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3 years ago

Every winter I fly to Michigan. It's a total distance of 3900km. It takes 5 hours. What is my average speed?

2 answers:

7 0

Your average speed is exactly 3900 km per 5 hours.

In more familiar units,

(3900 km) / (5 hr) = (3900/5) (km/hr) = 780 km/hr

olga_2 [115]3 years ago

4 0

$\frac{3900}{5} = 780\sf ~kilometer~per~hour$

Your average speed is 780 kilometer per hour.

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A car has a kinetic energy of 103kJ.

sergij07 [2.7K]

Answer: 1200kg

Explanation:

KE = (1/2)mv^2

103kJ = 103000J

103000J = (1/2) * m * (13.1m/s)^2

Solve for m

5 0

2 years ago

A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.13 s. W

vitfil [10]

Answer:

Part a)

T = 0.52 s

Part b)

$f = 1.92 Hz$

Part c)

$speed = 3.65 m/s$

Explanation:

As we know that the particle move from its maximum displacement to its mean position in t = 0.13 s

so total time period of the particle is given as

$T = 4\times 0.13 = 0.52 s$

now we have

Part a)

T = time to complete one oscillation

so here it will move to and fro for one complete oscillation

so T = 0.52 s

Part b)

As we know that frequency and time period related to each other as

$f = \frac{1}{T}$

$f = \frac{1}{0.52}$

$f = 1.92 Hz$

Part c)

As we know that

wavelength = 1.9 m

frequency = 1.92 Hz

so wave speed is given as

$speed = wavelength \times frequency$

$speed = 1.92 \times 1.9$

$speed = 3.65 m/s$

4 0

3 years ago

What is it called when sediment is dropped and comes to rest?

rosijanka [135]

It's called deposition.

7 0

3 years ago

A uniform plank 8.00 m in length with mass 50.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the

andreev551 [17]

To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.

According to Newton's second law we have to

$F = mg$

where,

m= mass

g = gravitational acceleration

For the balance to break, there must be a mass M located at the right end.

We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.

In this way, applying the static equilibrium equations, we have to sum up torques at point B,

$\sum \tau = 0$

Regarding the forces we have,

$3Mg-1mg=0$

Re-arrange to find M,

$M = \frac{m}{3}$

$M = \frac{50}{3}$

$M = 16.67Kg$

Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg

8 0

3 years ago

A unit of mass is ??

Effectus [21]

Hello,

Answer: kilogram

Further explaining: in science is used to measure weight of an object and used for accreditation.
Hope this helps!

3 0

3 years ago