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miss Akunina [59]
2 years ago
15

It took a squirrel 0.50\,\text s0.50s0, point, 50, start text, s, end text to run 5.0\,\text m5.0m5, point, 0, start text, m, en

d text leftward to a nearby tree while maintaining a constant acceleration. The squirrel was running 15\,\dfrac{\text m}{\text s}15 s m ​ 15, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction leftward when it reached the tree.
Physics
2 answers:
Alexandra [31]2 years ago
7 0

Answer:

-5

Explanation:

khan academy

STALIN [3.7K]2 years ago
3 0

Answer:

-5.0m/s

Explanation:

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A baseball is hit with a speed of 33.6 m/s. Calculate its height and the distance traveled if it is hit at angles of 30.0°, 45.0
Solnce55 [7]

Answer:

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m

Explanation:

Let ∝ be the angle with respect to the horizontal that the baseball is hit with.

The horizontal component of the velocity is vcos(∝) and the vertical component of the velocity is vsin(∝).

Ignore air resistant, only gravitational acceleration g = -9.81 m/s2 affect the ball vertically. We can use the following equation to calculate the time it takes to reach maximum height (at 0 speed)

vsin(\alpha) + gt = 0

t = \frac{-vsin(\alpha)}{g}

So the vertical distance it travels within time t is

y = vsin(\alpha)t + gt^2/2 = vsin(\alpha)\frac{-vsin(\alpha)}{g} + g\frac{(-vsin(\alpha))^2}{2g^2}

y = \frac{-v^2sin^2(\alpha)}{g} + frac{v^2sin^2(\alpha)}{2g}

y = \frac{-v^2sin^2(\alpha)}{2g}

Similarly the horizontal distance it travels within time t is:

x = vcos(\alpha)t = vcos(\alpha)\frac{-vsin(\alpha)}{g}

x = \frac{-v^2sin(2\alpha)}{2g}

We can pre-calcualte the quantity \frac{-v^2}{2g} = \frac{-33.6^2}{2*(-9.81)} = 57.54

So y = 57.54sin^2(\alpha)

x = 57.54sin(2\alpha)

From here we can plug-in the angles values of 30, 45 and 60 degrees

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m

8 0
2 years ago
A child on a 2.4 kg scooter at rest throws a 2.2 kg ball. The ball is given a speed of 3.1 m/s and the child and scooter move in
kykrilka [37]

Answer:

The child's mass is 14.133 kg

Explanation:

From the principle of conservation of linear momentum, we have;

(m₁ + m₂) × v₁ + m₃ × v₂ = (m₁ + m₂)  × v₃ - m₃ × v₄

We include the negative sign as the velocities were given as moving in the opposite directions

Since the child and the ball are at rest, we have;

v₁ = 0 m/s and v₂= 0 m/s

Hence;

0 = m₁ × v₃ - m₂ × v₄

(m₁ + m₂)× v₃ = m₃ × v₄

Where:

m₁ = Mass of the child

m₂ = Mass of the scooter = 2.4 kg

v₃ = Final velocity of the child and scooter = 0.45 m/s

m₃ = Mass of the ball = 2.4 kg

v₄ = Final velocity of the ball = 3.1 m/s

Plugging the values gives;

(m₁ + 2.4)× 0.45 = 2.4 × 3.1

(m₁ + 2.4) = 16.533

∴ m₁ + 2.4 = 16.533

m₁ = 16.533 - 2.4 = 14.133 kg

The child's mass = 14.133 kg.

3 0
2 years ago
Problem: An ice hockey player hits a puck of mass 0.15 kilograms with a force of 100 newtons in the horizontal direction. What i
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The acceleration produced in a body is always in the direction of the resultant force acting on the body. Therefore, we may determine the horizontal acceleration using the horizontal force applied. To do this, we may apply the mathematical form of Newton's second law:

Force = mass * acceleration
acceleration = force / mass

Substituting the values,
a = 100 / 0.15
a = 666.7 m/s²

The acceleration of the hockey puck is 670 m/s²
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