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Amiraneli [1.4K]
2 years ago
12

A basic physics question

Physics
1 answer:
Anna71 [15]2 years ago
3 0
<h3>1)</h3>

No,kinetic energy cannot be negative since its given by KE=mv²,mass cannot be negative and the square of speed cannot b negative.

<h3>2)</h3>

Yes,any force opposing motion or displacement does negative work. They are often referred to as resistive forces (friction,air resistance,drag...)

<h3>3)</h3>

Nope it does not, it just forces the object to move in a circular path known as a centripetal force. It can accelerate an object by changing it's direction but not it's speed.

<h3>4)</h3>

No it cannot,If an object is sliding on the table (assuming it is not an incline), then most probably that normal force cancels out the weights effect or assuming there is an incline, it cancels the weight's y component.

<h3>5)</h3>

w = f.d.cos \alpha  = f.d.cos90 = 0

The work done is zero

<h3>6)</h3>

w = f.d.cos \theta = f.d.cos0 = f.d

The work is just the product of the magnitude of the force exerted and the displacement of the object.

<h3>7)</h3>

0 <  \theta < 90 \\ 1 < cos \theta < 0

<h3>Work is decreasing but positive</h3>

<h3>8)</h3>

\theta = 90 \\ cos \theta = cos90 = 0

<h3>Work is zero</h3>

<h3>9)</h3>

90 <  \theta < 180 \\ 0 < cos \theta <  - 1

<h3>Work is negative</h3>
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Need help with two physics questions!
nikitadnepr [17]

1) The north component of the airplane velocity is 260 km/h.

2) The direction of the plane is 24^{\circ} north of east.

Explanation:

1)

In this problem, we have to resolve the velocity vector into its components.

Taking east as positive x-direction and north as positive y-direction, the components of the velocity along the two directions are given by:

v_x = v cos \theta

v_y = v sin \theta

where

v is the magnitude of the velocity

\theta is the angle between the direction of the velocity and the positive x-axis (the east direction)

For the airplane in this problem,

v = 750 km/h

\theta=20^{\circ}

So, the two components are

v_x = (750)(cos 20)=704.8 km/h

v_y = (750)(sin 20)=256.5 km/h

So, the component in the north direction is 256.5 km/h, so approximately 260 km/h.

2)

In this problem, we have to use vector addition.

In fact, the motion of the plane consists of two displacements:

- A first displacement of 220 km in the east direction

- A second displacement of 100 km in the north direction

Using the same convention of the same problem (x = east and y = north), we can write

d_x = 220 km

d_y = 100 km

Since the two vectors are perpendicular to each other, we can find their magnitude using Pythagorean's theorem:

d=\sqrt{d_x^2+d_y^2}=\sqrt{(220)^2+(100)^2}=241.7 km

And the direction is given by

\theta=tan^{-1}(\frac{d_y}{d_x})=tan^{-1}(\frac{100}{220})=24^{\circ} north of east.

Learn more about vectors here:

brainly.com/question/2678571

brainly.com/question/4945130

brainly.com/question/2678571

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

8 0
3 years ago
Estimate the force required to bind the two protons in the He nucleus together. (Hint: Model the protons as point charges. Assum
scoray [572]

Answer:

 F = 2.30 10⁴ N

Explanation:

The force required to link two gates must be equal to or greater than the electrostatic force of repulsion, because the protons have equal charges.

                 F = k q₁ q₂ / r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C²

   In this case the proton charge is 1.6 10⁻¹⁹ C and the distance between them is approximately the diameter of the core r = 10⁻¹⁵ m

Let's calculate

              F = 8.99 10⁹ (1.6 10⁻¹⁹)² / (10⁻¹⁵)²

              F = 2.30 10⁴ N

The bond strength must be equal to or greater than this value

3 0
3 years ago
Through what potential difference should electrons be accelerated so that their speed is 1.0 % of the speed of light when they h
omeli [17]

Answer:

Explanation:

Considering non - relativistic approach : ----

Speed of electron = 1 % of speed of light

= .01 x 3 x 10⁸ m /s

= 3 x 10⁶ m /s

Kinetic energy of electron = 1/2 m v²

= .5 x 9.1 x 10⁻³¹ x ( 3 x 10⁶ )²

= 40.95 x 10⁻¹⁹ J

Kinetic energy in electron comes from lose of electrical energy equal to

Ve where V is potential difference under which electron is accelerated and e is electronic charge .

V x e = kinetic energy of electron

V x 1.6 x 10⁻¹⁹ = 40.95 x 10⁻¹⁹

V = 25.6 Volt .

6 0
3 years ago
To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(
DerKrebs [107]

Answer:

0.0549 m

Explanation:

Given that

equation y(x,t)=Acos(kx−ωt)

speed  v = 8.5 m/s

amplitude A = 5.5*10^−2 m

wavelength λ   = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π/ λ

wave number =  2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~= 107 rad/sec

so

equation y(x,t)=Acos(kx−ωt)

y(x,t)= 5.5*10^−2 cos(12.568 x−107t)

when x =0 and and t = 0

maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))

maximum y(x,t)= 5.5*10^−2  m

and when x =  x = 1.52 m and t = 0.150 s

y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )

y(x,t)= 5.5*10^−2 × (0.9986)

y(x,t) = 0.0549 m

so the transverse displacement is  0.0549 m

5 0
3 years ago
Consider the plot below describing motion along a straight line with an initial position of 10 m. −4 −3 −2 −1 0 1 2 3 4 5 6 1 2
Ugo [173]

Answer:

+1 m/s^2

Explanation:

Acceleration is given by

a=\frac{\Delta v}{\Delta t}

where

\Delta v is the change in velocity

\Delta t is the time interval in which the change in velocity occurs

To find the acceleration at 1 second, we can take the data at t = 1 s and t = 2. We find:

\Delta t = -3 -(-4) = 1 s

\Delta v = 2 - (1) = +1 m/s

So, the acceleration is

a=\frac{+1}{1}=+1 m/s^2

4 0
3 years ago
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