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Lana71 [14]
3 years ago
11

Margaret wants to go for a swim, and decides to jump in using the diving board that measures 3-m long.

Physics
1 answer:
otez555 [7]3 years ago
6 0

Answer:

The horizontal component of her velocity is approximately 1.389 m/s

The vertical component of her velocity is approximately 7.878 m/s

Explanation:

The given question parameters are;

The initial velocity with which Margaret leaps, v = 8.0 m/s

The angle to the horizontal with which she jumps, θ = 80° to the horizontal

The horizontal component of her velocity, vₓ = v × cos(θ)

∴ vₓ = 8.0 × cos(80°) ≈ 1.389

The horizontal component of her velocity, vₓ ≈ 1.389 m/s

The vertical component of her velocity, v_y = v × sin(θ)

∴ v_y = 8.0 × sin(80°) ≈ 7.878

The vertical component of her velocity, v_y ≈ 7.878 m/s.

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Answer:

Final speed of the car, v = 24.49 m/s

Explanation:

It is given that,

Initial velocity of the car, u = 0

Acceleration, a=3.1\ m/s^2

Time taken, t = 7.9 s

We need to find the final velocity of the car. Let it is given by v. It can be calculated using first equation of motion as :

v=u+at

v=0+3.1\times 7.9

v = 24.49 m/s

So, the final speed of the car is 24.49 m/s. Hence, this is the required solution.

8 0
3 years ago
How far away is a cliff if an echo is heard 0.486 s after the original sound? Assume that sound traveled at 343 m/s on that day.
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Random errors can be reduced by taking repeated measurements.Error and uncertainty are interchangeable words that describe the s
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5 0
1 year ago
Which process is the most scientifically plausible explanation for how the organisms released oxygen into Earth's atmosphere? Gr
Stels [109]

Answer:

Photosynthesis

Explanation:

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8 0
3 years ago
Suppose that in a lightning flash the potential difference between a cloud and the ground is 0.96×109 V and the quantity of char
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(a) 2.98\cdot 10^{10} J

The change in energy of the transferred charge is given by:

\Delta U = q \Delta V

where

q is the charge transferred

\Delta V is the potential difference between the ground and the clouds

Here we have

q=31 C

\Delta V = 0.96\cdot 10^9 V

So the change in energy is

\Delta U = (31 C)(0.96\cdot 10^9 V)=2.98\cdot 10^{10} J

(b) 7921 m/s

If the energy released is used to accelerate the car from rest, than its final kinetic energy would be

K=\frac{1}{2}mv^2

where

m = 950 kg is the mass of the car

v is the final speed of the car

Here the energy given to the car is

K=2.98\cdot 10^{10} J

Therefore by re-arranging the equation, we find the final speed of the car:

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5 0
3 years ago
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