What is the net displacement of the particle between 0 seconds and 80 seconds

A vector has an x-component of length 10 and a y-component of length 3. What is the angle of the vector? (Hint: Use the inverse

Sonja [21]

The vector, the x-component and the y-component form a rectangle triangle where the vector is the hypothenuse and the x and y components are the two sides.
Calling $\alpha$ the angle between the vector and the horizontal direction (x), the two sides are related to $\alpha$ by
$\tan \alpha = \frac{v_y}{v_x}$
where vy and vx are the two components on the y- and x-axis. Using vx=10 and vy=3 we find
$\tan \alpha = \frac{3}{10} =0.3$
And so the angle is
$\alpha = \arctan (0.3)=16.7^{\circ}$

5 0

3 years ago

Which of the following is the most usable form of nitrogen to plants?

tresset_1 [31]

Ammonia because it doesn't have nitrogen

5 0

3 years ago

Biogenetically developed GMO foods are _____.

alex41 [277]

I think it might be D but I am not sure

7 0

3 years ago

Read 2 more answers

A)If your torsion balance deflects to an angle of 10° when your two spheres are 40cm apart, what angle will it deflect to when t

svp [43]

Answer:

Explanation:

Given

for $\theta=10^{\circ}$

Sphere are $d=40\ cm$

when sphere $d_2=10\ cm$ apart suppose deflection is $\theta _2$

We know

$F=k_t\cdot \theta$

Where F=force between charged particle

$\theta =$ Deflection

$F=\frac{kQ_1Q_2}{r^2}=k_t\cdot \theta$

$\theta =\frac{k}{k_t}\times \frac{Q_1Q_2}{r^2}----1$

thus $\theta \propto \frac{1}{r^2}$

for $\theta _2$

$\frac{\theta _1}{\theta _2}=(\frac{r_2}{r_1})^2$

$\theta _2=16\times \theta _1$

$\theta _2=160^{\circ}$

(b)for $10^{\circ}$ deflection Potential $v_1=8\ kV$

Electric Potential is $V=\frac{kQ}{r}$

$Q=\frac{V\cdot r}{k}$

where V=voltage

k=constant

r=distance between charges

Put value of Q in equation 1

$\theta =\frac{k}{k_t}\times \frac{V^2r^2}{k^2}$

$\theta =\frac{V^2r^2}{k\cdot k_t}$

thus $\theta \propto V^2$

therefore

$\frac{\theta _1}{\theta _2}=(\frac{V_1}{V_2})^2$

$\frac{10}{\theta _2}=(\frac{8}{4})^2$

$\theta _2=\frac{10}{4}=2.5^{\circ}$

5 0

3 years ago

So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th

shutvik [7]

Answer:

Answer is explained in the explanation section.

Explanation:

A)

Solution:

For this to find, we need to calculate the centripetal acceleration on the equator.

The centripetal acceleration of the equator:

a = 4 $\pi ^{2}$ RcosФ/ $T^{2}$ ,

where,

R is the radius of the earth

R = 6378 KM = 6.3 x $10^{6}$ and

T is the time period

T = 24 h = 86164.1 s

At Equator, Ф = 0°

So, CosФ = 1

Hence,

a = 4 $\pi ^{2}$ R/ $T^{2}$

By plugging in the values, we get:

a = 4 x ( $3.14^{2}$ ) x (6.3 x $10^{6}$ ) / $86164.1^{2}$

a = 0.03 m/ $s^{2}$

Hence, this is the centripetal acceleration on the equator. And we also know that, acceleration due to gravity is 9.8 m/ $s^{2}$ which is very higher than the centripetal acceleration on the equator.

B) Normal force exerted by chair will always be equal and opposite to the mass times gravitational acceleration (F = mg). Otherwise, I would be thrown away from chair in case the normal force is not equal and opposite or I would be drag down to the earth due to greater mass times gravitational acceleration. Hence, both are equal and opposite.

C) Of course, this is not a lie, it is true because the acceleration due to gravity is 9.8 m/ $s^{2}$ and as we calculated the acceleration on the equator is 0.03 m/ $s^{2}$ which way too low to experience.

For percentage difference,

9.8 - 0.03 = 9.77

So, % diff = (9.8 - 9.77)/9.8 x 100

% diff = 0.00306 x 100

% diff = 0.306%

Obviously, this is way too low to experience.

D) With the help of same formula as discussed above, we have:

a = 4 $\pi ^{2}$ RcosФ/ $T^{2}$ ,

Here, Ф = 44.4°

Just putting the values. we get

a = 0.0241 m/ $s^{2}$

Acceleration while sitting in my chair.

6 0

3 years ago