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2 years ago

5

A boat is traveling upstream at 14 km/h with respect to the water of a river. The water itself is flowing at 9 km/h with respect

to the ground. What is the velocity of the boat with respect to the ground

1 answer:

kirill [66]2 years ago

6 0

Answer:

16.64 km/h

Explanation:

Think of the x and y components as the a and b sides of a triangle and simply do the Pythagorean Theorem:

$a^2+b^2=c^2$

$14^2+9^2=c^2$

$\sqrt{196+81}=c$

$\sqrt{277}=c$

$16.64 km/h$

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A ball is thrown straight up into the air and passes a window 3 seconds after being released. It passes the same window on its w

melomori [17]

The initial velocity of the ball is 55.125 m/s.

<h3>Initial velocity of the ball</h3>

The initial velocity of the ball is calculated as follows;

During upward motion

h = vi - ¹/₂gt²

h = vi - 0.5(9.8)(3²)

h = vi - 44.1 ----------------- (1)

During downward motion

h = vi + ¹/₂gt²

h = 0 + 0.5(9.8)(1.5)²

h = 11.025 ----------- (2)

solve (1) and (2) together, to determine the initial velocity of the ball

11.025 = vi - 44.1

vi = 11.025 + 44.1

vi = 55.125 m/s

Thus, the initial velocity of the ball is 55.125 m/s.

Learn more about initial velocity here: brainly.com/question/19365526

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9 months ago

a sphere of diameter 6•0cm is moulded into a thin uniform wire of diameter 0•2mm.calculate the length of the wire in metres

Scilla [17]

The length of the wire is 36 m.

<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

We know that, radius can be found by taking the half in the diameter value. So,

$\text { sphere radius, } R=\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m}$

Similarly,

$\text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}$

We know the below formulas,

$\text {volume of sphere}=\frac{4}{3} \times \pi \times R^{3}$

$\text {volume of wire}=\pi \times r^{2} \times l$

When equating both the equations, we can find length of wire as below, where $\pi=\frac{22}{7}$

$\frac{4}{3} \times \pi \times R^{3}=\pi \times r^{2} \times l$

$\frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l$

The $\pi$ value gets cancelled as common on both sides, we get

$\frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l$

The $10^{-6}$ value gets cancelled as common on both sides, we get

$l=4 \times 9=36 m$

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2 years ago

A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 6.0s

pashok25 [27]

Answer:

109656.25 Nm

Explanation:

$\omega_f$ = Final angular velocity = 1.5 rad/s

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

t = Time taken = 6 s

m = Mass of disk = 29000 kg

r = Radius = 5.5 m

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}29000\times 5.5^2\times 0.25\\\Rightarrow \tau=109656.25\ Nm$

The torque specifications must be 109656.25 Nm

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no cap bro bro i thinks its poop Explanation:

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2 years ago

A child is sliding down a slide at the playground. is the mechanical energy conserved. why or why not.

aleksandrvk [35]

C: the mechanical energy isn't conserved. Some energy was lost to friction.

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2 years ago

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