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4 years ago

Which device in a car helps to protect riders from the effects of inertia in the event of an acident

2 answers:

5 0

The seatbelt, as it prevents you from flying out of your window.

if this was the appropriate answer make sure to mark as the brainliest!
-procklown

Oksi-84 [34.3K]4 years ago

5 0

The answer to the question is d

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I REALLY NEED HELP ON PHYSICS!!!<br> 10 POINTS

aleksley [76]

Answer:

25.021 sec

Explanation:

v=d/t

-2.33= - 58.3/t

t= - 58.3/-2.33

t=25.021 sec

6 0

3 years ago

Consider a motor that exerts a constant torque of 25.0 N⋅m to a horizontal platform whose moment of inertia is 50.0 kg⋅m2 . Assu

Step2247 [10]

To solve this exercise it is necessary to apply the concepts related to Work and Kinetic Energy. Work from the rotational movement is described as

$W=\tau \Delta\theta$

In the case of rotational kinetic energy we know that

$KE = \frac{1}{2}I\omega^2$

PART A) $\theta$ is given in revolutions and needs to be in radians therefore

$\theta = 12rev(\frac{2\pi rad}{1rev})$

$\theta = 24\pi rad$

Replacing in the work equation we have to

$W=\tau \Delta\theta$

$W= (25)(24\pi)$

$W = 1884.95J$

PART B) From the torque and moment of inertia it is possible to calculate the angular acceleration and the final speed, with which the kinetic energy can be determined.

$\tau = I \alpha$

Rearrange for the angular acceleration,

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{25}{50}$

$\alpha = 0.5rad/s$

From the kinematic equations of angular motion we have,

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

$\omega_f^2=0+2*0.5*24\pi$

$\omega_f=\sqrt{0+2*0.5*24\pi}$

$\omega_f = 8.68rad/s$

In this way the rotational kinetic energy would be given by

$KE = \frac{1}{2}I\omega_f^2$

$KE = \frac{1}{2}(50)(8.68)^2$

$KE = 1883.56J$

3 0

3 years ago

Read 2 more answers

A fluid moves through a tube of length 1 meter and radius r=0.002±0.0002 meters under a pressure p=4⋅105±1750 pascals, at a rate

yaroslaw [1]

Answer:

The maximum error is $\Delta \eta = 2032.9$

Explanation:

From the question we are told that

The length is $l = 1\ m$

The radius is $r = 0.002 \pm 0.0002 \ m$

The pressure is $P = 4 *10^{5} \ \pm 1750$

The rate is $v = 0.5*10^{-9} \ m^3 /t$

The viscosity is $\eta = \frac{\pi}{8} * \frac{P * r^4}{v}$

The error in the viscosity is mathematically represented as

$\Delta \eta = | \frac{\delta \eta}{\delta P}| * \Delta P + |\frac{\delta \eta}{\delta r} |* \Delta r + |\frac{\delta \eta}{\delta v} |* \Delta v$

Where $\frac{\delta \eta }{\delta P} = \frac{\pi}{8} * \frac{r^4}{v}$

and $\frac{\delta \eta }{\delta r} = \frac{\pi}{8} * \frac{4* Pr^3}{v}$

and $\frac{\delta \eta }{\delta v} = - \frac{\pi}{8} * \frac{Pr^4}{v^2}$

$\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v]$

substituting values

$\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ]$

$\Delta \eta = \frac{\pi}{8} [56 + 5120 ]$

$\Delta \eta = 647 \pi$

$\Delta \eta = 2032.9$

4 0

4 years ago

A Bear moves at a speed of 8 m/s for a distance of 300 m. How long does it take the Bear to travel this distance?

Lera25 [3.4K]

Answer:

2,400

Explanation:

Because when we use the formula D=st and we plug in our numbers it should look like D=8x300 . And when we do the math of multiplication it gives us the answer of 2,400 .

Hoped I helped <3

Sleepy~

8 0

3 years ago

A very long straight wire has charge per unit length 1.44×10-10C/m.

4vir4ik [10]

Answer:

Distance of the point where electric filed is 2.45 N/C is 1.06 m

Explanation:

We have given charge per unit length, that is liner charge density $\lambda =1.44\times 10^{-10}C/m$

Electric field E = 2.45 N/C

We have to find the distance at which electric field is 2.45 N/C

We know that electric field due to linear charge is equal to

$E=\frac{\lambda }{2\pi \epsilon _0r}$ , here $\lambda$ is linear charge density and r is distance of the point where we have to find the electric field

So $2.45=\frac{1.44\times 10^{-10} }{2\times 3.14\times 8.85\times 10^{-12}\times r}$

r = 1.06 m

So distance of the point where electric filed is 2.45 N/C is 1.06 m

3 0

3 years ago