Motion is a change in 1.Time 2.Speed 3.Position 4.Velocity

A 34n force is applied to a 213kg mass how much does the mass accelerate

motikmotik

We Know, F = m*a
Here, F = 34 N
m = 213 Kg

Substitute their values in the equation,
34 = 213 * a
a = 34/213
a = 0.159 m/s²

So, your final answer & the acceleration of the object would be 0.159 m/s²

Hope this helps!

3 0

3 years ago

Question 3 You see the coordinates 5 E longitude, 10 N latitude. You do not need to look at a map in order to deduce that this l

san4es73 [151]

Answer:

a. Near both the equator and the prime meridian.

Explanation:

The equator is at 0 degrees latitude and the prime meridian is 0 degrees longitude.

5 0

3 years ago

Read 2 more answers

On the way to school, Jed traveled 100m north, 300m east, 100 north, 100m east, and 100m north. A.) Find the total distance trav

Oduvanchick [21]

Answer:

Total distance = 700 m

Displacement = 500 m

Explanation:

Notice that Jed travelled a total of 3 x 100 m = 300 m in the North direction, and 300 m + 100 m = 400 m in the East direction. Therefore the total distance he travelled is: 300 + 400 = 700 m.

But the actual displacement is given by the Pythagorean theorem as the hypotenuse of a right angle triangle of legs 300 m and 400 m:

displacement = $\sqrt{300^2+400^2} =\sqrt{250000} =500\,m$

5 0

3 years ago

Select the correct statement to describe when a sample of liquid water vaporizes into water vapor

sweet-ann [11.9K]

Answer:

This procces is called evaporation.

Explanation:

When you have liquid water that is transformed into steam, a phase change is called evaporation. The temperature for the evaporation of water depends on the pressure, for example for water at atmospheric pressure the temperature of evaporation is equal to 100°C. as the pressure increases are achieved evaporation temperatures higher. When that happens, the phase change temperature of the water is not increasing, as the process that takes place is the transfer of latent heat and applies only to changes of phase, that is to say at atmospheric pressure when it has 100% of the steam this will be at 101°C.

8 0

3 years ago

A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.

kap26 [50]

Answer:

The frequencies are $(f, f_1) = (6615.4 \ Hz , 19846.2\ Hz)$

Explanation:

From the question we are told that

The length of the ear canal is $l = 1.3 \ cm =\frac{1.3}{100} = 0.013 \ m$

The speed of sound is assumed to be $v_s = 344 \ m/s$

Now taking look at a typical ear canal we see that we assume it is a closed pipe

Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

$f = \frac{v_s}{4 * l }$

substituting values

$f = \frac{344}{4 * 0.013 }$

$f = 6615.4 \ Hz$

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

$f_1 = \frac{3v_s}{4 * l}$

substituting values

$f_1 = \frac{3 * 344}{4 * 0.013}$

$f_1 = 19846.2 \ Hz$

Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies

6 0

3 years ago