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MA_775_DIABLO [31]
3 years ago
10

Titus drives his jetski a distance of 1000 meters in 7.045 seconds. How fast was he moving in meters per second? How fast was he

moving in miles per hour?
Physics
1 answer:
Ksenya-84 [330]3 years ago
5 0

Answer:

a) v = 141.9 m/s

b) v = 317.4 miles/h

Explanation:

a) How fast was he moving in meters per second?

v = \frac{d}{t} = \frac{1000 m}{7.045 s} = 141.9 m/s

Hence, the jet ski is moving at 141.9 meters per second.

b) How fast was he moving in miles per hour?

v = \frac{d}{t} = \frac{1000 m}{7.045 s} = 141.9 m/s*\frac{3600 s}{1 h}*\frac{1 mile}{1609.34 m} = 317.4 miles/h      

Therefore, the jet ski is moving at 317.4 miles per hour.

I hope it helps you!

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Recall that the spring constant is inversely proportional to the number of coils in the spring, or that shorter springs equate t
ruslelena [56]

Answer:

x_1= 0.0425m

Explanation:

Using the tension in the spring and the force of the tension can by describe by

T = kx

, T = mg

Therefore:

m*g = k*x

With two springs, let, T1 be the tension in each spring,  x1 be the extension of each spring.  The spring constant of each spring is 2k so:

T_1 = 2k*x_1

2T_1 = m*g=4k x_1

Solve to x1

x_1=\frac{m*g}{4k}

x_1=\frac{k*x}{4*k}

x_1=\frac{x}{4}

x_1 = 0.170 / 4

x_1= 0.0425m

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3 years ago
A rope.that is 1.50m long exhibits a standing wave pattern with 6 nodes if the frequency of the waves is 75.0 hz what is the vel
Sonja [21]
I think he answer is 34.32
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Compute the density in g/cm^3 of a piece of metal that had mass of 0.485 kg and a volume of 52cm^3
steposvetlana [31]

Answer:

9.3 g/cm³

Explanation:

First, convert kg to g:

0.485 kg × (1000 g / kg) = 485 g

Density is mass divided by volume:

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D = 9.33 g/cm³

Rounding to two significant figures, the density is 9.3 g/cm³.

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Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an
ddd [48]

Answer:

Al's mass is 102.92  kg  

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

F_{net} = 0

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

F_{net} = \frac{dp_{horizontal}}{dt} = 0

p_{horizontal_i }= p_{horizontal_f}

where the suffix i and f means initial and final respectively.

The initial momentum will be:

p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}

But, as they are at rest, initially

p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0

p_{horizontal_i} = 0

So, this means:

p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0

We know that the have an combined mass of 195 kg:

m_{total} = m_{Al} + m_{Jo} = 195 \ kg.

so:

m_{Jo} = 195 \ kg - m_{Al}.

m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0

m_{Al} \  v_{Al_f} - m_{Al} \  v_{Jo_f}= - 195 \ kg \  v_{Jo_f}

m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}

m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } {  v_{Al_f} - v_{Jo_f} }

m_{Al} = \frac{195 \ kg  \ v_{Jo_f} } {    v_{Jo_f} - v_{Al_f} }

Now, we can use the values:

v_{Al_f}= 10.2 \frac{m}{s}

v_{Jo_f}= - 11.4 \frac{m}{s}

where the minus sign appears as they are moving at opposite directions

m_{Al} = \frac{195 \ kg  ( - 11.4 \frac{m}{s} ) } {   (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }

m_{Al} = 102.92 \ kg

and this is the Al's mass.

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Answer: force is 0.42 N

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