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kozerog [31]
3 years ago
11

Two children, Ferdinand and Isabella, are playing with a water hose on a sunny summer day. Isabella is holding the hose in her h

and 1.0 meters above the ground and is trying to spray Ferdinand, who is standing 10.0 meters away. Will Isabella be able to spray Ferdinand if the water is flowing out of the hose at a constant speed v0 of 3.5 meters per second?
Physics
1 answer:
mojhsa [17]3 years ago
8 0

Answer:

Isabella will not be able to spray Ferdinand.

Explanation:

We'll begin by calculating the time taken for the water to get to the ground from the hose held at 1 m above the ground. This can be obtained as follow:

Height (h) = 1 m

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

Time (t) =.?

h = ½gt²

1 = ½ × 9.8 × t²

1 = 4.9 × t²

Divide both side by 4.9

t² = 1/4.9

Take the square root of both side

t = √(1/4.9)

t = 0.45 s

Next, we shall determine the horizontal distance travelled by the water. This can be obtained as follow:

Horizontal velocity (u) = 3.5 m/s

Time (t) = 0.45 s

Horizontal distance (s) =?

s = ut

s = 3.5 × 0.45

s = 1.58 m

Finally, we shall compare the distance travelled by the water and the position to which Ferdinand is located to see if they are the same or not. This is illustrated below:

Ferdinand's position = 10 m

Distance travelled by the water = 1.58 m

From the above, we can see that the position of the water (i.e 1.58 m) and that of Ferdinand (i.e 10 m) are not the same. Thus, Isabella will not be able to spray Ferdinand.

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marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force F_{1}=18\ N

Force F_{2}=26\ N

Force F_{3}=14\ N

We need to calculate the torque due to force F₁

Using formula of torque

\tau_{1}=-F_{1}d_{1}

\tau_{1}=-F_{1}\times\dfrac{a}{2}

Put the value into the formula

\tau_{1}=-18\times\dfrac{0.2}{2}

\tau_{1}=-1.8\ N-m

We need to calculate the torque due to force F₂

Using formula of torque

\tau_{2}=F_{2}d_{2}

\tau_{2}=F_{2}\times\dfrac{a}{2}

Put the value into the formula

\tau_{2}=26\times\dfrac{0.2}{2}

\tau_{2}=2.6\ N-m

We need to calculate the torque due to force F₃

Using formula of torque

\tau_{3}=F_{3}d_{3}

\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}

Put the value into the formula

\tau_{3}=0.1(14\sin45+14\cos45)

\tau_{3}=1.92\ N-m

We need to calculate the net torque on the square plate

\tau=\tau_{1}+\tau_{2}+\tau_{3}

\tau=-1.8+2.6+1.92

\tau=2.72\ N-m

Hence, The net torque on the square plate is 2.72 N-m.

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3 years ago
Which of the following statements concerning moral development is not true?
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Answer:

C- what is determined as right or wrong is universally defined and agreed upon.

Explanation:

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A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a
KatRina [158]

Answers:

a) 0.80 kg

b) 2.12 s

c) 1.093 m/s^{2}

Explanation:

We have the following data:

k=7 N/m is the spring constant

A=12.5 cm \frac{1 m}{100 cm}=0.125 m is the amplitude of oscillation

V=32 cm/s=0.32 m/s is the velocity of the block when x=\frac{A}{2}=0.0625 m

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

U_{o}+K_{o}=U_{f}+K_{f} (1)

Where:

U_{o}=k\frac{A^{2}}{2} is the initial potential energy

K_{o}=0  is the initial kinetic energy

U_{f}=k\frac{x^{2}}{2} is the final potential energy

K_{f}=\frac{1}{2} m V^{2} is the final kinetic energy

Then:

k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2} (2)

Isolating m:

m=\frac{k(A^{2}-x^{2})}{V^{2}} (3)

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m=0.80 kg (5)

<h3>b) Period</h3>

The period T is given by:

T=2 \pi \sqrt{\frac{m}{k}} (6)

Substituting (5) in (6):

T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}} (7)

T=2.12 s (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration a_{max} is when the force is maximum F_{max}, as well :

F_{max}=m.a_{max}=k.x_{max} (9)

Being x_{max}=A

Hence:

m.a_{max}=kA (10)

Finding a_{max}:

a_{max}=\frac{kA}{m} (11)

a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg} (12)

Finally:

a_{max}=1.093 m/s^{2}

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