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3 years ago

Jogger runs 400 m East in 120 s, then turns around and runs 250 m West in 90 seconds. Calculate the following values.

1 answer:

8 0

First, we must recall that distance is the total length traveled by an object in which the direction of the motion does not matter. Meanwhile, displacement is the distance of the object from its starting point which means the directions matters for displacement.

Now, speed is distance over time while velocity is displacement over time. Since we're talking about the same object's motion, the total time traveled is (120 + 90) = 210 seconds.

Now, the total distance traveled by the object is (400 + 250) = 650 m. Meanwhile the total displacement traveled by the object is 400 m, East + 250 m, West = 150 m, East.

Now, to find the speed and velocity, we just divide the values of distance and displacement, respectively, over time. Thus, we have

speed = 650 m / 210s ≈ 3.095 m/s
velocity = 150 m, East /210 s ≈ 0.714 m/s, East

Now, rounding the values up to 2 significant digits, we have

a) speed = 3.1 m/s
b) velocity = 0.71 m/s, East

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When Carbon -14 decays into Nitrogen -14, what kind of decay is this?

Nutka1998 [239]

Answer:

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Explanation:

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4 years ago

If 61.4 cm of copper wire (diameter = 1.08 mm, resistivity = 1.69 × 10-8Ω·m) is formed into a circular loop and placed perpendic

harkovskaia [24]

Answer:

the thermal energy generated in the loop = $6.64*10^{-4} \ W$

Explanation:

Given that;

The length of the copper wire L = 0.614 m

Radius of the loop r = $\frac{L}{2 \pi}$

r = $\frac{0.614}{2 \pi}$

r = 0.0977 m

However , the area of the loop is :

$A_L = \pi r^2$

$A_L = \pi (0.0977)^2$

$A_L = 0.02999 \ m^2$

Change in the magnetic field is $\frac{dB}{dt}= 0.0914 \ T/s$

Then the induced emf e = $A_L \frac{dB}{dt}$

e = $0.02999 * 0.0914$

e = 2.74 × 10⁻³ V

resistivity of the copper wire $\rho = 1.69* 10^{-8}$ Ω m

diameter of the wire = 1.08 mm

radius of the wire = 0.54 mm = 0.54 × 10⁻³ m

Thus, the resistance of the wire R = $\frac {\rho L}{\pi r^2}$

R = $\frac{(1.69*10^{-8})(0.614)}{ \pi (0.54*10^{-3})^2}$

R = 1.13× 10⁻² Ω

Finally, the thermal energy generated in the loop (i.e the power) = $\frac{e^2}{R}$

= $\frac{(2.74*10^{-3})^2}{1.13*10^{-2}}$

= $6.64*10^{-4} \ W$

8 0

3 years ago

when the armature of an ac generatr rotates at 15.0 rad/s, the amplitude of the induced emf is 27.0 V. What is the amplitude of

nata0808 [166]

To solve this problem we will apply the concepts related to the electric field. This is defined as the product between the angular frequency, the number of turns of the body (solenoid in this case) the magnetic field and the sine of the angular frequency and time. Mathematically this can be described as

$E = \omega NBA |sin \omega t|$