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rosijanka [135]

2 years ago

10

Dave is moving 3 m/s when he crashes his bike into a wall, which stops him in 0.6 seconds. If Dave and his bike have a mass of 9

0 kg,
what was the force applied by the wall?

1 answer:

Arisa [49]2 years ago

7 0

Force = change of momentum / time taken
Force = (90x3)/0.6

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~~~NEED HELP ASAP~~~<br>Please solve each section and show all work for each section.

anastassius [24]

Explanation:

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for <em>N</em> as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

or

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension <em>T</em>, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

2 years ago

A 120 kg tackler moving at 3.0 m/s meets head-on (and tackles) a 91 kg halfback moving at 7.5 m/s. What will be their mutual vel

Vesna [10]

Explanation:

It is given that,

Mass of the tackler, m₁ = 120 kg

Velocity of tackler, u₁ = 3 m/s

Mass, m₂ = 91 kg

Velocity, u₂ = -7.5 m/s

We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,

$v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}$

$v=\dfrac{120\ kg\times 3\ m/s+91\ kg\times (-7.5\ m/s)}{120\ kg+91\ kg}$

v = -1.5 m/s

Hence, their mutual velocity after the collision is 1.5 m/s and it is moving in the same direction as the halfback was moving initially. Hence, this is the required solution.

3 0

2 years ago

What distance is required for a train to stop if its initial velocity is 23 m/s and its

Varvara68 [4.7K]

-- The train starts at 23 m/s and slows down by 0.25 m/s every second.

So it'll take (23/0.25) = 92 seconds to stop.

-- Its average speed during that time will be (1/2)(23+0) = 11.5 m/s

-- Moving at an average speed of 11.5 m/s for 92 sec, the train will cover

(11.5 m/s) x (92 sec) = <em>1,058 meters</em> .

7 0

2 years ago

Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19

Alisiya [41]

Answer:

The workdone is $W = 9.28 * 10^{3} J$

Explanation:

From the question we are told that

The height of the cylinder is $h = 0.588\ m$

The face Area is $A = 4.19 \ m^2$

The density of the cylinder is $\rho = 0.346 * \rho_w$

Where $\rho_w$ is the density of freshwater which has a constant value

$\rho_w = 1000 kg/m^3$

Now

Let the final height of the device under the water be $= h_f$

Let the initial volume underwater be $= V_n$

Let the initial height under water be $= h_i$

Let the final volume under water be $= V_f$

According to the rule of floatation

The weight of the cylinder = Upward thrust

This is mathematically represented as

$\rho_c g V_n = \rho_w gV_f$

$\rho_c A h = \rho A h_f$

So $\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}$

=> $\frac{h_f}{h_c} = 0.346$

Now the work done is mathematically represented as

$W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh$

$= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.$

$= \frac{g A \rho}{2} [h^2 - h_f^2]$

$= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]$

Substituting values

$W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)$

$W = 9.28 * 10^{3} J$

4 0

2 years ago

A force of 100 newtons was necessary to lift a rock. a total of 150 joules of work was done. how far was the rock lifted

Makovka662 [10]

Answer:

1.5m

Explanation:

150÷100=1.5

Work done ÷force

4 0

2 years ago