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marishachu [46]

2 years ago

11

Can someone explain how exactly to calculate the net force? Do you subtract the smaller number from the larger number? Any help

would be appreciated!

1 answer:

aleksley [76]2 years ago

3 0

Explanation:

Let's say right is positive and left is negative.

F₁ = -150 N

F₂ = 220 N

Fnet = F₁ + F₂

Fnet = -150 N + 220 N

Fnet = 70 N

The magnitude of Fnet is 70 N, and since it's positive, the direction is to the right.

And since Fnet isn't 0, the force is unbalanced and the motion is changing.

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A. Because they reflect their color and absorb all the others

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A rock is thrown from the top of a 20-m building at an angle of 53Â° above the horizontal. If the horizontal range of the throw

NemiM [27]

Answer:

28.5 m/s

18.22 m/s

Explanation:

h = 20 m, R = 20 m, theta = 53 degree

Let the speed of throwing is u and the speed with which it strikes the ground is v.

Horizontal distance, R = horizontal velocity x time

Let t be the time taken

20 = u Cos 53 x t

u t = 20/0.6 = 33.33 ..... (1)

Now use second equation of motion in vertical direction

h = u Sin 53 t - 1/2 g t^2

20 = 33.33 x 0.8 - 4.9 t^2 (ut = 33.33 from equation 1)

t = 1.17 s

Put in equation (1)

u = 33.33 / 1.17 = 28.5 m/s

Let v be the velocity just before striking the ground

vx = u Cos 53 = 28.5 x 0.6 = 17.15 m/s

vy = uSin 53 - 9.8 x 1.17

vy = 28.5 x 0.8 - 16.66

vy = 6.14 m/s

v^2 = vx^2 + vy^2 = 17.15^2 + 6.14^2

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6 0

2 years ago

A schoolbus accelerates to 65 mph and enters the freeway. It travels for 2.3 hours at that speed while on the freeway. What's th

PtichkaEL [24]

Answer:

The distance travelled on the freeway is 149.5 miles.

Explanation:

The school bus travels on the freeway at constant speed. According to the statement, we need to calculate the distance travelled by the vehicle by means of the following formula:

$x = v\cdot t$ (1)

Where:

$x$ - Traveled distance, in miles.

$v$ - Speed, in miles per hour.

$t$ - Time, in hours.

If we know that $v = 65\,\frac{mi}{h}$ and $t = 2.3\,h$ , then the distance travelled by the school bus is:

$x = v\cdot t$

$x = \left(65\,\frac{mi}{h} \right)\cdot (2.3\,h)$

$x = 149.5\,mi$

The distance travelled on the freeway is 149.5 miles.

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2 years ago

Evaporating water produces gaseous water. What has occurred during this process?

Katyanochek1 [597]

Answer:

Heat has accelerated water atoms enough to break the surface tension which leads the liquid to turn into a gas

Explanation:

The state of a substance depends on the distribution of its atoms, therefore any atmosphere change (in this case heat) enough to change the atoms Distribution results in a change of state.

brainliest please ;)

3 0

2 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

3 0

2 years ago