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Tems11 [23]
3 years ago
11

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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The temperature decreases as you go up. 
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3 years ago
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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
Lisa [10]

Answer:

1.379 m/s2

Explanation:

F=ma

There is a negative force from the elevator...

F = (715 kg) (-9.81 m/s2)

F = -7014.15 N

Cable + elevator = 8000 N + (-7014.15 N)

F = 985.85 N

a = F/m

985.85 N/715 kg = 1.379 m/s2

3 0
3 years ago
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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
PIT_PIT [208]

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

l is the length of the ruler in moving condition which is given as 6 inches c is the speed of the light and v is the velocity of the ruler.

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

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

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

1-(\frac{6}{12})^2=\frac{v^2}{c^2}

0.866=\frac{v}{c}

v=0.866\times 3\times 10^8=2.598\times 10^8 m/sec

3 0
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)
  • a=\frac{-5}{2} =-2.5\ \frac{m}{s^{2}}

<u>2. Find distance traveled:</u>

  • v_0=10\ \frac{m}{s}
  • v=0
  • a=-2.5\ \frac{m}{s^{2} }
  • v^2-v_0^2=2ad (Kinematic equation)
  • -100=-5d
  • d=20\ m

3. Find work done by friction:

  • W=Fd (Work formula when angle between Force and Displacement vectors are 0°)
  • W=-5\times20=-100\ J
4 0
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)

V1 = velocity before the collision [m/s]

V2 = velocity after the collision [m/s]

m1 and m2 = mass of the vehicles [kg]

Therefore:

(240*2.3) - (260*2.7) = - (240*2.9) + (260*2.1)

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
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