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3 years ago

A car is driven 110 km west and then 40 km southwest, how far is the car from the point of origin?

Physics

1 answer:

Margaret [11]3 years ago

8 0

Answer:

141.147 km

Explanation:

$sin45=\frac{BD}{BC}\\\Rightarrow BD=BCsin45\\\Rightarrow BD=40\times sin45\\\Rightarrow BD=28.28\ km$

Now, AE=BD also CD=BD as the angle B and C are equal

So, from Pythagoras theorem

AC² = AE²+(CD+DE)²

$AC=\sqrt{AE^2+(CD+DE)^2}\\\Rightarrow AC=\sqrt{(40\times sin45)^2+(110+40\times sin45)^2}\\\Rightarrow AC=141.147\ km$

Displacement of the car is 141.147 km

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An elevator is raised by a cable, as shown in the diagram. If the mass of the elevator is 715 kg and the tension in the

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Answer:

1.379 m/s2

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When at rest, a ruler has a length of 12.0 inches. How fast (in m/s) must the ruler fly past you in order for it to shorten to 6

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Answer:

2.598× $10^8 m/sec$

Explanation:

The length of the contraction is given by $l=l_0\sqrt{1-\frac{v^2}{c^2}}$ here $l_0$ is length of the ruler at rest which is given as 12 inches

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$l=l_0\sqrt{1-\frac{v^2}{c^2}}$

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$1-(\frac{l}{l_0})^2=\frac{v^2}{c^2}$

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3 years ago

Work done of frictional force from instant

Liono4ka [1.6K]

Answer:

$-100\ J$

Step-by-step explanation:

<u>1. Find acceleration:</u>

$m=2\ kg$
$F=-5\ N$
$a=\frac{F}{m}$ (Newton's second law)
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<u>2. Find distance traveled:</u>

$v_0=10\ \frac{m}{s}$
$v=0$

$a=-2.5\ \frac{m}{s^{2} }$
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$-100=-5d$
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3. Find work done by friction:

$W=Fd$ (Work formula when angle between Force and Displacement vectors are 0°)
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3 years ago

A bumper car with a mass of 240 kg is moving to the right with a velocity of 2.3

Sauron [17]

Answer:

See the explanation below

Explanation:

To solve this problem we will take the definition of linear momentum, which tells us that momentum is equal to the product of mass by the velocity vector. Since velocity is a vector, we will take the right-hand movement as positive and the left-hand movement as negative, the left-hand members are taken as before the collision and the right-hand members as after the collision

ΣM1 = ΣM2

(m1*v1) + (m2*v1) = (m1*v2) + (m2*v2)

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Therefore:

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Resolving this arithmetic operation we will have:

720 - 702 = - 696 + 546

- 150 = - 150

We can see that before the crash and after the crash the momentum is preserved

7 0

3 years ago