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Gnoma [55]
3 years ago
14

A bicycle traveled 150 meters west from point A to point B. Then it took the same route and came back to point A. It took a tota

l of 2 minutes for the bicycle to return to point A. What is the average speed and average velocity of the bicycle?
Physics
2 answers:
Zinaida [17]3 years ago
8 0

Answer:

speed=distance÷time

150÷0. 033

Evgesh-ka [11]3 years ago
7 0

Answer:

2.5 m/s

0 m/s

Explanation:

Average speed is distance over time.

Average velocity is displacement over time.

The bicycle travels 150 meters from A to B, then 150 meters from B to A.  So the distance is 300 meters and the displacement is 0 m.

s = 300 m / 120 s = 2.5 m/s

v = 0 m / 120 s = 0 m/s

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A major artery with a cross sectional area of 1.00cm^2 branches into 18 smaller arteries, each with an average cross sectional a
dybincka [34]

Here we can say that rate of flow must be constant

so here we will have

A_1v_1 = 18 A_2v_2

now we know that

A_1 = 1 cm^2

A_2 = 0.4 cm^2

now from above equation

1 cm^2 v_1 = 18(0.400 cm^2)v_2

\frac{v_2}{v_1} = \frac{1}{18\times 0.4}

\frac{v_2}{v_1} = 0.14

so velocity will reduce by factor 0.14

3 0
3 years ago
If an object is being pulled by two forces, one 4 N to the left and the other 2 N to the right, what is the net
nika2105 [10]

Answer:

2N

Explanation:

subtract  rthe two forces to see which is greater

4-2=2

6 0
3 years ago
The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming a
Vera_Pavlovna [14]

Answer:

The radius of the gold nucleus is 7.1x10⁻¹⁵m

Explanation:

The nearest distance is:

r=\frac{kze^{2} }{mpV^{2} } (eq. 1)

Where

z = atomic number of gold = 79

e = electron charge = 1.6x10⁻¹⁹C

k = electrostatic constant = 9x10⁹Nm²C²

energy of the particle = 32 MeV = 5.12x10⁻¹²J

At the potential energy is zero, all the energy will be kinetic energy:

E_{k} =\frac{1}{2} m V^{2}

Where

m = 4 mp = mass of proton

5.12x10^{-12} =\frac{1}{2} *4*m_{p}* V^{2} \\m_{p}* V^{2} = 2.56x10^{-12}

Replacing in equation 1

r=\frac{9x10^{9}*79*(1.6x10^{-19})^{2}   }{2.56x10^{-12} } =7.1x10^{-15} m

8 0
3 years ago
Express the kinetic energy K in terms of the potential energy U.<br><br><br> K=GMm/2R
max2010maxim [7]

Answer:

K = -½U

Explanation:

From Newton's law of gravitation, the formula for gravitational potential energy is;

U = -GMm/R

Where,

G is gravitational constant

M and m are the two masses exerting the forces

R is the distance between the two objects

Now, in the question, we are given that kinetic energy is;

K = GMm/2R

Re-rranging, we have;

K = ½(GMm/R)

Comparing the equation of kinetic energy to that of potential energy, we can derive that gravitational kinetic energy can be expressed in terms of potential energy as;

K = -½U

7 0
3 years ago
A car travels at a speed of 40 m/s for 29.0 s;<br>what is the distance traveled by the car?
lianna [129]

Answer: 1160 m

Explanation:

Speed = distance / time. Plug in 40 m/s for speed and 29 s for time in order to get the distance, 1160 m.

8 0
2 years ago
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