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Zina [86]
2 years ago
4

A giraffe trots 3.40 km due north, then 2.60 km due west. what is the magnitude of its displacement?

Physics
2 answers:
vivado [14]2 years ago
5 0
R=(x^2+y^2)^(1/2)
R= 4.28
goldenfox [79]2 years ago
4 0
First, draw a picture to help you visualize what is happening. I attached my picture below. 

Before you do the Pythagorean theorem make sure to convert the lengths to meters by multiplying by 1000.

(in case my handwriting is too messy to read)
Here is what I wrrote on my paper poiting at the blue displacement vector: " The magnitude of this displacement vector will be a combination of length and direction of the hypotenuse(blue vector) or this triangle. 

To solve for the length use the Pythagorean theorem(where "c" is the hypotenuse):
a= 3400 meters
b= 2600 meters
c^{2} = a^{2} + b^{2}
c=  \sqrt{( a^{2} + b^{2} )}
c= \sqrt{( (3400^{2}) +(2600^{2}) )}
c= \sqrt{( (11,560,000) +(6,760,000) )}
c= \sqrt{ (18,320,000) }
c=4,280.2 meters Northwest

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Apollo 14 astronaut Alan B. Shepard Jr. used an improvised six-iron to strike two golf balls while on the Fra Mauro region of th
Citrus2011 [14]

Answer:

(a) Rm = 268.4 m

(b) f = 6

Explanation:

The horizontal range of a projectile is given by the following formula:

R = V₀² Sin 2θ/g

(a)

For moon:

R = Range on moon = Rm

V₀ = Launch Speed = 28 m/s

θ = Launch Angle = 17°

g = acceleration due to gravity on moon = (9.8 m/s²)/6 = 1.63 m/s²

Therefore,

Rm = (28 m/s)²Sin (2*17°)/(1.63 m/s²)

<u>Rm = 268.4 m</u>

(b)

For Earth:

R = Range on Earth = Re

V₀ = Launch Speed = 28 m/s

θ = Launch Angle = 17°

g = acceleration due to gravity on Earth = 9.8 m/s²

Therefore,

Re = (28 m/s)²Sin (2*17°)/(9.8 m/s²)

Re = 44.7 m

Therefore.

f = Rm/Re = 268.4 m/44.7 m

<u>f = 6</u>

3 0
2 years ago
Superman is standing 393 m horizontally away from Lois Lane. A villain drops a rock from 4.00 m directly above Lois. If Superman
Sergio039 [100]

Answer:

-963.93 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 4=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4\times 2}{9.81}}\\\Rightarrow t=0.903\ s

s=ut+\frac{1}{2}at^2\\\Rightarrow 393=0\times 0.0903+\frac{1}{2}\times a\times 0.903^2\\\Rightarrow a=\frac{393\times 2}{0.903^2}\\\Rightarrow a=963.93\ m/s^2

The acceleration of Superman would be -963.93 m/s² from Lois' perspective

6 0
2 years ago
A semi is traveling down the highway at a velocity of v = 26 m/s. The driver observes a wreck ahead, locks his brakes, and begin
Dovator [93]

Answer:

fcosθ + Fbcosθ  =Wtanθ

Explanation:

Consider the diagram shown in attachment

fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)

Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)

Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)

sum of x-direction forces = 0

fx+ Fbx=Wx

fcosθ + Fbcosθ  =Wtanθ

7 0
2 years ago
A child's toy consists of a m = 36 g monkey suspended from a spring of negligible mass and spring constant k. When the toy monke
kolezko [41]

Answer:

Part A - 3N/m

Part B - see attachment

Part C - 4.9 × 10-³J

Part D - E = 1/2kd² + 1/2mv² + mgh

Explanation:

This problem requires the knowledge of simple harmonic motion for cimplete solution. To find the spring constant in part A the expression relating the force applied to a spring and the resulting stretching of the spring (hooke's law) is required which is F = kx.

The free body diagram can be found in the attachment. Fp(force of pull), Ft(Force of tension) and W(weight).

The energy stored in the pring as a result of the stretching of d = 5.7cm is 1/2kd².

Part D

Three forces act on the spring-monkey system and they do work in different forms: kinetic energy 1/2mv² , elastic potential

energy due to the restoring force in the spring or the tension force 1/2kd², and the gravitational potential energy mgh of the position of the system. So the total energy of the system E = 1/2kd² + 1/2mv² + mgh.

8 0
2 years ago
Explain why it takes more energy to remove the second electron from a lithium atom than it does to remove the fourth electron fr
slava [35]
It takes more energy to remove the second electron from a lithium atom than it does to remove the fourth electron from a carbon atom because its inner core e, not valence e. C's 4th removed e is still a valence e. And also <span>because more nuclear charge acting on the second electron, it is more close to the nucleus, thus the the protons attract it more than the 4th electron.</span>
8 0
2 years ago
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