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ser-zykov [4K]
3 years ago
13

A toy car with an initial velocity of 5 m/s slows to a stop with an acceleration of -1.5 m/s^2.

Physics
1 answer:
enot [183]3 years ago
4 0

Car's Initial velocity (u) = 5 m/s

Car's Final velocity (v) = 0 m/s (Stop)

Acceleration of the car (a) = -1.5 m/s²

Equation used to solve this problem:

\boxed{ \bf{ {v}^{2}  =  {u}^{2}  + 2as}}

By substituting values in the equation, we get:

\rm \longrightarrow  {0}^{2}  =  {5}^{2}  +2  \times ( - 1.5) \times s \\  \\   \rm \longrightarrow 0 = 25 - 3s \\  \\  \rm \longrightarrow 25 - 3s = 0 \\  \\  \rm \longrightarrow 25 - 25 - 3s =0  - 25 \\  \\  \rm \longrightarrow  - 3s =  - 25 \\  \\  \rm \longrightarrow 3s = 25 \\  \\  \rm \longrightarrow  \dfrac{3s}{3}  =  \dfrac{25}{3}  \\  \\  \rm \longrightarrow s = 8.3 \: m

\therefore Displacement of the toy car (s) = 8.3 m

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Answer:

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7 0
3 years ago
What is the magnitude of your displacement when you follow directions that tell you to walk 100.0m north, then 25.0m East?
atroni [7]

Answer:

Displacement from the starting position is 103.21m

Explanation:

If you draw these directions, it will create the two legs of a triangle.

Using this method, you can visualize why your displacement is what it is.

Using the pythagorean theorem

{a}^{2}  +  {b}^{2}  =  {c}^{2}

Plug in both values

{100m}^{2}  +  {25m}^{2}  =  {c}^{2}

c =  \sqrt{10652}

c = 103.2085

c= 103.21

6 0
3 years ago
Light travels at 3 × 108 m/s, and it takes about 8 min for light from the sun to travel to Earth. Based on this, the order of ma
N76 [4]

Answer:

The order of magnitude of the distance from the sun to Earth is 10⁸ km.

Explanation:

The order of magnitude of the distance from the sun to Earth can be calculated as follows:

c = \frac{x}{t}

Where:

c: is the speed of light = 3x10⁸ m/s

t: is the time = 8 min

Hence, the distance is:

x = c*t = 3 \cdot 10^{8} m/s*8 min*\frac{60 s}{1 min} = 1.44 \cdot 10^{11} m = 1.44 \cdot 10^{8} km

Therefore, the order of magnitude of the distance from the sun to Earth is 10⁸ km.

I hope it helps you!

5 0
3 years ago
A student pushes a 0.2 kg box against a spring causing the spring to compress 0.15 m. When the spring is released, it will launc
german

Answer:

The maximum height the box will reach is 1.72 m

Explanation:

F = k·x

Where

F = Force of the spring

k = The spring constant = 300 N/m

x  = Spring compression or stretch = 0.15 m

Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N

Mass of box = 0.2 kg

Work, W, done by the spring = \frac{1}{2} kx^2 and the kinetic energy gained by the box is given by KE = \frac{1}{2} mv^2

Since work done by the spring = kinetic energy gained by the box we have

\frac{1}{2} mv^2 =  \frac{1}{2} kx^2  therefore we have v = \sqrt{\frac{kx^2}{m} } = x\sqrt{\frac{k}{m} } = 0.15\sqrt{\frac{300}{0.2} } = 5.81 m/s

Therefore the maximum height is given by

v² = 2·g·h or h = \frac{v^2}{2g} = \frac{5.81^{2} }{2*9.81} = 1.72 m

6 0
3 years ago
If the force on a hammer is 24 N and its mass is 1.6 kg, then the
Gnesinka [82]

Answer:

15 m/s²

Explanation:

F = ma

make "a" the subject

a = F/m

6 0
3 years ago
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