different bodies but the magnitude of the force is same.

As the ball is hitted by the bat, the bat exerts the force on the ball and the same force is exerted on the bat by the ball according to the Newton's third law.

So, the force of the ball on the bat is same as the force of the bat on the ball but the direction of force is opposite.

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3 years ago

If the hiker starts climbing at an elevation of 350 ft, what will their change in gravitational potential energy be, in joules,

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Answer:

352,088.37888Joules

Explanation:

Complete question;

A hiker of mass 53 kg is going to climb a mountain with elevation 2,574 ft.

A) If the hiker starts climbing at an elevation of 350 ft., what will their change in gravitational potential energy be, in joules, once they reach the top? (Assume the zero of gravitational potential is at sea level)

Chane in potential energy is expressed as;

ΔGPH = mgΔH

m is the mass of the hiker

g is the acceleration due to gravity;

ΔH is the change in height

Given

m = 53kg

g = 9.8m/s²

ΔH = 2574-350 = 2224ft

since 1ft = 0.3048m

2224ft = (2224*0.3048)m = 677.8752m

Required

Gravitational potential energy

Substitute the values into the formula;

ΔGPH = mgΔH

ΔGPH = 53(9.8)(677.8752)

ΔGPH = 352,088.37888Joules

Hence the gravitational potential energy is 352,088.37888Joules

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2 years ago

A 5.00-V battery charges the parallel plates in a capacitor, with a plate area of 865 mm2 and an air-filled separation of 3.00 m

Westkost [7]

Answer:

W = 3.21x10⁻¹¹ J

Explanation:

The work required to separate the plates can be calculated using the following equation:

$W = U_{2} - U_{1} = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2})$

<u>Where</u>:

U₂: is the final stored energy

U₁: is the initial stored energy

C₂: is the final capacitance

C₁: is the initial capacitance

V₁: is the initial potential difference = 5.00 V

V₂: is the final potential difference

The initial and final capacitance is:

$C_{1} = \epsilon_{0}*\frac{A}{d_{1}}$

<u>Where</u>:

ε₀: is the vacuum permittivity = 8.85x10⁻¹² C²/(N*m²)

d: is the initial distance = 3.00 mm = 3.00x10⁻³ m

A: is the plate area = 865 mm² = 8.65x10⁻⁴ m²

$C_{1} = \epsilon_{0}*\frac{A}{d_{1}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 \cdot 10^{-3} m} = 2.55 \cdot 10^{-12} F$

Similarly, C₂ is:

$C_{2} = \epsilon_{0}*\frac{A}{d_{2}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 + 3.00 \cdot 10^{-3} m} = 1.28 \cdot 10^{-12} F$

Now, V₂ can be calculated by finding the initial charge (q₁):

$q_{1} = C_{1}V_{1} = 2.55 \cdot 10^{-12} F*5.00 V = 1.28 \cdot 10^{-11} C$

Since, q₁ is equal to q₂, V₂ is:

$V_{2} = \frac{q_{2}}{C_{2}} = \frac{1.28 \cdot 10^{-11} C}{1.28 \cdot 10^{-12} F} = 10 V$

Finally, we can find the work:

$W = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2}) = \frac{1}{2}(1.28 \cdot 10^{-12} F*(10 V)^{2} - 2.55 \cdot 10^{-12} F(5.00 V)^{2}) = 3.21 \cdot 10^{-11} J$

Therefore, the work required to separate the plates is 3.21x10⁻¹¹ J.

I hope it helps you!

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At 15°C air is transmitted at 340 m/s. Express this speedin Kilometers per hour.​

At 15°C air is transmitted at 340 m/s. Express this speedin Kilometers per hour.