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Mandarinka [93]

3 years ago

13

A 1500 kg car traveling east at 40 km/hr turns a corner and speeds up to a velocity of 50 km/hr due north. What is the change in

the car's momentum?

1 answer:

pantera1 [17]3 years ago

5 0

Answer:

96046 Ns.

Explanation:

We shall represent velocity in vector form considering east direction as + ve x axis and north as + y direction.

40 km/h in the east

V₁ = 40 i

V₂ = 50j

momentum p₁ = mV₁

= 1500 X 40 i

= 60000 i

Momentum p₂ = mV₂

= 1500 X 50j

= 75000 j

Change in momentum

p₂ - p₁

75000j - 60000i

Magnitude of change

= $\sqrt{(750000)^2 +(60000)^2$

= 96046 Ns.

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What is the significance of the nose end marking on a rocket or missile?

Gemiola [76]

If you're referring to the different colors that usually occur at the tip of missles, rockets and some other aircraft, it either a) signifies the end of a particular plate of metal, fabricated specifically to be for the nose. Sometimes these can even be a different alloy or metal all together. or b) this shows where the curved surface begins, so in the case of damage or imperfections due to wear, they can be repaired and measured more easily. The shape of the nose is extremely important for smooth flight, and a dent or bump formed on it can make the aircraft unstable. If you can measure from where the curve starts by the difference in color, it makes repairing or re-fabricating the part much easier. Many of these curves aren't as simple as they appear.

5 0

3 years ago

Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

a.

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

b.

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

7 0

3 years ago

A rectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. Suppose that a 410-kg crate of auto parts is loaded onto

sesenic [268]

<h2>Answer:</h2><h2>The depth of barge float=3 cm</h2><h2>Explanation:</h2>

Length of rectangular barge=5.2 m

Width of rectangular barge=2.4m

Mass of crate=410 kg

Let h be the height of barge float

Volume of barge float= $l\times b\times h=5.2\times 2.4\times h=12.48h$

Density of water= $10^3kg/m^3$

Weight of water displaced by barge=Buoyant force=-Weight of horse

$Volume\;of\;water\times density\;of\;water\times g=410\times g$

$12.48h\times 1000=410$

$h=\frac{410}{12.48\times 1000}=0.03 m$

1 m=100 cm

$0.03 m=0.03\times 100=3$ cm

Hence, the depth of barge float=3 cm

<h2 />

4 0

3 years ago

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The following molecule can be synthesized from an epoxide and alkyllithium reagent, followed by aqueous workup. However, there a

Assoli18 [71]

X. X

I. I
I. I
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8 0

3 years ago

A group of students performed a compression experiment where they placed weights on top of a cylinder of material and measured t

kolbaska11 [484]

The material that the cylinder is made from is Butyl Rubber.

<h3>What is Young's modulus?</h3>

Young's modulus, or the modulus of elasticity in tension or compression, is a mechanical property that measures the tensile or compressive strength of a solid material when a force is applied to it.

<h3>Area of the cylinder</h3>

A = πr²

$A = \pi \times (0.02)^2 = 0.00126 \ m^2$

<h3>Young's modulus of the cylinder</h3>

$E = \frac{stress}{strain} \\\\E = \frac{F/A}{e/l} \\\\E = \frac{Fl}{Ae} \\\\$

Where;

e is extension

When 5 kg mass is applied, the extension = 10 cm - 9.61 cm = 0.39 cm = 0.0039 m.

$E = \frac{(5\times 9.8) \times 0.1}{0.00126 \times 0.0039} \\\\E = 9.97 \times 10^5 \ N/m^2\\\\E = 0.000997 \times 10^9 \ N/m^2\\\\E = 0.000997 \ GPa\\\\E \approx 0.001 \ GPa$

When the mass is 50 kg,

extension = 10 cm - 7.73 cm = 2.27 cm = 0.0227 m

$E = \frac{(50\times 9.8) \times 0.1}{0.00126 \times 0.0227} \\\\E = 1.7 \times 10^6 \ N/m^2\\\\E = 0.0017 \times 10^9 \ N/m^2\\\\E = 0.0017 \ GPa\\\\E \approx 0.002 \ GPa$

The Young's modulus is between 0.001 GPa to 0.002 GPa

Thus, the material that the cylinder is made from is Butyl Rubber.

Learn more about Young's modulus here: brainly.com/question/6864866

3 0

3 years ago

Read 2 more answers