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OLEGan [10]
3 years ago
5

Sylvia and Jadon now want to work a problem. Imagine a puck of mass 0.5 kg moving as in the simulation. Suppose that the tension

in the string is 1.0 N, and that the radius of its circular path is 0.7 m. What will Jadon and Sylvia find for the tangential speed of the puck? m/s
Physics
1 answer:
Pavlova-9 [17]3 years ago
5 0

Answer:

1.2 m/s

Explanation:

The puck in this problem is moving with uniform circular motion, so the net force acting on it (the tension in the string) must be equal to the centripetal force.

So we can write:

T=m\frac{v^2}{r}

where:

T is the tension in the string

m is the mass of the puck

v is its tangential speed

r is the radius of the circular path

For the puck in this problem, we have:

m = 0.5 kg

T = 1.0 N

r = 0.7 m

Substituting and solving for v, we find the tangential speed:

v=\sqrt{\frac{Tr}{m}}=\sqrt{\frac{(1.0)(0.7)}{0.5}}=1.2 m/s

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Ulleksa [173]
Distance traveled by him = circumference of that circular path = 2πr = 2π(3.5)
= 7π = 7×3.14 = 21.98 m
time = 8.9 s  [ Given ]

Now, Average speed = distance / time
s = 21.98 / 8.9
s = 2.46 m/s

Hope this helps!
7 0
3 years ago
What is the resistance (R) when voltage is 179V and current is 5 Amps?
Evgesh-ka [11]

Answer:

R = 35.8 Ω

Explanation:

Recall Ohm's Law:

V = I * R

then R = V / I

in our case:

R = 179 V / 5 A = 35.8 Ω

3 0
2 years ago
An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo
Ymorist [56]

Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =<u>16321.0769 N/C</u>

The electric field strength at point (x,y) = (0cm, 20 mm) is =<u>35321.58999 N/C</u>

Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

= 0.135 ÷ (8.271513x10^-6)

=<u>16321.0769 N/C</u>

 Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷  0.000000152881

=<u>35321.58999 N/C</u>

8 0
2 years ago
A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is at rest at a stopligt. The car comes to a
Arlecino [84]

Answer:

<em><u>M</u></em><em><u>a</u></em><em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u>m</u></em><em><u>a</u></em><em><u>t</u></em><em><u>i</u></em><em><u>c</u></em><em><u>a</u></em><em><u>l</u></em><em><u>l</u></em><em><u>y</u></em><em><u>:</u></em>

That will be

<em>=</em><em> </em><em>1</em><em>5</em><em>0</em><em>0</em><em> </em><em>x</em><em> </em><em>1</em><em>5</em><em> </em><em>x</em><em> </em><em>4</em><em>5</em><em>0</em><em>0</em>

<em>=</em><em> </em><em><u>1</u></em><em><u>0</u></em><em><u>1</u></em><em><u>,</u></em><em><u>2</u></em><em><u>5</u></em><em><u>0</u></em><em><u>,</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>

5 0
2 years ago
At a particular instant, a proton at the origin has velocity &lt; 5e4, -2e4, 0&gt; m/s. You need to calculate the magnetic field
vesna_86 [32]

Answer:

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Explanation:

Velocity =5\times 10^4i-2\times 10^4j

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v=\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}

We know that

B=\frac{mv}{qr}

Where q=1.6\times 10^{-19} C

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Using the formula

B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}

B=9.7\times 10^{-5} T

3 0
3 years ago
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