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AURORKA [14]
3 years ago
11

Winner of five swimming gold medals at the 1988 olympics

Physics
1 answer:
Fed [463]3 years ago
6 0
Matt Biondi..?
(I don’t know if it’s right, sorry if it is wrong)
:)
You might be interested in
Two UFPD are patrolling the campus on foot. To cover more ground, they split up and begin walking in different directions. Offic
miss Akunina [59]

Answer:

0.256 hours

Explanation:

<u>Vectors in the plane </u>

We know Office A is walking at 5 mph directly south. Let X_A be its distance. In t hours he has walked

X_A=5t\ \text{miles}

Office B is walking at 6 mph directly west. In t hours his distance is

X_B=6t\ \text{miles}

Since both directions are 90 degrees apart, the distance between them is the hypotenuse of a triangle which sides are the distances of each office

D=\sqrt{X_A^2+X_B^2}

D=\sqrt{(5t)^2+(6t)^2}

D=\sqrt{61}t

This distance is known to be 2 miles, so

\sqrt{61}t=2

t =\frac{2}{\sqrt{61}}=0.256\ hours

t is approximately 15 minutes

3 0
3 years ago
The potential difference between points A and B in an electric
Flauer [41]

Answer:

i hope  the answear is D becuase  went over this long time ago when i was like you

Explanation:

3 0
3 years ago
A 0.03 kg golf ball is hit off the tee at a speed of 34 m/s. The golf club was in contact with the ball for 0.003 s. What is the
Liula [17]

Answer:

The average force on ball by the golf club is 340 N.

Explanation:

Given that,

Mass of the golf ball, m = 0.03 kg

Initial speed of the ball, u = 0

Final speed of the ball, v = 34 m/s

Time of contact, \Delta t=0.003\ s

We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :

F=\dfrac{\Delta p}{\Delta t}\\\\F=\dfrac{mv-mu}{\Delta t}\\\\F=\dfrac{mv}{\Delta t}\\\\F=\dfrac{0.03\ kg\times 34\ m/s}{0.003\ s}\\\\F=340\ N

So, the average force on ball by the golf club is 340 N.

4 0
4 years ago
X = v xo t. x = (10.0 m/s)(3.53 s) x = ????
LenaWriter [7]

x = V<em>x</em> * t

given V<em>x</em> = 10m/s n t = 3.53s

x = 10 * 3.53

= 35.3m


7 0
3 years ago
Read 2 more answers
What is the resistance of the coil A at 600 kelvin if its resistance at 300 kelvin is 50 ohms? (Assume the temperature coefficie
leonid [27]

155Ω

Explanation:

R = R ref ( 1 + ∝ ( T - Tref)  

where R = conduction resistance at temperature T

R ref = conductor resistance at reference temperature

∝ = temperature coefficient of resistance for conductor

T = conduction temperature in degrees Celsius

T ref = reference temperature that ∝ is specified at for the conductor material

T = 600 k - 273 k = 327 °C

Tref = 300 - 273 K = 27 °C

R = 50 Ω ( 1 + 0.007 ( 327 - 27) )

R = 155Ω

8 0
3 years ago
Read 2 more answers
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