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3 years ago

7

It can take some of those large , industrial vehicles up to ____ feet to stop when traveling only 60 MPH, and therefore , you sh

oukd be mindful to not be jn the traffic right in front of them
A. 100 feet
B. 200 feet
C. 335 feet
D. 425 feet

1 answer:

kherson [118]3 years ago

4 0

Answer:

a 200 feet, and trains go a whole mile even after hitting the brakes

Explanation:

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In general, metalloids are slightly reactive.

algol [13]

Yes, in general, metalloids are slightly reactive.

6 0

3 years ago

Read 2 more answers

If 270 watts of power is used in 42 seconds, how much work was done<br>

navik [9.2K]

Answer: W = 11340J

Explanation:

Hey there! I will give the following steps, if you have any questions feel free to ask me in the comments below.

So this is the Formula: Power = Work / Time.

<u>Step 1:</u><em><u> Find the Formula</u></em>

P = W / T

<em><u> </u></em>

<u>Step 2: </u><u><em>Make W the subject of the equation.</em></u>

W = PT

<u>Step 3:</u><u> </u><u><em>Given.</em></u>

P = 270 Watts, T = 42 seconds

<u>Step 4:</u><u><em> Substitute these values into equation 2 .</em></u>

W = 270(42)

<u>Step 5:</u><u> </u><u><em>Simplify.</em></u>

W = 11340J

The amount of work done was 11340.

~I hope I helped you! :)~

4 0

3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

a 1 gram spiders sits on a platform rotating at 78 rpm. the spider is 15 cm from the centre disk. find the speed of the spider

olga nikolaevna [1]

The spider is traveling in a circle with radius = 15cm

The circumference of any circle = <em>2 pi (radius)</em>
The circumference of the spider's path = 2 pi (15 cm) = 30 pi cm

The spider completes a trip around this path 78 times per minute.
Its speed, relative to you, is

   (78) x (30 pi) cm/min =

   2,340 pi cm/min = 7,351.33 cm/min =

   <em> 73.5133 meter/min =</em>

   <em>4.411 km/hr =</em>

   <em>2.74 miles/hour

</em>(After the last appearance of pi,
all numbers are rounded.)<em>

</em>

8 0

3 years ago

A 50.0-kg projectile is fired at an angle of 30 degrees above thehorizontal with an initial speed of 1.20 x 102 m/s fromthe top

Advocard [28]

Answer:

a) Em₀ = 42.96 104 J , b) $W_{fr}$ = -2.49 105 J , c) vf = 3.75 m / s

Explanation:

The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has

Em = K + U

a) Let's look for the initial mechanical energy

Em₀ = K + U

Em₀ = ½ m v2 + mg and

Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142

Em₀ = 36 104 + 6.96 104

Em₀ = 42.96 104 J

b) The work of the friction force is equal to the change in the mechanical energy of the body

$W_{fr}$ = Em₂ -Em₀

Em₂ = K + U

Em₂ = ½ m v₂² + m g y₂

Em₂ = ½ 50 85 2 + 50 9.8 427

Em₂ = 180.625 + 2.09 105

Em₂ = 1,806 105 J

$W_{fr}$ = Em₂ -Em₀

$W_{fr}$ = 1,806 105 - 4,296 105

$W_{fr}$ = -2.49 105 J

The negative sign indicates that the work that force and displacement have opposite directions

c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job

We have that the work of friction is equal to the change of mechanical energy

$W_{fr}$ = ΔEm

$W_{fr}$ = Emf - Emo

-1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴

½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵

½ 50.0 vf² = 0.561

vf = √ 0.561 25

vf = 3.75 m / s

6 0

3 years ago