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bonufazy [111]
3 years ago
10

Steven carefully places a m = 1.85 kg wooden block on a frictionless ramp so that the block begins to slide down the ramp from r

est. The ramp makes an angle of θ = 51.3 ° up from the horizontal. Which forces do nonzero work on the block as it slides down the ramp?
Physics
1 answer:
Helen [10]3 years ago
4 0

Answer:

Gravity force

Explanation:

We are given that

Mass of wooden block,m=1.85 kg

\theta=51.3^{\circ}

We have to find the force which do non zero work on the block as it slides down the ramp.

When a block slides down the ramp on a frictionless surface .

Then, weight act on the wooden block in downward direction.

We know that weight is that force which act on the body due to gravity.

Then, we can say that the force which do nonzero work on the block as it slides down the ramp is gravity force.

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A transverse wave on a string is described by the wave functiony(x, t) = 0.350 sin (1.25x + 99.6t)where x and y are in meters an
ella [17]

The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

<h3>How to solve for the time interval</h3>

We have y = 0.175

y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.5

99.62 = pi/6

t1 = 5.257 x 10⁻³

99.6t = pi/6 + 2pi

= 0.0683

The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

b. we have k = 1.25, w = 99.6t

v = w/k

99.6/1.25 = 79.68

s = vt

= 79.68 * 0.0683

= 5.02

Read more on waves here

brainly.com/question/25699025

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complete question

A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?

7 0
1 year ago
Does Radiation Affects this question? If it doesn't, is the answer C?​
kotegsom [21]
The answer will be C
5 0
3 years ago
A bird carries a 25 g oyster to a height of 11m what is the gravitational potential energy of the oyster
garik1379 [7]

Given data:

Mass of oyster (m) = 25 g,

                              = 0.025 kg            (<em> since 1 Kg = 1000 g)</em>

Height of oyster (h) = 11 m,

determine gravitational potential energy (GPE) = ?

Gravitational potential energy is energy of an object possesses because of its position on the earth. The amount of gravitational potential energy of an object on earth depends on its<em> mass</em> and<em> height </em>from the earth.  

Mathematically,

        GPE = m. g. h <em>Joules</em>

<em>         Where, </em>

<em>               </em>GPE =<em> </em>gravitational potential energy,

<em>         </em>          m = mass of object in Kg,

                  g = gravitational force in m/s²,     g = 9.8 m/s²

                  h = height of the object in meters,

     Now substituting all the values in the equation GPE

           GPE = 0.025 Kg × 9.8 m/s² × 11 m

                   = 2.695 J

Gravitational potential energy of the oyster is 2.695 J


7 0
3 years ago
What currents are responsible for powering the movement of tectonic plates ? A. Platonic B. Tectonic C. Convection D. Plutonium
Ivan
The currents responsible for powering the movement of tectonic plates is convection currents which occur in the mantle. As the hotter, less dense liquid rises it displaces the cooler more dense liquid which moves the tectonic plates out of allignment
3 0
3 years ago
Read 2 more answers
Which of these charges is experiencing the electric field with the largest magnitude? A 2C charge acted on by a 4 N electric for
Pavlova-9 [17]

Answer:

The highest electric field is experienced by a 2 C charge acted on by a 6 N electric force. Its magnitude is 3 N.

Explanation:

The formula for electric field is given as:

E = F/q

where,

E = Electric field

F = Electric Force

q = Charge Experiencing Force

Now, we apply this formula to all the cases given in question.

A) <u>A 2C charge acted on by a 4 N electric force</u>

F = 4 N

q = 2 C

Therefore,

E = 4 N/2 C = 2 N/C

B) <u>A 3 C charge acted on by a 5 N electric force</u>

F = 5 N

q = 3 C

Therefore,

E = 5 N/3 C = 1.67 N/C

C) <u>A 4 C charge acted on by a 6 N electric force</u>

F = 6 N

q = 4 C

Therefore,

E = 6 N/4 C = 1.5 N/C

D) <u>A 2 C charge acted on by a 6 N electric force</u>

F = 6 N

q = 2 C

Therefore,

E = 6 N/2 C = 3 N/C

E) <u>A 3 C charge acted on by a 3 N electric force</u>

F = 3 N

q = 3 C

Therefore,

E = 3 N/3 C = 1 N/C

F) <u>A 4 C charge acted on by a 2 N electric force</u>

F = 2 N

q = 4 C

Therefore,

E = 2 N/4 C = 0.5 N/C

The highest field is 3 N, which is found in part D.

<u>A 2 C charge acted on by a 6 N electric force</u>

3 0
3 years ago
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