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Nataly_w [17]
2 years ago
12

A Turtle and a Snail are 360 meters apart, and they start to move towards each other at 3 p.m. If the Turtle is 11 times as fast

as the Snail, and they met at 3:40 p.m., find the speed of each.A Turtle and a Snail are 360 meters apart, and they start to move towards each other at 3 p.m. If the Turtle is 11 times as fast as the Snail, and they met at 3:40 p.m., find the speed of each.A Turtle and a Snail are 360 meters apart, and they start to move towards each other at 3 p.m. If the Turtle is 11 times as fast as the Snail, and they met at 3:40 p.m., find the speed of each.A Turtle and a Snail are 360 meters apart, and they start to move towards each other at 3 p.m. If the Turtle is 11 times as fast as the Snail, and they met at 3:40 p.m., find the speed of each.
Physics
2 answers:
Neko [114]2 years ago
8 0

Answer:

Snail's speed = \frac{30m}{2400s} = 0.0125m/s

Turtle's speed =  \frac{330m}{2400s} = 0.1375m/s

Explanation:

Let the snail's speed be x m/s

The turtle's speed then is 11x m/s

Speed = Distance ÷ Time

Since speed and distance are directly proportional;

The ratio of the distances snail and turtle cover before they meet is x:11x respectively.

Simplified, the ratio of snail distance : turtle distance = 1:11

So snail covers a distance of \frac{1}{12} × 360 = 30m

And turtle covers a distance of \frac{11}{12} × 360 = 330m

The time each took before they met is 40 × 60 = 2400 seconds

Snail's speed = \frac{30m}{2400s} = 0.0125m/s

Turtle's speed =  \frac{330m}{2400s} = 0.1375m/s

Delvig [45]2 years ago
6 0

Answer:

Turtle: 495

Snail: 45

Explanation:

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The correct answer is C) Technician A and B are both correct.

Explanation:

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Cheers!

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2 years ago
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An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and ind
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The answer is d.a capacitor
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2 years ago
If a current of 2.4 a is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?
Gnom [1K]

The average current density in the wire is given by:

J=\frac{I}{A}

where I is the current intensity and A is the cross-sectional area of the wire.


The cross-sectional area of the wire is given by:

A=\pi r^2

where r is the radius of the wire. In this problem, r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m, so the cross-sectional area is

A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2


and the average current density is

J=\frac{I}{A}=\frac{2.4 A}{3.14 \cdot 10^{-6} m^2}=7.64 \cdot 10^5 A/m^2

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A segment of wire of total length 2.0 m is formed into a circular loop having 5.0 turns. If the wire carries a 1.2-A current, de
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Answer:

Magnetic field at the center of the loop B=5.89\times 10^{-5}\ T.

Explanation:

It is given that total length of wire is 2 m and number of circular loop is 5 turns.

Therefore ,

5\times ( 2\pi r)=2 \ m .\\\\r=\dfrac{1}{5 \pi}=0.064\ m.

We know , magnetic field at the center of loop is given by :

B=N\dfrac{\mu_o i}{2r}

Putting all values in above equation we get :

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8 0
2 years ago
A BMX bicycle rider takes off from a ramp at a point 2.4 m above the ground. The ramp is angled at 40 degrees from the horizonta
adoni [48]

Answer:

The BMX lands 5.4 m from the end of the ramp.

Explanation:

Hi there!

The position of the BMX is given by the position vector "r":

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t

x0 = initial horizontal position

v0 = initial velocity

α = jumping angle

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

Please, see the attached graphic for a better understanding of the situation. At final time, when the bicycle reaches the ground, the vector position will be "r final" (see figure). The y-component of the vector "r final" is - 2.4 m (placing the origin of the frame of reference at the jumping point). With that information, we can use the equation of the y-component of the vector "r" (see above) to calculate the time of flight. With that time, we can then obtain the x-component (rx in the figure) of the vector "r final". Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

-2.4 m = 0 m + 5.9 m/s · t · sin 40° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 5.9 m/s · t · sin 40° + 2.4 m

Solving the quadratic equation:

t = 1.2 s

Now, we can calculate the x-component of the vector "r final" that is the horizontal distance traveled by the bicycle:

x = x0 + v0 · t · cos α

x = 0 m + 5.9 m/s · 1.2 s · cos 40°

x = 5.4 m

The BMX lands 5.4 m from the end of the ramp.

Have a nice day!

8 0
2 years ago
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