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2 years ago

What force is acting on the rainwater in the model?

Physics

2 answers:

Vikki [24]2 years ago

8 0

The force is gravitational because when something is falling is call gravitational

iren [92.7K]2 years ago

6 0

It’s showing gravity

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For nitrogen feel like with its temperature must be within 12.78 Fahrenheit of -333.22 Fahrenheit which equation can be used to

photoshop1234 [79]

Answer:

The following equation can be used.

(32°F − 32) × 5/9=C

7 0

2 years ago

Read 2 more answers

A plane increases its speed from 450 km/hr to 750 km/hr in 0.30 hour. What is its acceleration?

ryzh [129]

Answer:

The acceleration is found as:

a = 1000 km/h²

Explanation:

Initial speed of the plane = 450 km/h

Final speed of the plane = 750 km/h

Time taken = 0.3 h

Acceleration can be defined as the change of speed of the object divided by the time taken to bring that change

Acceleration is given as:

$a=\frac{v_f-v_i}{t}$

where v(f) = 750 km/h , v(i) = 450 km/h , t = 0.3 h

Substitute the values in the equation of acceleration

$a= \frac{750-450}{0.3}\\a=\frac{300}{0.3}\\a=1000 km/h^2$

This is the found result

5 0

2 years ago

Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an a

serious [3.7K]

To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

$\omega = 1.25 rad/s \rightarrow$ The angular speed

$\alpha = 0.745 rad/s2 \rightarrow$ The angular acceleration

$r = 4.65 m \rightarrow$ The distance

The relation between the linear velocity and angular velocity is

$v = r\omega$

Where,

r = Radius

$\omega =$ Angular velocity

At the same time we have that the centripetal acceleration is

$a_c = \frac{v^2}{r}$

$a_c = \frac{(r\omega)^2}{r}$

$a_c = \frac{r^2\omega^2}{r}$

$a_c = r \omega^2$

$a_c = (4.65 )(1.25 rad/s)^2$

$a_c = 7.265625 m/s^2$

Now the tangential acceleration is given as,

$a_t = \alpha r$

Here,

$\alpha =$ Angular acceleration

r = Radius

$\alpha = (0.745)(4.65)$

$\alpha = 3.46425 m/s^2$

Finally using the properties of the vectors, we will have that the resulting component of the acceleration would be

$|a| = \sqrt{a_c^2+a_t^2}$

$|a| = \sqrt{(7.265625)^2+(3.46425)^2}$

$|a| = 8.049 m/s^2 \approx 8.05 m/s2$

Therefore the correct answer is C.

7 0

2 years ago

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V. What are (a) the total energy stored i

Debora [2.8K]

Answer:

(A) Total energy will be equal to $0.044\times 10^{-5}J$

(b) Energy density will be equal to $0.0175J/m^3$

Explanation:

We have given diameter of the plate d = 2 cm = 0.02 m

So area of the plate $A=\pi r^2=3.14\times 0.02^2=0.001256m^2$

Distance between the plates d = 0.50 mm = $0.50\times 10^{-3}m$

Permitivity of free space $\epsilon _0=8.85\times 10^{-12}F/m$

Potential difference V =200 volt

Capacitance between the plate is equal to $C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 0.001256}{0.50\times 10^{-3}}=0.022\times 10^{-9}F$