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jeyben [28]
4 years ago
13

A cat is running at 24 m/s. It then accelerates at 7 m/s2. How long will it take the cat to reach a speed of 49 m/s?

Physics
1 answer:
kozerog [31]4 years ago
8 0

Answer:

t should be 3.57 second

Explanation:

Formula used is v = u+at

In which v is final velocity, u is initial velocity, a is acceleration and t is time.

Substitute each of the info given into the formula and calculate.

49 = 24 + (7)t

t = 3.57s

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6 0
3 years ago
The tidal lung volume of human breathing, representing the amount of air inhaled and exhaled in a normal breath, is 500 cm3. (As
Darina [25.2K]

Answer:

Explanation:

Temperature of air = 18°C = 273 + 18 = 291 K .

volume = 500 cc = 0 .5 litre .

pressure = one atmosphere ( atm) .

From gas equation , we can calculate this volume at NTP  as follows.

volume = .5 x ( 273 / 291  ) litre

= 0.469 litre .

In any gas at NTP , 22.4 litre contains 6.02 x 10²³ molecules

.469 litre will contain  6.02 x 10²³ x .469 / 22.4 molecules

= 126 x 10²⁰ molecules .

b )

one mole = 6.02 x 10²³ molecules

6.02 x 10²³ molecules  has weight of 28.96 grams

126 x 10²⁰ molecules has weight of 28.96 x 126 x 10²⁰ / 6.02 x 10²³ grams

= .606 gram .

c )

volume of all the air in the atmosphere = volume of sphere

=  4 / 3 x π  x R³

= ( 4 / 3) x 3.14 x (999.5 x 10³ )³ m³

= 4.18 x 10¹⁸ m³

density of air = 1.225 kg / m³

mass of air = 1.225 x 4.18 x 10¹⁸ kg

= 5.12 x 10¹⁸ kg

d )

volume of air inhaled by 7 billion people

= . 5 x 7 x 10⁹ litre

= 3.5 x 10⁶ m³ .

Total volume of air in atmosphere = 4.18 x 10¹⁸ m³

required percentage

= 3.5 x 10⁶ x 100 /  4.18 x 10¹⁸

= .8373 x 10⁻¹⁰ % .

6 0
3 years ago
Does wax paper, aluminum foil, or plastic wrap cause more friction
Levart [38]
Yes i does. they cause more friction <span />
4 0
3 years ago
Read 2 more answers
A(n) 131 g ball is dropped from a height
larisa [96]

Answer:

26.59 N/m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 131 g

Extention (e) = 4.82755 cm

Acceleration due to gravity (g) = 9.8 m/s²

Spring constant (K) =?

Next, we shall convert 131 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

131 g = 131 g × 1 Kg / 1000 g

131 g = 0.131 Kg

Thus, 131 g is equivalent to 0.131 Kg.

Next, we shall the force exerted by the ball on the spring. This can be obtained as follow:

Mass (m) = 0.131 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = ma

F = 0.131 × 9.8

F = 1.2838 N

Next, we shall convert 4.82755 cm to metre (m)

This can be obtained as follow:

100 cm = 1 m

Therefore,

4.82755 cm = 4.82755 cm × 1 m / 100 cm

4.82755 cm = 0.0482755 m

Thus, 4.82755 cm is equivalent to 0.0482755 m

Finally, we shall determine the spring constant as follow:

Force (F) = 1.2838 N

Extention (e) = 0.0482755 m

Spring constant (K) =?

F = Ke

1.2838 = K × 0.0482755

Divide both side by 0.0482755

K = 1.2838 / 0.0482755

K = 26.59 N/m

Thus the spring constant is 26.59 N/m

7 0
3 years ago
What is the kinetic energy of a 150 kg bear running at 3 m/s?
OverLord2011 [107]

Explanation:

KE = ½ mv²

KE = ½ (150 kg) (3 m/s)²

KE = 675 J

5 0
3 years ago
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