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3 years ago

The________is a measure of the number of waves that pass a point in a given amount of time

Physics

2 answers:

madreJ [45]3 years ago

5 0

The ______ is a measure of the number of waves that pass a point in a given amount of time is called Frequency.

cricket20 [7]3 years ago

4 0

Frequency measures how many waves pass a certain point in a given amount of time.
High frequency means the waves are closer together and more frequent, making for a lower wave length.
Low frequency means the waves are farther apart and less frequent, making for a higher wave length.

So the answer is frequency.

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Which four equations can be used to solve for acceleration

emmainna [20.7K]

The four equations for acceleration are obtained from the three equations of motion and from second law of motion.

Explanation:

Acceleration is defined as the rate of change of velocity with respect to time. So the change in velocity with respect to time can be determined using the three equations of motions.

So from the first equation of motion, v = u + at , we can determine the value of acceleration if time taken, final and initial velocity is known. The equation can be re-written as $a = \frac{v-u}{t}$

Similarly, from the second equation of motion, s = ut + 1/2 at², we can determine the equation for acceleration as $a = 2*\frac{s-ut}{t^{2} }$

So this is second equation for acceleration.

Then from the third equation of motion, $v^{2}- u^{2} = 2* a *s$

the acceleration equation is determined as $a = \frac{v^{2}-u^{2} }{2s}$

In addition to these three equation, another equation is present to determine the acceleration with respect to force from the Newton's second law of motion. F = Mass × acceleration. From this, acceleration = Force/mass.

So, these are the four equations for acceleration.

8 0

3 years ago

You push the ball with aforce of 22.8N which induces a -2.3N frictional force. What is the net force while you push?

inna [77]

The net force is 22.8-2.3 or 20.5 N

5 0

3 years ago

The portion of earth underneath the crust containing the asthenosphere and the mesosphere is the

Hatshy [7]

Answer:

The portion of earth underneath the crust containing the asthenosphere and mesosphere is outer core and inner core .

Explanation:

The outer core of the earth is ball of very hot metals . Its temperature is hot enough to melt all metals into liquid state . This is mainly composed of the melted metals nickel and iron . This has got got temperature between 4000 degrees to 9000 degrees .

Inner core is underneath the outer core ,Here the pressure is too high and this makes the metals squeezed and not even able to move like liquid . They are forced to vibrate just like a solid . This has got a thickness of 800 miles . The pressure here is comparably 3,000,000 times the air pressure at sea level.

6 0

3 years ago

What information do you think the temperatures of stars give us?

Paul [167]

Explanation:

One way of classifying stars is by their temperature .

Science strives to be able to describe how stars and planets form and evolve. This requires theories to describe the processes which include:

Star and planet formation

Star and planet composition

Stellar and solar system evolution

The nuclear processes happening inside stars

The scientific method means that all theories are put to the test. By measuring or calculating the temperature, age and composition of other planets and stars the theories can be tested. If observed values of these parameters are not predicted by theories, then the theories are wrong and need to be revised or replaced.

6 0

3 years ago

Read 2 more answers

Problem 3: Thermal expansionThe steel rod has the length 2 m and cross-section area 200 cm2at the room temperature 20◦C. Weapply

zalisa [80]

Answer:

a) 2.00024 m

b) 0.036%

c) 436.67°C

Explanation:

Given

Initial length = L₀ = 2 m

Initial cross sectional Area = A₀ = 200 cm² = 0.02 m²

We can obtain initial volume = V₀ = A₀L₀ = 0.02 × 2 = 0.04 m³

Initial Temperature = T₀ = 20°C

Coefficient of linear expansivity = α = (2 × 10⁻⁶) (°C)⁻¹

a) New length of the rod after heating to 80°C

Linear expansion is given as

ΔL = L₀ × α ×ΔT

ΔL = 2 × 2 × 10⁻⁶ × (80 - 20) = 0.00024 m = 0.24 mm

New length = old length + expansion = 2 + 0.00024 = 2.00024 m

b) The percentage of the volume change of the rod.

Volume expansion is given by

ΔV = V₀ × (3α) × ΔT

Volume expansivity ≈ 3 × (linear expansivity)

ΔV = 0.04 × (3×2×10⁻⁶) × (80 - 20) = 0.0000144 m³

Percentage change in volume = 100% × (ΔV/V₀) = 100% × (0.0000144/0.04) = 0.036%

c) The maximal temperature we can allow if the volume should not increase by more than half percent.

For a half percent increase in volume, the corresponding change in volume needs to be first calculated.

Percentage change in volume = 100% × (ΔV/V₀)

0.5 = 100% × (ΔV/0.04)

(ΔV/0.04) = 0.005

ΔV = 0.0002 m³

Then we now investigate the corresponding temperature that causes this.

ΔV = V₀ × (3α) × ΔT

0.0002 = 0.04 × (3×2×10⁻⁶) × ΔT

ΔT = (0.0002)/(0.04 × 3 × 2 × 10⁻⁶) = 416.67°C

Maximal temperature = T₀ + ΔT = 20 + 416.67 = 436.67°C

4 0

3 years ago