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dimaraw [331]
3 years ago
7

how fast is a ball going when it hits the ground after being dropped from a height of 16m the acceleration of gravity is 9.8

Physics
1 answer:
Aneli [31]3 years ago
8 0

Answer:

17.7 m/s

Explanation:

Given:

y₀ = 0 m

y = 16 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2 (9.8 m/s²) (16 m − 0 m)

v = 17.7 m/s

The ball is moving at a speed of 17.7 m/s when it hits the ground.

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The Lamborghini Huracan has an initial acceleration of 0.75g. Its mass, with a driver, is 1510 kg.
Cerrena [4.2K]

Answer:

11109.825 N

Explanation:

Given Data:

total mass =m=1510 kg

initial acceleration (a) =0.75g                  ( g=9.81 m/s² )

F=ma

  = (1510)*( 0.75*9.81)

  = 11109.825 N

4 0
4 years ago
is the rate at which ____________________ changes. (Velocity or Acceleration, velocity or acceleration)
Troyanec [42]

<u>Acceleration</u> is the rate at which <u>velocity</u> changes.

8 0
3 years ago
A car accelerates from rest for 8.0 s, and reaches a speed of 66 m/s. How far will the
PSYCHO15rus [73]

Answer: C

Explanation:

Find the acceleration using this kinematic equation:

v_f=v_i+at\\66=0+a*8\\a=8.25m/s^2

Now use this kinematic equation to find the displacement:

v_f^2=v_i^2+2a*d\\66^2=0^2+2(8.25)d\\4356=16.5d\\d=264m

3 0
3 years ago
A circle has an initial radius of 50ft when the radius begins decreasing at a rate of 2ft/s. what is the rate of change of the a
valkas [14]
The area of the circle with radius r is
A = πr²

The rate of change of area with respect to time is
\frac{dA}{dt} = \frac{dA}{dr} . \frac{dr}{dt} =2 \pi r. \frac{dr}{dt}

The rate of change of the radius is given as
\frac{dr}{dt} =-2 \,  \frac{ft}{s}
Therefore
\frac{dA}{dt} =-4 \pi r \,  \frac{ft^{2}}{s}

When r = 10 ft, obtain
\frac{dA}{dt}|_{r=10 \, ft} = -40 \pi  \,  \frac{ft^{2}}{s}

Answer: - 40π ft²/s (or - 127.5 ft²/s)
7 0
4 years ago
A blow-dryer and a vacuum cleaner each operate with a voltage of 120 V. The current rating of the blow- dryer is 11 Amps, and th
AURORKA [14]

Answer:

(a) 1320 W

(b) 480 W

(c) E':E ≈ 11:2

Explanation:

(a) Applying,

P' = VI'................. Equation 1

Where P' = Power of the blow-dryer, V = Voltage, I = current rating of the blow-dryer.

From the question,

Given: V = 120 V, I' = 11 A

Substitute these values into equation 1

P = (120×11)

P = 1320 W

(b) Similarly,

P = VI................... Equation 2

Where P = Power of the vacuum cleaner. I = current rating of the vacuum cleaner.

Also Given: I = 4 A,

Therefore

P = 4(120)

P = 480 W

(c)

E' = P'/t'............. Equation 3

E = P/t................ Equation 4

Where E' = Energy of the blow-dryer, t' = time of use of the blow-dryer, E = Energy of the vacuum cleaner, t = time of use of the vacuum cleaner

From the question,

Given: t' = 15 minutes = (15×60) = 900 seconds, t = 30 minutes = (30×60) = 1800 seconds

Substitute these values into equation 3 and 4

E' = 1320/900

E' = 1.47 J,

E = 480/1800

E = 0.267

Therefore,

E':E = 1.47:0.267

E':E ≈ 11:2

5 0
3 years ago
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